Test Bank for Quantitative Chemical Analysis 10th Edition by Daniel C. Harris, Charles A. Lucy

1 Chapter 0 – The Analytical Process Multiple-Choice 1. substance is present. a sample.

A) B) C) D) E)

chemical analysis is the measurement of how much of a chemical chemical analysis is the determination of what is present in

Quantitative; Qualitative Stoichiometric ; Qualitative Qualitative; Quantitative Stoichiometric ; Identification Quantitative; Identification

Answer: A Easy 0-1

The Analytical Chemist’s Job

2.

A sample with a uniform chemical composition is a

A) B) C) D) E)

homologous homogeneous uniform consistent heterogeneous

sample.

Answer: B Easy 3.

A(n) place to place.

A) B) C) D) E)

variable homogeneous random inconsistent heterogeneous

Answer: E Easy

sample is a sample where the chemical composition differs from

2 4.

When extracting a sample with a liquid, the liquid is

A) B) C) D) E)

transferred drained decanted effused dispensed

from the sample.

Answer: C Easy 5.

A(n)

A) B) C) D) E)

orbital shaker vortexer mixer mortar and pestle centrifuge

is used to grind solids into smaller particles.

Answer: D Easy 6.

A) B) C) D) E)

is the substance being measured during a chemical analysis.

Bulk Lot Sample Analyte Aliquot

Answer: D Easy 7.

The liquid above the packed solid following a centrifugation is the

A) B) C) D) E)

solvonatant analyte serum decanted supernatant

Answer: E Easy

.

3 8.

A(n)

A) B) C) D) E)

slurry colloid gel supernatant allotrope

is a suspension of a solid in a liquid.

Answer: A Easy 9.

A) B) C) D) E)

is the series of procedures applied to a sample prior to analysis.

Pre-analysis clean up Sample preparation Filler elimination Matrix removal Lot cleaning

Answer: B Easy 10.

For separations, the plot of detector response versus time is a(n) and the area under the peak is to the quantity of compound passing through the detector.

A) B) C) D) E)

column plot; proportional column plot; inversely proportional chromatogram; proportional chromatogram; inversely proportional absorbance spectrum; proportional

Answer: C Easy

4 11.

A(n) is a plot of detector response as a function of analyte concentration. The curve is constructed using containing known concentrations of the analyte of interest.

A) B) C) D) E)

analyte curve; response solutions response curve; standard solutions analyte curve; analyte solutions calibration curve; standard solutions response curve; response solutions

Answer: D Easy 12.

A) B) C) D) E)

is the process of procuring a representative sample to analyze.

Inspection Examination Representation Sampling Partaking

Answer: D Easy 13.

Solid-phase extraction is a sample preparation technique. Which of the following statements is(are) NOT true for an aqueous solid-phase extraction?

I Solid-phase extraction separates analyte from the sample matrix using a chromatography stationary phase.

II An aliquot of the aqueous sample is applied to the solid-phase extraction tube, washing the sample with additional sample and then a water wash.

III The analyte sticks to the column and the water wash removes all species that do not adhere to the column.

IV An organic solvent is used to wash the analyte from the column.

V The organic phase containing the analyte is evaporated to dryness and the solid is dissolved in water, ready for analysis.

A) B) C) D) E)

III I II and III IV II

Answer: E Easy

5 0-2

General Steps in a Chemical Analysis

14. occurs when a species other than analyte increases or decreases the analytical signal and makes it appear that the concentration is greater or less than the real concentration.

A) B) C) D) E)

Interference Masking Aliquots Disruption Intervention

Answer: A Easy 15. not detected.

A) B) C) D) E)

is the transformation of an interfering species into a form that is

Interference Masking Obscurance Cloaking Camouflaging

Answer: B Easy 16.

Which is NOT a general step in the analytical process?

A) B) C) D) E)

Sample Preparation Selecting analytical procedures Make policy Reporting and interpretation Analysis

Answer: C Easy

6 17.

Which statement is NOT true?

A) For a random heterogeneous material differences in composition are random and on a fine scale.

B) A representative random sample is collected from randomly selected portions of the sample for a given number of times.

C) A segregated heterogeneous material has large regions with obviously different compositions.

D) A representative composite sample is collected from a segregated material by taking portions from each region, the number of collected portions proportional to the area of the region.

E) All are true statements.

Answer: E Intermediate 18. are repeated measurements to assess the variability in the analysis and to guard against a gross error in the analysis of a single aliquot.

A) B) C) D) E)

Replicate measurements Aliquots Sampling Analysis Error measurements

Answer: A Easy 19.

A) B) C) D) E)

is any chemical of interest.

Analyte Species Replicate Aliquot Bulk

Answer: B Easy

7 Calculated 0-1

The Analytical Chemist’s Job

20.

A calibration curve for the determination of aspirin is constructed from known concentration aspirin solutions, mg/mL, and the HPLC peak area’s for each standard. If the equation is y = 12.565x − 0.71, what is the concentration for an unknown that has a peak area of 83.5?

Answer: 6.70 mg/mL Easy Substitute y = 83.5, m = 12.565 and b = −0.71 into y = mx + b and solve for x.

Multiple Choice 1-1

SI Units

1.

A satellite in low Earth orbit with a circular orbit has an orbital speed of 7.3 km/s relative to the Earth’s surface. Calculate the satellite’s speed in miles per hour.

A) B) C) D) E)

1.6  104 mile per hour 1.3  10−3 mile per hour 4.2  104 mile per hour 3.3  10−3 mile per hour 3.1  102 mile per hour

Answer: A Easy 2.

The planet Mars orbits 2.279  1011 m from the Sun. Express the distance using the

appropriate prefix.

A) B) C) D) E)

227.9 Gm 227.9 mM 2.279 km 22.79 nm None of these is correct.

Answer: A Easy 3.

The calorie content of a candy bar is 230 calories per serving (1 bar). Calculate the specific energy (kJ/g) of the candy bar. (1 candy bar = 52.7 g, 1 calorie = 4.184 J) A)

2.90  103 kJ/g

8 B) C) D) E)

18.3 kJ/g 1.83  10−2 kJ/g 5.07  104 kJ/g 9.59  10−1 kJ/g

Answer: B Intermediate

9 4.

The gas mileage for a new car model destined for sale in Europe must be determined for regulatory and promotional purposes. If the car uses 10.5 gallons to travel 250 miles, what is the gas mileage in km/L? A) B) C) D) E)

174 km/L 3.25 km/L 67.3 km/L 14.9 km/L 10.1 km/L

Answer: E Easy 1-2

Chemical Concentrations

5.

Calculate the mass NaCH3CO2 contained in 500.0 mL of a 0.1500 M NaCH3CO2 solution.

(NaCH3CO2 = 82.0343 g/mol) A) B) C) D) E)

914.3 µg 283.4 g 24.61 µg 6.378 g 24.61 g

Answer: D Easy 6.

Which of the statements below are TRUE for expressing the concentration of a 54.9-ppm Fe solution in terms of molarity?

I II III IV

Iron’s molar mass must be known to calculate the moles iron in solution.

Iron’s density must be known to calculate the mass iron in solution.

The solution density must be known to calculate the solution volume.

The type of glassware used to prepare the solution must be known.

A) B) C) D) E)

I, III, and IV I and II I and III II and III None of these is true.

Answer: C Intermediate

10 7.

Calculate the molarity of a 30.0% hydrogen peroxide solution. The reported density of 30% hydrogen peroxide is 1.135 g/cm3. (H2O2 = 34.0147 g/mol) A) B) C) D) E)

7.77 M 0.0100 M 0.100 M 10.0 M 8.82 M

Answer: D Intermediate 8.

Arrange the four microscopic views below of four different solutions in order of increasing concentration.

A) B) C) D) E)

I, II, III, and IV II, III, I, and IV IV, II, III, and I IV, I, III, and II III, IV, II, and I

I

II

III

IV

11 Answer: D Intermediate 1-3

Preparing Solutions

9.

Calculate the mass of Na2CO3 needed to prepare a 15.00 mM solution with a volume of 500.0 mL. (Na2CO3 = 105.9888 g/mol) A) B) C) D) E)

1.258 g 3.180 g 0.7949 g 7.076 g 0.0141 g

Answer: C Intermediate 10.

What volume of 12.1 M HCl must be diluted to prepare a 0.2500 M HCl solution with a volume of 2.000 L? A) B) C) D) E)

41.3 mL 96.8 mL 10.3 mL 24.2 mL 6.05 mL

Answer: A Easy 11.

What volume of a 36% HCl solution must be diluted to prepare 1.000 L of a 0.1000 M HCl solution? The density of 36% HCl is 1.18 g/mL. A) B) C) D) E)

11.7 mL 8.6 mL 1.2 mL 10.1 mL 64.6 mL

Answer: B Intermediate

12 12.

A student must prepare 500.0 mL of solution containing 0.999 grams of solid copper(II) sulfate. Which of the following statements are FALSE regarding the proper procedure to prepare this solution?

I II III

A) B) C) D) E)

The 0.999 grams of solid copper(II) sulfate is dissolved in a 500.0-mL volumetric flask containing 500.0 mL of water.

The 0.999 grams of solid copper(II) sulfate is dissolved in a 500.0-mL volumetric flask containing 400 mL of distilled water before dilution to 500.0-mL. The 0.999 grams of solid copper(II) sulfate is placed in an empty 500.0-mL volumetric flask, diluted to 500.0 mL and allowed to dissolve.

I and II II and III I and III I, II, and III None are false.

Answer: C Intermediate 1-4

Stoichiometry Calculations for Gravimetric Analysis

13.

The sulfur content of an ore is determined gravimetrically by reacting the ore with concentrated nitric acid and potassium chlorate, converting all sulfur to sulfate. The excess nitrate and chlorate is removed by reaction with concentrated hydrochloric acid and the sulfate is precipitated using barium cation.

Ba2+ (aq) + SO42- (aq)  BaSO4 (s) Analysis of 10.1830 grams of a sulfur containing ore yielded 13.0221 grams of BaSO4. What is the percent by mass sulfur in the ore? (BaSO4 = 233.43 g/mol) A) B) C) D) E)

32.18% 52.63% 10.74% 17.57% The answer cannot be calculated with available data.

Answer: D Hard

13 14.

When solutions of Pb2+ and CrO42− are mixed the precipitate PbCrO4 is produced. What volume of 0.1750 M CrO42− removes all lead from 50.00 mL of a 0.3400 M Pb2+ solution?

A) B) C) D) E)

97.14 mL 25.74 mL 48.57 mL 194.29 mL 75.00 mL

Answer: A Easy 15.

A mixture of 50.00 grams propane, C3H8, and 100.00 grams oxygen, O2, is combusted. is the limiting reactant and grams of is in excess. (C3H8 = 44.10

g/mol) C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g) A) B) C) D) E)

Propane; 22.44; oxygen Oxygen; 63.71; propane Oxygen; 22.42; propane Propane; 27.44; oxygen None of these answers is correct.

Answer: C Intermediate Calculated 1-1

SI Units

16.

On average one gallon of kerosene contains 135,000 BTUs of heat energy per gallon combusted. Convert the energy content of kerosene to SI units. (1 BTU = 1055 J) Answer: 37.4 MJ/L Intermediate 1-2

Chemical Concentrations

17.

An NaCl solution is labeled as having a concentration of 33.5% by mass. If the NaCl solution has a density of 1.0492 g/mL, what is the molarity of the solution? (NaCl= 58.44 g/mol) Answer: 6.01 M Intermediate

14 18.

The maximum contaminant level for arsenic is 0.010 ppm for drinking water per EPA regulation. The arsenic concentration for the drinking water of a municipality was measured to be 4.92  10−6 M arsenic. What is the arsenic concentration of the water sample in ppm? Does the water sample meet EPA guidelines? Assume the drinking water sample has a density of 1.0000 g/mL. Answer: 0.369 ppm, exceeds EPA regulation Hard 1-3

Preparing Solutions

19.

What volume of a 50.0% by mass NaOH solution must be diluted to prepare a 0.3500 M NaOH solution with a volume of 500 mL? The density of 50% NaOH solution is 1.515 g/mL at 25°C. (NaOH = 40.00 g/mol) Answer: 9.24 mL Intermediate 1-4

Stoichiometry Calculations for Gravimetric Analysis

20.

The recommended daily allowance of calcium for men between the ages of 19 to 50 is 1 000 mg Ca. Three multivitamin tablets are analyzed for calcium gravimetrically with the precipitation of calcium cation by oxalate anion. If the mass dry calcium oxalate obtained is 2.0136 g, how many tablets must a man take in a given day to meet the recommended daily allowance?

Ca2+ (aq) + C2O42− (aq)  CaC2O4 (s) Answer: five vitamin tablets Hard 21.

Aqueous lead precipitates when mixed with aqueous carbonate.

Pb2+ (aq) + CO32− (aq)  PbCO3 (s) If 5.000 g Pb(NO3)2 and 2.500 g Na2CO3 are mixed in water, which ion is the limiting reactant?

What mass PbCO3 is precipitated? (PbCO3 = 267.21 g/mol, Pb(NO3)2 = 331.2 g/mol, Na2CO3 = 105.9888 g/mol) Answer: Pb2+ is the limiting reactant and 4.035 g PbCO3 is precipitated.

Intermediate

Multiple Choice 2-1

Safe, Ethical Handling of Chemicals and Waste

1.

Which of the following statements are TRUE?

15

I Organic solvents, concentrated acids, and concentrated ammonia should be handled in a fume hood.

II A respirator should be worn when handling organic solvents.

III All containers should be labeled to indicate what they contain.

IV Contact lenses are adequate to protect eyes from liquids and gases in the lab.

A) B) C) D) E)

I and II II and IV I and III III and IV II and III

Answer: C Easy 2.

A) B) C) D) E)

is a set of principles intended to help sustain a habitable planet.

Environmental chemistry Analytical chemistry Biological chemistry Atmospheric chemistry Green chemistry

Answer: E Easy 2-2

The Lab Notebook

3.

The fulfills the critical function of reporting what a researcher has done and what they observed, and allows another researcher to repeat their work.

A) B) C) D) E)

lab report lab notebook MSDS project report technical note

Answer: B Easy 4.

Which of the following is NOT a good practice when keeping a laboratory notebook?

A) B) C)

Use complete sentences when writing notes.

Write a balanced chemical equation for every reaction used.

Paste hard copies of important data into the notebook.

16 D) E)

Record the names of computer files where programs and data are stored.

All of the above are good practices when keeping a laboratory notebook.

Answer: E Easy 2-3

Analytical Balance

5.

The mass of the empty receiving vessel used with an analytical balance is the:

A) B) C) D) E)

linearity. buoyancy. readability. tare. tolerance.

Answer: D Easy 6. is the upward force exerted on an object in a gaseous or liquid fluid. The mass measured by an analytical balance in air is its actual mass.

A) B) C) D) E)

buoyancy; heavier than buoyancy; lighter than electromagnetic force; heavier than electromagnetic force; lighter than tare; equal to

Answer: B Intermediate

17 7.

Which of the responses below are sources of weighing error?

I II III IV

Weighing a sample that is warmer than ambient temperature.

Cooling a sample in a desiccator prior to weighing.

Periodically calibrating the balance.

The temperature of the balance changing over time.

A) B) C) D) E)

I and II I and III II and IV II and III I and IV

Answer: E Intermediate 2-4

Burets

8.

Which scenario below has the lowest relative uncertainty?

A) B) C) D) E)

Delivering 35.50 mL of titrant with a 50 +/− 0.05 mL class A buret.

Delivering 15.40 mL of titrant with a 50 +/− 0.05 mL class A buret.

Delivering 18.50 mL of titrant with a 25 +/−0.03 mL class A buret.

Delivering 5.40 mL of titrant with a 25 +/−0.03 mL class A buret.

Delivering 97.30 mL of titrant with a 100 +/−0.10 mL class A buret.

Answer: E (0.10% relative uncertainty) Hard 9.

A student prepares a solution using a 1 L volumetric flask. When he finishes, the meniscus is above the calibration mark on the flask neck. The concentration of the solution is: A) B) C) D) E)

less than calculated. greater than calculated. unchanged. irrelevant. impossible to predict.

Answer: A Intermediate

18 10.

A student titrated extracted chloride from a soil sample with 0.1 M silver nitrate. During the titration he performed the following actions while operating the buret.      

Washed the buret with silver nitrate solution Drained titrant slowly Delivered a fraction of a drop near end point Read bottom of concave meniscus Avoided parallax Accounted for graduation thickness in readings

His instructor notes on the student’s lab report that the student forgot to buret.

A) B) C) D) E)

when operating his

eliminate air bubbles estimate reading the buret to 1/10 of a division fill the buret to exactly 0.00 mL eliminate air bubbles and fill the buret to exactly 0.00 mL. eliminate air bubbles and estimate reading the buret to 1/10 of a division.

Answer: E Hard 2-5

Volumetric Flasks

11.

Which of the following is FALSE regarding volumetric flasks?

A) B) C)

Volumetric flasks are calibrated to obtain a particular volume at 20oC. Volumetric flasks are calibrated to deliver their indicated volume.

To properly use a volumetric flask, dissolve reagent in less than final volume and then dilute to volume.

The volume of the flask changes with temperature because liquid and glass expand when heated.

To obtain the calibrated volume, the bottom of the meniscus is aligned to the center of the neck of flask.

D) E)

Answer: B Intermediate

19 12.

Which of the statements below is(are) TRUE regarding collecting and storing samples for trace analysis?

I II III

A) B) C) D) E)

Trace ionic analytes stored in glass are lost by adsorption or contaminated by metals leaching from the glass surface.

Plastic bottles are recommended to collect and store ionic analyte samples.

Amber glass bottles are best for collecting and storing aqueous samples of organic materials.

I and II II and III I and III I, II, and III III

Answer: D Intermediate 2-6

Pipets and Syringes

13. than

are calibrated to deliver one fixed volume and are .

A) B) C) D) E)

Measuring pipets; more accurate; transfer pipets Transfer pipets; less accurate; measuring pipets Measuring pipets; more precise; transfer pipets Transfer pipets; more accurate; measuring pipets None of these answers is correct.

Answer: D Intermediate

20 14.

On a lab quiz, a student listed the steps to properly use a pipet. Which step is incorrect?

A)

Use a rubber bulb to twice pull up a volume of liquid past the calibration mark and discard the contents into a waste container.

Pull up a third volume past the calibration mark and quickly replace the bulb with the index finger.

Touch the tip of the pipet to the side of a beaker and use the index finger to drain the liquid until the meniscus reaches the center of the calibration mark.

Transfer the pipet to the receiving vessel, touch the tip of the pipet to the side of the vessel, and allow the pipet to drain by gravity.

Use the rubber bulb to blow any remaining liquid from the pipet.

B) C) D) E)

Answer: E Intermediate 2-7

Filtration

15.

A) B) C) D) E)

is the liquid from which a substance precipitates or crystallizes.

Filtrate Eluate Effluent Mother liquor Slurry

Answer: D Easy 16.

A) B) C) D) E)

is used to convert a precipitate to a known, constant composition.

Ashless filter paper A fritted-glass funnel Ignition A rubber policeman A dessicator

Answer: C Easy

21 2-8

Drying

17.

Drying to constant mass is a common gravimetric analysis technique. Which of the statements describe the source of false weights?

I II III IV

A warm crucible Touching the crucible with bare fingers Using a microwave oven to dry reagents and crucibles Using a desiccator

A) B) C) D) E)

I, II, and III II, III, and IV I and II II and IV III and IV

Answer: C Intermediate 2-9

Calibration of Volumetric Glassware

18.

A 0.1500 M HCl solution was prepared on a day when the temperature was 20oC. What is concentration of the solution when used the next day at 27oC? The density of water is 0.9982071 g/mL at 20oC and 0.9965162 g/mL at 27oC. A) B) C) D) E)

0.1503 M 0.1497 M 0.1508 M 6.653 M 6.632 M

Answer: B Intermediate

22 19.

A researcher dispenses distilled deionized water from a 20-mL volumetric pipet into an empty 8.4376 g weighting bottle. If the total mass of water and weighting bottle is 28.5845 g, what is the volume of the water delivered by the 20-mL pipet? The density of water is 0.9967867 g/mL. A) B) C) D) E)

20.21 mL 20.08 mL 28.68 mL 19.94 mL 19.90 mL

Answer: A Intermediate Calculated 2-5

Volumetric Flasks

20) Describe how to prepare a 20.00 µg/mL iron solution from a 1 000 mg/mL iron standard solution using 10-mL and 50-mL volumetric pipets and 500-mL and 1 000-mL volumetric flasks.

Answer: 10 mL 1 000 mg/mL iron solution diluted to 500 mL to give a 20 000 µg/mL iron solution. 10 mL of 20 000 µg/mL iron solution diluted to 1 000 mL to give a 200 µg/mL iron solution. 50 mL of the 200 µg/mL solution is diluted to 500 mL to give a 20 µg/mL solution.

Hard

Multiple-Choice 3-1

Significant Figures

1.

For which measurement has the number of significant digits been incorrectly determined?

A) B) C) D) E)

12.9830 g 1920.3 m 0.0043 g 1.003 L 59.0 km

six significant digits five significant digits four significant digits four significant digits three significant digits

Answer: C Easy 2.

The distance between two cities is measured and the reported value with ambiguous significant digits is: A)

740. km.

23 B) C) D) E)

7.40  102 km.

740.0 km.

740 km.

7.4  102 km.

Answer: D Easy

3-2

Significant Figures in Arithmetic

3.

Calculate the mass of a heterogeneous mixture containing 139.32 grams sand, 34.99 grams gravel, and 9.372 grams salt.

A) B) C) D) E)

184 g 183.7 g 183.68 g 184.0 g 183.684 g

Answer: C Intermediate

24 4. tall.

Calculate the volume of an aquarium that is 73.1 cm long, 30.1 cm deep, and 25.44 cm

A) B) C) D) E)

55975 cm3 55980. cm3 5.6  104 cm3 5.60  104 cm3 55975.9 cm3

Answer: D Intermediate

5.

Calculate the mass of a concrete slab that is 51.0 cm long, 17.34 cm deep, and 6.2 cm tall with a density of 5.3 g/cm3.

A) B) C) D) E)

1.0  103 g 2.9  104 g 2.91  104 1.03  103 g 2.906  104 g

Answer: B Intermediate 6.

For significant digits and calculations, the following statements are true, EXCEPT for which one(s)?

I II III

A) B) C) D) E)

Addition and subtraction―round off the answer according to the number of decimal places in the number with the most decimal places.

Multiplication and division―the answer is limited to number of digits in the number with the fewest significant digits.

Logarithms and antilogarithms ―the number of significant figures in the antilogarithm should equal the number of digits in the mantissa.

I II III I and II I and III

Answer: A Intermediate

25 3-3

Types of Error

7.

A) B) C) D) E)

is a consistent error that can be detected and corrected.

User error Internal error Systematic error Random error Indeterminate error

Answer: C Intermediate 8.

A) B) C) D) E)

cannot be eliminated but it may be reduced by a better experiment.

User error Internal error Systematic error Random error Determinate error

Answer: D Intermediate 9.

A) B) C) D) E)

expresses the margin of uncertainty associated with a measurement.

Nominal uncertainty Maximum uncertainty Relative uncertainty Minimum uncertainty Absolute uncertainty

Answer: E Intermediate 10.

For a fixed absolute uncertainty, as the magnitude of the measurement the percent relative uncertainty .

A) B) C) D) E)

increases; decreases increases; increases decreases; decreases decreases; increases increases; remains constant

Answer: A Intermediate

,

26

11.

Two students are tasked with determining the milligrams of chloride in a simulated soil sample. Both students extract chloride into aqueous solution and perform triplicate titration with silver nitrate solution and dichlorofluorescein indicator. The students report the following results to their laboratory instructor.

Trial 1 2 3 Average

Student #1 Student #2 (mg/g) (mg/g) 35.98 35.99 30.11 36.40 32.88 36.29 32.88 36.23

If the accepted value is 36.90 mg chloride/g sample, the laboratory instructor infers that student #1’s work exhibits student #2.

A) B) C) D) E)

higher accuracy and higher precision than higher accuracy and lower precision than lower accuracy and lower precision than lower accuracy and higher precision than results impossible to differentiate from

Answer: C Intermediate 12.

Two students are tasked with determining the milligrams of chloride in a simulated soil sample. Both students extract chloride into aqueous solution and perform triplicate titration with silver nitrate solution and dichlorofluorescein indicator. The students report the following results to their laboratory instructor.

Student #1 Student #2 Trial (mg/g) (mg/g) 1 35.98 35.99 2 30.11 36.40 3 32.88 36.29 Average 32.88 36.23 If the accepted value is 36.90 mg chloride/g sample, the laboratory instructor infers that student #2’s work exhibits student #1.

A) B) C) D) E)

higher accuracy and higher precision than higher accuracy and lower precision than lower accuracy and lower precision than lower accuracy and higher precision than results impossible to differentiate from

27 Answer: A Intermediate 13.

For the paired statements below, which are TRUE for the detection of an analytical methods systematic error?

I

Analyze a known sample.

II

Analyze blank samples.

III Analyze the sample using a different analytical technique.

IV Analyze the sample by different people in different laboratories with the same or different techniques.

A) B) C) D) E)

No systematic error if known answer reproduced.

Non-zero answer indicates systematic error.

If results do not agree, systematic error in one or both methods.

Disagreement beyond estimated random error is systematic error.

I and II II and IV I and IV II, III, and IV I, II, III, and IV

Answer: E Intermediate 3-4

Propagation of Uncertainty from Random Error

14.

A mass of approximately 10 grams is measured using weight by difference. If a vial containing calcium carbonate powder has an initial mass of 87.36 ± 0.03 g and a final mass of 76.99 ± 0.03 g, what mass of calcium carbonate was taken?

A) B) C) D) E)

10.37 ± 0.04 g 10.37 ± 0.03 g 10.370 ± 0.042 g 10.370 ± 0.004 g 10.370 ± 0.005 g

Answer: A Intermediate

28 15.

The density of a solution is determined using a class A 25-mL volumetric flask and an analytical balance. If the mass of the empty volumetric flask is 26.9872 ± 0.0003 g, the mass of the flask filled with solution is 53.9820 ± 0.0003 g, and the tolerance for the flask is ± 0.03 mL, what is the density of the solution?

A) B) C) D) E)

1.07979 ± 0.00001 g/mL 1.0798 ± 0.0004 g/mL 1.0798 ± 0.0003 g/mL 1.080 ± 0.001 g/mL 1.0798 ± 0.0005 g/mL

Answer: D Hard 3-5

Propagation of Uncertainty from Systematic Error

16.

When calculating the uncertainty associated with molar mass, which of the following statements are TRUE? I II III

Uncertainty in the mass of n identical atoms = n  (uncertainty in atomic mass for one atom.)

Uncertainties for different elements are dependent.

The uncertainty for the sum of the masses of different elements is calculated using the equation %e =

(%e1 ) + (%e2 ) + ..... .

2

2

IV

When summing the masses of different elements, the uncertainties for the mass of different elements are propagated as random errors.

A) B) C) D) E)

I, II, and III I and III II, III, and IV I and IV II and III

Answer: D Easy 17.

Calculate the error for the molar mass of acetic acid, CH3CO2H. The atomic masses for each element are C = 12.0106 ± 0.0010, H = 1.00798 ± 0.00014, and O = 15.9990 ± 0.0008.

A) B) C) D) E)

± 0.010 g/mol ± 0.0026 g/mol ± 0.0013 g/mol ± 0.0042 g/mol ± 0.021 g/mol

29 Answer: B Hard Calculated 3-2

Significant Figures in Arithmetic

18.

To the correct number of significant digits, what is the molarity of a sodium hydroxide solution when 12.9 grams of NaOH is dissolved in enough water to prepare 0.500 L of solution?

(NaOH = 40.00 g/mol) Answer: 0.654 M Intermediate 3-4

Propagation of Uncertainty from Random Error

19.

An irregular solid with a mass of 25.9863 ± 0.0003 g displaces 20.5 ± 0.2 mL of water.

Calculate the density of the solid.

Answer: 1.27 ± 0.01 g/mL Intermediate 3-5

Propagation of Uncertainty from Systematic Error

20.

Calculate the molar mass of diethyl ether, CH3CH2OCH2CH3. (C = 12.0106 ± 0.0010, H = 1.00798 ± 0.00014, and O = 15.9990 ± 0.0008).

Answer: 74.121 ± 0.0043 g/mol Hard Multiple-Choice 4-1

Gaussian Distribution

1.

The Gaussian distribution is characterized by the

A) B) C) D) E)

mean; precision accuracy; standard deviation mean; standard deviation accuracy; precision precision; standard deviation

and the

.

Answer: C Easy 2.

Which of the statements below are TRUE regarding the mean and standard deviation?

30 I II III IV

As the number of measurements increases, ̅ approaches  if there is no random error.

The square of the standard deviation is the average deviation.

The mean is the center of the Gaussian distribution.

The standard deviation measures the width of the Gaussian distribution.

A) B) C) D) E)

II, III, and IV I, III, and IV I and II I, II, and IV III and IV

Answer: E Intermediate 3.

Calculate the mean and standard deviation for the results below for the concentration of lead in a soil sample.

23.2 ppm, 20.1 ppm, 24.7 ppm, 19.9 ppm, 21.8 ppm A) B) C) D) E)

21.9  3.4 (n = 5) ppm 21.9  2.0 (n =5) ppm 27.4  2.0 (n = 5) ppm 22.0  1.8 (n = 5) ppm 27.4  4.2 (n = 5) ppm

Answer: B Hard

4-2

Comparison of Standard Deviations with the F Test

4.

A technician determines the concentration of calcium in milk using two instrumental methods. If Fcalculated > Ftable for the two sets of calcium data, what conclusion(s) can the technician make?

I II III IV

The difference in standard deviations for the two instrumental methods is significant.

The difference in standard deviations for the two instrumental methods is not significant.

The data comes from populations with the same standard deviation.

The data does not come from populations with the same standard deviation.

A) B) C) D) E)

I and III I and IV II and III II and IV Only II

31 Answer: B Hard

5.

A technician determines the concentration of calcium in milk using two instrumental methods. If Fcalculated < Ftable for the two sets of calcium data, what conclusion(s) can the technician make?

I II III IV

The difference in standard deviations for the two instrumental methods is significant.

The difference in standard deviations for the two instrumental methods is not significant.

The data comes from populations with the same standard deviation.

The data does not come from populations with the same standard deviation.

A) B) C) D) E)

I and III I and IV II and III II and IV Only II

Answer: C Hard

32 4-3

Confidence Intervals

6.

For the statements below, which is(are) TRUE for confidence intervals?

I II

As the percentage confidence increases, the confidence interval range decreases.

Confidence intervals are calculated using the calculated mean and standard deviation of a set of n measurements; and the results of the F test.

The 95% confidence interval will include the true population mean for 95% of the sets of n measurements.

III

A) B) C) D) E)

I only II only I and III I and II III only

Answer: E Intermediate

33 Note to Instructor: This question requires a small table, listing the 95% confidence interval student t values for the first 10 degrees of freedom. Reference Table 4-4 page 72.

7.

The density of a solution is measured six times with the results of 1.098, 1.100, 1.089, 1.095, 1.097 and 1.101 g/mL. Calculate the 95% confidence interval for the density.

A) B) C) D) E)

1.0967  0.0043 g/mL 1.0967  0.0038 g/mL 1.0967  0.0041 g/mL 1.0967  0.0045 g/mL 1.0967  0.0039 g/mL

Answer: D Intermediate 4-4

Comparisons of Means with Student’s t

8.

The null hypothesis for the comparison of two means using the t test is:

A) B) C)

the two sets of measurements come from populations with the same standard deviations. the two sets of measurements come from populations with different population means. the two sets of measurements come from populations with the same population mean and different population standard deviations. the two sets of measurements come from populations with different standard deviations. the two sets of measurements come from populations with the same population mean.

D) E)

Answer: E Difficulty: Intermediate 9.

A biosensor was developed to measure aqueous arsenic concentration. To determine the accuracy of the sensor, the sensor was tested against a standard reference sample. The biosensor results agree with the standard reference sample when: A) B) C) D) E)

the accepted value falls outside the 95% confidence interval for biosensor results. the accepted value falls within the 95% confidence interval for biosensor results. the standard deviations for the biosensor and accepted value are not significantly different. the standard deviations for the biosensor and accepted value are not significantly different and tcalculated < ttable. the standard deviations for the biosensor and accepted value are significantly different and tcalculated < ttable.

Answer: B Intermediate

34 10.

To determine if two methods result in statistically different results for calcium, a technician analyzes a calcium sample five times with each method. If the F test determines the standard deviations are significantly different, which set of equations must the technician use to calculate t.

A)

B) C)

s pooled

s12 n1 + s 22 n 2   n1n 2 , where s pooled = 2 2 n1 +n 2 s12 n1  + s22 n 2  n1 -1 n 2 -1

t calculated =

x1− x2 s pooled

n1n 2 , where s pooled = n1 +n 2

t calculated =

s n  s n  , where degrees of freedom = s n  + s n  s n + s n 

t calculated =

2

x1− x-2

s12 n1 −1+s 22 n 2−1 n1 +n 2 −2 2 1

x1− x 2

2 1

1

2 2

2

2 1

1

2

1

n1−1

D)

t calculated =

E)

None of these.

x1−x 2

, where degrees of freedom = 2 2 s n + s n  1 1  2 2 

2

2 2

2

2 2

2

2

n 2−1

s12 n1−1+s22 n 2 1 n1 +n 2 −2

Answer: C Hard

4-6

Grubbs Test for an Outlier

11.

A pair of students standard a sodium hydroxide solution using KHP and obtain the following concentrations: 0.1028 M, 0.1031 M, 0.1200 M, 0.1030 M and 0.1026 M. The students suspect that 0.1200 M is an outlier. Is 0.1200 M an outlier and what is the average concentration of the sodium hydroxide solution?

A) B) C) D) E)

0.1200 M is an outlier; ̅ = 0.1029 M 0.1200 M is not an outlier; ̅ = 0.1063 M 0.1200 M is not an outlier; ̅ = 0.1029 M 0.1200 M is an outlier; ̅ = 0.1063 M Tha answer is impossible to calculate.

Answer: A Intermediate

35 12.

A data point may be excluded when:

I II III IV

Gcalculated is greater than Gtable.

Gcalculated is less than Gtable. your TA or supervisor says to disregard the datum. the datum is the result of a faulty procedure.

A) B) C) D) E)

II and IV I, III, and IV I and IV II, III, and IV I, II, III, and IV

Answer: C Intermediate 4-7

The Method of Least Squares

13.

Which of the following are true for the method of least squares?

I

IV

The method of least squares draws the ―best‖ straight line through experimental data points.

The equation for a straight line is y = bx + m.

Deviations in y are minimized because the uncertainty in y is assumed to be greater than the uncertainty in x.

The slope and y-intercept are calculated using determinants.

A) B) C) D) E)

I, II, and III I, III, and IV I, II, and IV II, III, and IV III and IV

II III

Answer: B Intermediate

36 14.

Doubling the number of calibration curve data points decreases the standard uncertainty in the slope and y-intercept by a factor of: A) B) C) D) E)

2. ½.

4. √ . √

Answer: E Hard 4-8

Calibration Curves

15.

Which term is INCORRECTLY defined?

A)

linear range: the range over which the response of an analytical method is proportional to analyte concentration corrected absorbance: the average absorbance of the blanks is subtracted from each measured absorbance blank solution: solution containing all reagents and solvents used in the analysis to which no analyte is deliberately added standard solutions: solutions for which the standard deviation is known dynamic range: concentration range over which there is a measurable response to analyte, though the response is not linear

B) C) D) E)

Answer: D Intermediate

37 16.

The absorbance for a dye sample of unknown concentration is measured and the corrected absorbance is 0.5482. Calculate the concentration of the dye sample.

A) B) C) D) E)

0.00065 g/mL 0.00052 g/mL 0.00092 g/mL 0.0011 g/mL 0.00079 g/mL

Answer: A Intermediate Calculated 4-1

Gaussian Distribution

Note to Instructor: This question requires the use of Table 4-1 on page 67 of the textbook.

17.

The average percent purity for a batch of acetic acid manufactured by Acetic Acid Corp is 97.5% with a standard deviation of 0.2%. The minimum percent purity a buyer requires is 97.0%. What percentage of batches will have a percent purity unacceptable to the customer?

Answer: 0.62% The value of Z is −2.5 with an area of 0.4938. The percentage below 97% is 0.50 − 0.4938 = 0.0062 or 0.62%.

Hard

38 4-2

Comparison of Standard Deviations with the F Test

Note to Instructor: This question requires the use of Table 4-3 on page 70 of the textbook.

18.

A technician determines the concentration of calcium in milk using two instrumental methods. For method A, ̅ = 95.03 ppm, s = 0.40 (n = 8) and for method B, ̅ = 94.96 ppm, s = 0.72 (n = 6). Is the difference in standard deviation significant at the 95% confidence level? Show all work.

Answer: Fcalc = (0.72 × 0.72)/(0.40 × 0.40) = 3.24 Ftable = 3.97 for n1−1 = 5 and n2−1 = 7 Fcalc (= 3.24) < Ftable (=3.97), reject hypothesis that s1 (= 0.72) is significantly larger than s2 (= 0.40).

Not significantly different.

Difficultly: Hard 4-3

Confidence Intervals

Note to Instructor: This question requires the use of Table 4-4 on page 72 of the textbook.

19.

The molarity of a sodium hydroxide solution was determined by titration against KHP to be 0.1025 M, 0.1087 M, 0.1100 M, 0.0951 M, and 0.0997 M. Calculate the 95% confidence level for the molarity sodium hydroxide.

Answer: 0.1032  0.0077 M Intermediate Calculate the average and standard deviation, 0.1032 and 0.00622.

At the 95% confidence level with 4 degrees of freedom, t = 2.776 ts/sqrt(n) = 0.0077 4-4

Comparisons of Means with Student’s t

Note to Instructor: This question requires the use of Table 4-4 on page 72 of the textbook.

20.

A researcher has developed a new analytical method to determine the percent by mass iron in solids. To test the new method, the researcher purchases a standard reference material sample that is 2.85% iron by mass. Analysis of the iron standard returns values of 2.75%, 2.89%, 2.77%, 2.81%, and 2.87%. Are the results of the new method significantly different from the standard reference sample?

Answer: The 95% confidence level for the new method is 2.798  0.097% by mass Fe or 2.70% to 2.90%. The accepted value falls within the 95% confidence level. x = 2.798, s = 0.078, t = 2.776 for 4 degrees freedom and 95% confidence, and n = 5 Hard

39 4-6

Grubbs Test for an Outlier

Note to Instructor: This question requires the use of Table 4-6 on page 81 of the textbook.

21.

The molarity of a sodium hydroxide solution was determined by titration against KHP to be 0.1025 M, 0.1087 M, 0.1100 M, 0.1052 M, and 0.0997 M for pooled class data. The TA suspects that 0.0997 M is an outlier. Can the TA throw out the data point? Show all work.

Answer: The data point cannot be thrown out because Gcalc (= 1.29) is smaller than Gtable (=1.67). ̅ = 0.1052, s = 0.0043, n = 5 Intermediate 4-8

Calibration Curves

22.

The concentration and absorbance data below was collected for compound X. The least squares line for concentration versus absorbance is y = 6.83x + 0.0881. If a sample of compound X of unknown concentration has a corrected average absorbance of 0.6521 (three replicate measurements) calculate the concentration compound X and the uncertainty in the concentration.

Concentration (mg/mL) Absorbance 0.050 0.4426 0.075 0.5744 0.100 0.7841 Answer: 0.0826  0.0039 mg/mL compound X Hard Multiple-Choice 5-1

Basics of Quality Assurance

1.

A is a sample containing all components except analyte, and it is taken through all steps of the analytical procedure.

A) B) C) D) E)

method blank reagent blank field blank spike blank matrix blank

Answer: A Easy 2.

A is a sample that has been exposed to the site of sampling, containing all components except analyte, and it is taken through all steps of the analytical procedure.

40 A) B) C) D) E)

method blank reagent blank field blank spike blank matrix blank

Answer: C Easy 3.

The

A) B) C) D) E)

blind sample performance test sample spike matrix assessment

Answer: D Easy

is everything in the sample, except analyte.

41 4.

The

state(s) what steps will be taken and how they will be carried out.

A) B) C) D) E)

assessment calibration check control charts specifications standard operating procedure

Answer: E Easy 5.

A is an analytical result that indicates the sample concentration is below a stated limit when the concentration is actually above the limit.

A) B) C) D) E)

positive result false negative negative result false positive negative

Answer: B Easy

5-2

Method Validation

6.

There are many studies required to validate an analytical method. Which of the studies below is not required for method validation?

A) B) C) D) E)

method specificity linearity range calibration limit of detection

Answer: D Easy

42 7.

Which of the following is(are) not a way to demonstrate the accuracy of a method?

I II III IV

Determine the limit of detection for the method.

Analysis of a certified reference material in a matrix similar to the sample.

Compare the results between two or more different analytical methods.

Analyze a blank sample with a known amount of analyte in the same matrix as the sample.

A) B) C) D) E)

Only IV II and III I and III Only III Only I

Answer: E Easy 8.

There are several types of precision involved with method validation.

A) B) C) D) E)

instrument precision final precision intermediate precision interlaboratory precision intra-assay precision

Answer: B Easy 9. response.

A) B) C) D) E)

is the concentration range over which there is measurable instrument

Accuracy range Dynamic range Linear range Detection range Range

Answer: B Easy

43 5-3

Standard Addition

10.

Which statement regarding standard addition to a single solution is FALSE?

A B) C)

Analyte signal is measured after each standard addition.

The volume standard added must be kept small to minimize altering the matrix.

Detector response is corrected to take in account dilution before being plotted on the yaxis.

The absolute value of the y-intercept for the best fit line is the unknown analyte concentration.

Volume diluted standard is plotted on the x-axis.

D) E)

Answer: D Intermediate 11.

Which statement regarding the multiple solutions with constant volume standard addition method is FALSE? A) B) C) D) E)

Equal volumes of unknown are pipetted into a set of volumetric flasks with identical volume.

Equal volumes of standard are pipetted into each volumetric flask.

Analytical reagents are added to each volumetric flask and diluted to volume.

Detector response versus diluted standard is plotted.

The absolute value of the x-intercept for the best-fit line is the diluted unknown concentration.

Answer: B Difficulty: Intermediate 12.

A set of five solutions are prepared by delivering 10 mL of unknown sample and increasing volumes of standard (0, 5, 10, 15, and 20 mL) into 50-mL volumetric flasks.

Analytical reagents are then added and diluted to volume. The x-intercept is −3.90 mM. What is the unknown concentration?

A) B) C) D) E)

39.0 mM 156 mM 19.5 mM 0.390 mM 195 mM

Answer: C Intermediate

44 5-4

Internal Standards

13.

A(n) is a known amount of compound, different from the analyte, that is added to the unknown.

A) B) C) D) E)

standard addition internal standard external standard self-contained standard known standard

Answer: B Easy 14.

Which are TRUE statements regarding internal standards?

I

The response factor is calculated by using a mixture containing known amounts of analyte and internal standard and by using detector response to the analyte and the internal standard.

Internal standards are useful where instrument response or quantity of sample analyzed varies over time.

Detector response is inversely proportional to concentration.

Detector response to analyte and internal standard is assumed to be constant over a range of concentrations.

II III IV

A) B) C) D) E)

I, II, and III II, III, and IV I and II I and III I, II, and IV

Answer: E Intermediate 15.

A solution is 2.0% by mass compound A and 3.5% by mass compound I. Separation via GC returns peak areas of 38234 for compound A and 23880 for compound I. Calculate the response factor.

A) B) C) D) E)

0.664 2.03 1.50 0.491 1.17

Answer: B Intermediate

45

16. The response factor, F, for the analysis of pentane with the internal standard decane is 3.5.

Which statements below are INCORRECT for the analysis?

I II III IV

The pentane concentration is 3.5 times greater than the decane concentration.

The decane concentration is 3.5 times greater than the pentane concentration.

Detector response to pentane is 3.5 times greater than detector response to decane.

Detector response to decane is 3.5 times greater than detector response to pentane.

A) B) C) D) E)

I and III II and III II and IV I and IV None of these is incorrect.

Answer: A Intermediate Calculated 5-1

Basics of Quality Assurance

17.

A water sample collected downstream from a gold-mining operation returns a cyanide concentration of 6.18 g/L CN−. A 10-mL aliquot is spiked with 100 L of 500.0 g/L sodium cyanide and analyzed. Analysis returns a cyanide concentration of 10.5 g/L CN−. What is the percent recovery for the spike?

Answer: 13.1% recovery without accounting for dilution of the original sample; 11.5% recovery when accounting for dilution of the original sample.

Intermediate

46 5-2

Method Validation

18.

The detection limit for caffeine analysis must be calculated. Eight absorbance readings from a caffeine sample close to the estimated detection limit are 0.1239, 0.1098, 0.1182, 0.1150, 0.1219, 0.1251, 0.1199, and 0.1099. The average absorbance for eight blanks is 0.0182. The slope of a calibration curve is 0.59 /mM. What is the detection limit?

Answer: 0.030 mM Intermediate 5-3

Standard Addition

19.

A chemist is tasked with determining the concentration of nickel for a water sample drawn from a well next to a nickel mine. To analyze the sample using atomic absorbance spectroscopy (AAS), she prepares six samples (0-5) by adding 10.00 mL well water to a 50.00mL volumetric flask, then adding an increasing volumes of 25.00-ppm nickel solution, diluting the resulting solution to 50.00 mL with 0.1 M HNO3. Following AAS analysis, a graph of absorbance versus ppm nickel standard gave a best-fit line y = 0.0428x + 0.0487. Calculate the ppm nickel in the well water sample.

Answer: 5.69 ppm Hard 5-4

Internal Standards

20.

The percent by mass octane is quantified for an organic sample via GC using decane as the internal standard. A GC vial containing 0.5023 grams organic sample and 0.2384 grams octane is prepared and analyzed, returning peak areas of 119,993 for octane and 83755 for decane. If a response factor of 2.50 from previous work analyzing octane with a decane internal standard is used, what is the percent by mass octane in the organic sample?

Answer: 27.3% octane Hard Multiple-Choice 6-1

The Equilibrium Constant

1.

For the reaction aA (aq) + bB (s)  cC (l) + dD (g), all of the statements below are TRUE, except: A) B) C)

concentrations of solutes are expressed in moles per liter. concentrations of gases are expressed in bars.

47 D) E)

concentrations of pure solids, pure liquids and solvents are omitted because they are negligible. quantities in the equilibrium expression are the ratio of the concentration of the species to the concentration in its standard state.

Answer: D Easy 2.

When the direction for a reaction is reversed the new value of K is:

A) B) C) D) E)

K = log(K).

K = ln(K).

K = 10K. K = 1/K. K = eK.

Answer: D Easy 6-2

Equilibrium and Thermodynamics

3.

Which statements below are TRUE for enthalpy and entropy?

I II

Enthalpy is the heat flow, q, under constant applied pressure.

Standard conditions for entropy and enthalpy changes are 1 M concentration, 1 bar pressure and 298.15 K. The entropy change for a system is the amount of energy, at a given temperature, that is dispersed into motion of the molecules in the system.

When the value of H is negative, heat is released. When the value of S is negative, entropy is increased.

III IV

A) B) C) D) E)

I, II and III I, III and IV II and IV III and IV Another combination is true.

Answer: A Intermediate 4.

For a reaction with a Ho = +50 kJ and a So = +100.0 J/K, Go is the equilibrium constant is .

A) B) C)

negative; less than 1 negative; greater than 1 positive; less than 1

and

48 D) E)

positive; greater than 1 zero; 1

Answer: C Easy 5.

For the reaction HCN (aq) + H2O (l)  CN- (aq) + H3O+ (aq), all of the following will shift the equilibrium toward the products, except: I II III IV

adding HCN. adding H2O. removing CN−. removing H3O+.

A) B) C) D) E)

I II I, III, and IV II and IV III

Answer: B Easy

49 6-3

Solubility Product

6.

The molar solubility of a saturated iron(II) carbonate solution derived from the Ksp value is 5.6 x 10−6 M. The molar solubility is greater than 5.6 x 10−6 M when accounting for additional reactions. Which of the reactions below will not increase the solubility of iron(II) carbonate?

A) B) C) D) E)

Fe2+ (aq) + OH− (aq)  FeOH+ (aq) CO32− (aq) + H+ (aq)  HCO3− (aq) HCO3− (aq) + H+ (aq)  H2 CO3 (aq) H2CO3 (aq)  CO2 (g) + H2O (l) FeOH+ (aq) + OH−  Fe(OH)2 (s)

Answer: E Intermediate 7.

In which solution is solubility of silver chloride the largest?

A) B) C) D) E)

Pure water 0.05 M CaCl2 0.10 M NaCl 0.10 M FeCl3 0.15 M AgNO3

Answer: A Difficulty: Intermediate 8.

A student attempts to separate Cl− and I− from a solution that is 0.05 M Cl− and 0.10 M I− with Pb(NO3)2. As Pb(NO3)2 is added, which anion will precipitate first and what will the concentration of the anion be when the second anion begins to precipitate? Ksp = 1.7 x 10−5 for PbCl2 and Ksp = 9.8 x 10−9 for PbI2.

A) B) C) D) E)

iodide precipitates first; 0.0068 M iodide iodide precipitates first; 0.0012 M iodide iodide precipitates first; 0.05 M iodide chloride precipitates first; 0.0012 chloride chloride precipitates first; 0.10 M chloride

Answer: B Hard

50 6-4

Complex Formation

9.

Nickel forms three complexes with hydroxide, NiOH+ (aq), Ni(OH)2 (aq) and Ni(OH (aq). Which statement(s) regarding the impact of complex formation on the solubility of nickel hydroxide are FALSE? I II III

Hydroxide behaves as a Lewis acid when forming a complex with nickel.

The total nickel concentration is [Ni2+]tot = [Ni2+] + [NiOH+] + [Ni(OH)2] + [Ni(OH ].

As [OH-] increases, the solubility of Ni(OH)2 initially decreases but then increases.

IV A) B) C) D) E)

Only I II and III Only III Only IV I, II, III, and IV

Answer: A Intermediate 6-5

Protic Acids and Bases

10.

Which of the following is INCORRECT for Brønsted-Lowry acids and bases?

A) B) C) D) E)

Acids are proton donors.

NH4+ is the conjugate acid of the weak base NH3.

When an acid and a base react, the acid and base neutralize each other and form a salt.

Water reacts with itself to form hydronium and hydroxide, a process called autoprotolysis.

CH3CN is a protic solvent.

Answer: E Intermediate

51 6-6

pH

11.

What is the pH for a 5 x 10−5 M OH− solution?

A) B) C) D) E)

5.70 4.30 9.70 5.00 9.00

Answer: B Difficulty: Easy 12.

The pH of a solution is 9.65. The solution is concentration is than the hydronium concentration.

A) B) C) D) E)

acidic; less acidic; greater neutral; equal basic; greater basic; less

Answer: D Easy 6-7

Strengths of Acids and Bases

13.

All of the following are strong acids, except:

A) B) C) D) E)

HCl. HNO3.

HBr. HClO4.

HF.

Answer: E Easy

and

the

hydroxide

52 14.

Which of the classes of acids and bases and strength have been incorrectly matched?

A) B) C) D) E)

RCO2H: weak acid R3N: weak base RCO2−: weak base R3NH+: weak acid R4NOH: weak base

Answer: E Intermediate 15.

For a triprotic acid, which of the following is NOT true?

A)

Ka1Kb3=Kw

B) C) D)

Ka1Kb2=Kw Ka3Kb1=Kw

E) Answer: C Intermediate Calculated 6-2

Equilibrium and Thermodynamics

16.

A reaction has a  = −167.5 kJ/mol and  equilibrium constant for the reaction.

Answer: K = 2.19 x 1032 Easy

= 57.3 J/mol K. Calculate the

53 6-3

Solubility Product

17.

Silver ion is being considered as a reagent for separating IO3− from CO32− in a solution that is 0.060 M in K2CO3 and 0.070 M in NaIO3. Which anion will precipitate first and what will its concentration be when the second anion begins to precipitate? Can the two ions be separated with 99.99% efficiency?

Ksp = 3.1 x 10−8 for AgIO3 Ksp = 8.1 x 10−12 for Ag2CO3 Answer: Iodate precipitates first, [IO3−] = 2.67 × 10−3M when carbonate begins to precipitate.

Percentage iodate removed is 96.2%, so they cannot be successfully separated.

Hard 18.

Calculate the molar solubility for calcium fluoride in a 0.10 M NaF solution. Ksp = 3.2 × 10−11 for CaF2.

Answer: 3.2 x 10−9 M Intermediate 6−4

Complex Formation

19.

Calculate the total nickel concentration for a solution prepared by dissolving Ni(OH)2 (s) in a 0.10 M OH− solution. Ksp = 6 × 10−16, 1 = 1.26 × 104, 2 = 1 × 109, and 3 = 1 × 1012.

Answer: 6.06 x 10-5 M Hard [Ni2+]tot = [Ni2+] + [NiOH+] + [Ni(OH)2] + [Ni(OH ] = Ksp[OH−]2 + 1[Ni2+][ OH−] + 2[Ni2+][ OH−]2 + 3[Ni2+][ OH−]3 Solve for Ni2+ from Ksp when [OH−] = 0.10 M, then substitute for 1, 2, and 3.

6-6

pH

20.

Calculate the pH for a 0.293 M OH− solution.

Answer: 13.467 Easy Calculate pOH and subtract from 14.

Multiple-Choice 7-1

Titrations

1.

Which of the following terms is INCORRECTLY defined?

54 A) B) C) D) E)

titration error: the difference between the end point and the equivalent point direct titration: titrant is added to analyte until reaction is complete equivalent point: volume of titrant added in excess of the end point to change a physical property of the analyte solution blank titration: titration performed without analyte to calculate titration error standardization: titration of a known amount of analyte to determine the concentration of the titrant

Answer: C Easy 2.

A compound available in high purity, does not decompose under ordinary storage conditions and stable while being dried by heating or vacuum is a: A) B) C) D) E)

secondary standard. primary standard. standard solution. indicator. titrant.

Answer: B Easy 3.

A(n) the equivalence point.

A) B) C) D) E)

primary standard titrant analyte indicator masking agent

Answer: D Easy

is a compound with a physical property that changes abruptly near

55 7-2

Titration Calculations

4.

A Na2CO3 standard solution is prepared by transferring 2.4817 grams of primary standard-grade sodium carbonate to a 250.0-mL volumetric flask, dissolving the sample in ~100 mL of distilled deionized water and diluted to the mark. A 25.00-mL aliquot is taken and titrated with 42.65 mL of HCl solution. Calculate the concentration of the HCl solution.

A) B) C) D) E)

0.05490 M 0.1598 M 0.3196 M 0.1098 M 0.1997 M

Answer: D Intermediate 5.

A 1.9450 g sample containing only Na2CO3 and NaHCO3 is titrated with 35.31 mL of 0.8724 M HCl. Calculate the percentage sodium carbonate and sodium bicarbonate.

A) B) C) D) E)

43.53% Na2CO3; 56.47% NaHCO3 44.45% Na2CO3; 55.55% NaHCO3 55.55% Na2CO3; 44.45% NaHCO3 71.62% Na2CO3; 28.38% NaHCO3 56.47% Na2CO3; 43.53% NaHCO3

Answer: E Hard 6.

The key step to titration calculations is to relate:

A) B) C) D) E)

moles titrant to moles analyte. volume titrant to volume analyte. grams titrant to grams analyte. conductivity titrant to conductivity analyte. reduction titrant to oxidation analyte.

Answer: A Easy

56 7-3

Precipitation Titration Curves

7.

For the precipitation titration of analyte X− by titrant M+, which of the following statements is NOT true?

A) B) C)

Concentration M+ at the equivalence point is the molar solubility of MX or [M+] = √ .

Concentration M+ past the equivalence point is the excess M+ after reacting with all of the X−.

Concentration M+ before the equivalence point is the allowed M+ for the excess [X−] or [M+] =

D) E)

.

The plot of the titration curve is [M+] vs VM+.

The titration reaction is M+ (aq) + X− (aq)  MX (s).

Answer: D Intermediate 8.

Calculate pPb when 25.00 mL of 0.100 M CO32− is titrated with 21.00 mL of 0.100 M Pb2+. Ksp = 7.4 x 10−14 for PbCO3.

A) B) C) D) E)

11.07 2.06 11.41 1.72 11.99

Answer: A Hard 9.

Calculate pPb when 25.00 mL of 0.100 M CO32− is titrated with 28.00 mL of 0.100 M 2+ Pb . Ksp = 7.4 x 10−14 for PbCO3.

A) B) C) D) E)

11.21 10.88 6.57 2.25 1.92

Answer: A Hard

57 10.

Calculate pPb at the equivalence point when 25.00 mL of 0.100 M CO32- is titrated with 0.100 M Pb2+. Ksp = 7.4 x 10−14 for PbCO3.

A) B) C) D) E)

5.17 7.96 6.57 4.87 8.26

Answer: C Hard 11.

For the graphical equivalence point determination for a precipitation reaction, which of the following is(are) TRUE? I II III

A) B) C) D) E)

The steepest slope of the titration curve is the equivalence point.

The equivalence point occurs where the first derivative of the titration curve reaches its largest value.

The equivalence point occurs where the second derivative of the titration curve equals zero.

I only II and III II only III only I, II, and III

Answer: E Intermediate 7-4

Titration of a Mixture

12.

The end point of some precipitation titrations for mixtures can be up to 3% high. The error is attributed to: A) B) C) D) E)

ion pairs. coprecipitation. turbidity. solubility of the precipitates. redox chemistry.

Answer: B Easy

58 7-6

End Point Detection

13.

Common indicators are anionic dyes that are attracted to positively charged particles produced immediately after the equivalence point.

A) B) C) D) E)

redox absorption acid-base adsorption complexometric

Answer: D Easy 14.

Argentometric titrations are titrations using:

A) B) C) D) E)

Au+.

Ag+.

Ar+.

Al3+.

As3−.

Answer: B Easy 15.

The end point for the Volhard titration is indicated by:

A) B) C) D) E)

adsorption of a colored indicator on the precipitate at the end point. reduction of a colored indicator at the end point. deprotonation of a colored acid-base indicator at the end point. formation of a soluble, colored complex at the end point. protonation of a soluble, colored complex at the end point.

Answer: D Intermediate Calculated 7-2

Titration Calculations

16.

A NaOH solution is standardized using the monoprotic primary standard potassium hydrogen phthalate, KHP (204.22 g/mol.) If 0.6986 g of KHP requires 43.92 mL of NaOH, what is the NaOH concentration?

Answer: 0.07789 M NaOH Easy

59 7-3

Precipitation Titration Curves

17.

Calculate pPb when 25.00 mL of 0.150 M CO32− is titrated with 17.00 mL of 0.150 M 2+ Pb . Ksp = 7.4 x 10−14 for PbCO3.

Answer: 11.59 Hard 18.

Calculate pPb when 25.00 mL of 0.200 M CO32− is titrated with 28.00 mL of 0.200 M Pb2+. Ksp = 7.4 x 10−14 for PbCO3.

Answer: 1.95 Hard 19.

Calculate pPb at the equivalence point when 25.00 mL of 0.120 M CO32− is titrated with 0.120 M Pb2+. Ksp = 7.4 x 10−14 for PbCO3.

Answer: 6.57 Hard 7-4

Titration of a Mixture

20.

An aqueous mixture that is 0.100 M Ni2+ and 0.100 M Co2+ M is titrated with a 0.100 M 2− CO3 . Which cation precipitates first and what is pCO32− at the first and second equivalence points. Ksp = 1.0 x 10−10 for CoCO3 and Ksp = 1.3 x 10−7 for NiCO3.

Answer: Co2+ precipitates first, pCO32− = 5.00, Ni2+ precipitates second, pCO32− = 3.44 Hard Multiple-Choice 8-1

The Effect of Ionic Strength on Solubility of Salts

1.

Which statement below is INCORRECT for the increased solubility of a sparingly soluble salt, AgCl, in an inert salt solution, NaNO3?

A) B) C) D) E)

The ionic atmosphere around the silver cations becomes increasingly negative as the nitrate anions are attracted to the silver cations.

The ionic strength around the chloride anions becomes increasingly positive as the sodium cations are attracted to the chloride anions.

The ionic atmosphere increases the attraction between silver cations and chloride anions, increasing solubility by creating more AgCl (aq) in solution.

Ions continually diffuse into and out of the ionic atmosphere.

The negative ionic atmosphere around silver cations results in an overall positive charge less than the positive charge of the silver cations.

60 Answer: C Intermediate 2.

A) B) C) D) E)

is a measure of the total concentration of ions in solution.

Ionic radius Ionic density Ionic degree Ionic strength Ionic atmosphere

Answer: D Easy 3.

What is the ionic strength of a 2:1 electrolyte with a concentration of 0.100 M?

A) B) C) D) E)

0.100 0.200 0.300 0.400 0.500

Answer: C Intermediate

61 8-2

Activity Coefficients

4.

To correct for the effect of ionic strength on chemical reactions, concentrations are replaced by in equilibrium expressions.

A) B) C) D) E)

activity coefficients ion activities molecular activities activities active strength

Answer: D Easy 5.

All of the following are TRUE for activities and activity coefficients, except:

A) B) C) D) E)

activity for a chemical species is the product of concentration and activity coefficient. the activity coefficient corrects for non-ideal behavior due to ionic strength. as ionic strength increases, the value of the activity coefficient increases. for ions, the activity coefficient approaches unity as the ionic strength approaches 0. the activity coefficient for neutral molecules is approximately unity when the ionic strength is less than 0.1 M.

Answer: C Intermediate 6.

For Ag+ when  = 0.01 M,  = 0.898 and when  = 0.05,  = 0.80. What is the activity coefficient when  = 0.024 M? A) B) C) D) E)

0.804 0.845 0.853 0.864 0.894

Answer: D Intermediate

62 8-3

pH Revisited

7.

Which of the following is NOT true for pH?

A) B) C) D) E)

pH = −log[H+]H+. pH = −log[H+] when  = 0.

The pH = pOH for pure water.

As ionic strength increases, so does the pH of pure water.

The ionic strength of pure water is 1.0 x 10−7 M.

Answer: D Intermediate 8.

Which statement below is NOT correct for the pH of a 0.01 M NaCl solution versus the pH of a 0.01 M in FeSO4 solution?

A) B) C) D) E)

The activity coefficient for H+ will be larger in 0.01 M NaCl than in 0.01 M FeSO4.

The activity coefficient for OH− will be larger in 0.01 M NaCl than in 0.01 M FeSO4.

The hydronium concentration in 0.01 M NaCl will equal the hydronium concentration in 0.01 M FeSO4.

The H+ activity in 0.01 M NaCl will not equal the H+ activity in 0.01 M FeSO4.

The OH− activity in 0.01 M NaCl will not equal the OH− activity in 0.01 M FeSO4.

Answer: C Intermediate 9.

Calculate the pOH of a 0.010 M HCl solution with an ionic strength of 0.10 M. H+ = 0.83, OH− = 0.76.

A) B) C) D) E)

11.92 11.80 11.88 12.00 12.20

Answer: A Intermediate

63 8-4

Systematic Treatment of Equilibrium

10.

A buffer is prepared from NaH2PO4 and Na2HPO4. The charge balance equation for the buffer is which of the following? 

A) B) C) D)

3 [Na+ ] + [H+ ] = [H2 PO4 ] + 2 [HPO24 ] + [OH] 3 [Na+ ] + [H+ ] = [H2 PO4 ] + [HPO24 ] + [OH] 

E)



[Na+ ] + [H+ ] = [H2 PO4 ] + 2 [HPO24 ] + [OH]



[Na+ ] + [H+ ] = [H2 PO4 ] + [HPO24 ] + [OH]

Answer: A Intermediate 11.

A) B) C) D) E)

The mass balance equation for a 0.10 M HO2CCO2H solution is which of the following?

0.10 M + [H+ ] = [HO2CCO2 ] + [  O2CCO2 ] + [OH] 0.10 M = [HO2CCO2 ] + [  O2 CCO2 ] + [OH]

Answer: D Intermediate 12.

Which of the statements below are INCORRECT for mass balance and charge balance?

I II

The mass balance is a statement of the conservation of matter.

Mass balance is satisfied when the quantity of all species in a solution containing a particular atom (or group of atoms) equals the amount of that atom (or group) reacted.

The sum of the positive charges in solution equals the sum of the negative charges in solution.

The coefficient in front of each species in a charge balance always equals the magnitude of the charge on the counter ion.

III IV

A) B) C) D) E)

I, II, III and IV I, III and IV I, II and III I and IV II and III

64

Answer: E Intermediate 8-5

Applying the Systematic Treatment of Equilibrium

13.

HA?

Which equation is NOT required to determine the pH of 0.10 M solution of weak acid,

A) B) C) D) E)

HA + H2O ⇋ A− + H3O+ H2O + H2O ⇋ H3O+ + OH− 0.10 = [HA] + [A−] [H3O+] = [OH−] + [A−] A− + H2O ⇋ HA + OH−

Ka = 2.4 × 10−6 Kw = 1 × 10−14 Kb = 4.2 × 10−9

Answer: E Intermediate 14.

Which equation is NOT required to determine the molar solubility of AgCN?

A) B) C) D) E)

AgCN ⇋ Ag+ + CN− H2O + H2O ⇋ H3O+ + OH− [Ag+] = [CN−] + [HCN] CN− + H2O ⇋ HCN + OH− HCN + H2O ⇋ CN− + H3O+

Ksp = 2.2 × 10−16 Kw = 1 × 10−14 Kb = 1.6 × 10−5 Ka = 6.2 × 10−10

Answer: E Intermediate 15.

Calculate the pH of a 0.010 M CH3CO2H solution. Ka = 1.8 × 10−5 for acetic acid.

A) B) C) D) E)

9.26 3.39 2.02 10.61 4.74

Answer: B Intermediate Calculated 8-2

Activity Coefficients

16.

Calculate the ionic strength of a 0.210 M FeCl2 solution.

65 Answer: 0.630 Intermediate A straightforward ionic strength calculation.

17.

Calculate the activity coefficient for Fe2+ for an ionic strength of 0.034 M. For  = 0.01 M,  = 0.675 and for  = 0.05 M,  = 0.485.

Answer:  = 0.561 for Fe2+ when  = 0.034 M Intermediate A straightforward interpolation problem.

8-3

pH Revisited

18.

Calculate the pH of a 0.040 M HCl solution that is also 0.010 M in NaNO3.

Answer: pH = 1.46 Intermediate This problem requires Table 8.1 for the students to look up the activity coefficients for H+ and Cl−. H+ = 0.86, Cl− = NO3− = 0.805 and Na+ = 0.82 when  = 0.05 M. Since HCl is a strong base, the hydronium concentration is 0.04. Calculate the pH using the activity coefficient for H+.

8-5

Applying the Systematic Treatment of Equilibrium

19.

Calculate the pH of a 0.001 M HCN solution. Ka = 6.2 × 10−10

Answer: pH = 6.10 Intermediate This question is a simple weak-acid equilibria problem.

20.

Calculate the pH of a 0.002 M (CH3)3N solution. Kb = 6.6 × 10−5

Answer: 10.62 Hard This question is a simple weak-base equilibria problem.

Multiple-Choice 9-1

Strong Acids and Bases

1.

There are seven common strong acids. Which of the following statements is NOT true for strong acids?

A) B)

Dilute solutions are completely ionized.

HNO2 is a strong acid.

66 C) D) E)

For strong acid solutions with a concentration between 10−6 and 10−8 M, the pH is determined using systematic equilibrium.

For strong acid solutions with concentrations  10−6 M, the pH is calculated from the concentration of the strong acid.

For strong acid solutions with concentrations  10−8 M, the pH is always 7.

Answer: B Intermediate 2.

There are two types of metal hydroxides, soluble metal hydroxides and insoluble hydroxides, both of which are strong bases. Which of the following statements is NOT true for strong bases?

A) B) C) D) E)

Dilute solutions are completely dissociated.

Fe(OH)2 is a strong soluble base.

For strong base solutions with a concentration between 10−6 and 10−8 M, the pOH is determined using systematic equilibrium.

For strong base solutions with concentrations  10−6 M, the pOH is calculated from the concentration of the strong base.

For strong base solutions with concentrations  10−8 M, the pOH is always 7.

Answer: B Intermediate 3. and

For a 0.0005 M HBr solution, the dissociation of water produces OH−.

A) B) C) D) E)

1.00E-07 M; 2.00E-11 M 5.00E-04 M; 5.00E-04 M 2.00E-11 M; 2.00E-11 M 1.00E-07 M; 1.00E-07 M 5.00E-04 M; 2.00E-11 M

Answer: C Difficulty: Intermediate

H3O+

67 9-2

Weak Acids and Bases

4.

Which of the following statements is NOT true for weak acids?

A) B)

Weak acids are partially ionized.

The conjugate base is A− and its strength is weak.

C)

[H+ ][A] Ka = [HA]

D) E)

for the conjugate base.

HA + H2O ⇋ A− + H3O+

Answer: D Intermediate 5.

Which of the following statements is NOT true for weak bases?

A) B)

Weak bases partially hydrolyze water.

The conjugate acid is BH+ and its strength is strong.

C) D) E)

for the conjugate acid.

B + H2O ⇋ BH+ + OH−

Answer: B Intermediate 6.

Which of the following is NOT true for weak acids and weak bases?

A)

Kw = KaKb

B) C) D) E)

pKa = −log Ka pKb = −log Kb

Answer: B Intermediate

68 9-3

Weak-Acid Equilibria

7.

The pH of a weak-acid solution is calculated using systematic treatment. Which of the following is NOT an equation needed to solve for pH? A) B) C)

F = [HA] + [A−] [H+] = [A−] + [OH−]

D) E)

Kw = [H+][OH−]

Answer: D Intermediate 8.

Calculate the percentage dissociation for a 0.010 M hypochlorous acid solution. The Ka for hypochlorous acid is 3.0 x 10−8.

A) B) C) D) E)

0.35% 0.30% 0.17% 0.99% 0.65%

Answer: C Intermediate 9.

The rule of thumb for neglecting the change in the formal concentration of a weak acid or a weak base, x, during an equilibrium problem is valid: A) B) C) D) E)

if the change in x is greater than 1% of the formal concentration. if the change in x is less than 0.1% of the formal concentration. if the change in x is equal to 0.1% of the formal concentration. if the change in x is less than 10% of the formal concentration. if the change in x is less than 1% of the formal concentration.

Answer: E Intermediate

69 9-4

Weak-Base Equilibria

10.

The pH of a weak-base solution is calculated using systematic treatment. Which of the following is NOT an equation needed to solve for pOH? A) B) C) D) E)

[B][H + ] [BH + ] F = [BH+] + [B] [OH−] = [BH+] + [H+] [BH + ][OH- ] Kb = [B] + Kw = [H ][OH−] Ka =

Answer: A Intermediate 11.

A student calculates the pOH for a 0.01 M triethanolamine solution to be 4.12. To simplify the math she decided to neglect the change in the formal concentration. Was she justified in her decision? Kb = 5.78 × 10−7 A) B) C) D) E)

She was justified, as the change in ×, 7.59 × 10−5, is less than 1% of the formal concentration.

She was not justified, as the change in ×, 9.92 × 10−3, is greater than 1% of the formal concentration.

She was justified, as the change in ×, 1.32 × 10−10, is less than 1% of the formal concentration.

She was not justified, as the change in ×, 1.00 × 10−2, is greater than 1% of the formal concentration.

She was justified, as the change in ×, 5.78 × 10−5, is less than 1% of the formal concentration.

Answer: A Difficulty: Intermediate 12.

A 0.01 M solution of a weak base has a percent association of 18.32%. Calculate the K b for the weak base.

A) B) C) D) E)

1.83 × 10−3 2.43 × 10 −11 5.46 × 10−12 4.11 × 10−4 1.22 × 10−12

Answer: D Hard

70 9-5

Buffers

13.

Which is NOT a property of buffers?

A) B) C) D) E)

Buffers resist changes in pH when acids or bases are added or when dilution occurs.

Buffers are a mixture of weak acid and conjugate base.

The pH of a buffer is independent of ionic strength.

The pH of a buffer is dependent on temperature.

Buffers are a mixture of weak bases and conjugate acid.

Answer: C Intermediate 14.

The buffer needed for a capillary electrophoresis experiment must have a pH of 10.0.

Which weak acid is the best choice for the buffer? Assume  = 0.

A) B) C) D) E)

ammonium chloride disodium phosphate 3-(Cyclohexylamino)propanesulfonic acid sodium dihydrogen borate cyclohexylaminoethanesulfonic acid

pKa = 9.24 pKa = 12.38 pKa = 10.40 pKa = 12.74 pKa = 9.39

Answer: C Intermediate 15.

A buffer is prepared by mixing 200.0 mL of 0.1500 M NaOH with 200.0 mL of 0.200 M CH3CO2H in a 1-L volumetric flask and then diluted to volume with distilled deionized water.

Calculate the pH of the buffer. Ka = 1.75 × 10−5 A) B) C) D) E)

5.23 9.72 7.00 8.77 4.28

Answer: A Intermediate

71 Calculated 9-1

Strong Acids and Bases

16.

Calculate the pH for a 4.3 × 10−8 M HCl solution with an ionic strength of  = 0.1. H+ = 0.83, OH− = 0.76 and Cl− = 0.755.

Answer: pH = 6.59 (When activities are neglected pH = 6.69.)

Hard The concentration of HCl is less than the hydronium concentration from the auto-ionization of water, so [OH−] is calculated from Kw and the concentration of HCl, then keeping in mind that for each OH− in water, a H+ must also be generated. Add [H+] from Kw and from HCl to get total [H+]. Use activity coefficients where needed, and when calculating the pH. 9-3

Weak-Acid Equilibria

17.

Calculate the pH of 0.01 M HCN ( = 0.10) using activity coefficients, neglecting activity coefficients, and the percent difference between the two. Ka = 6.2 x 10−10 for HCN. H+ = 0.83, OH− = 0.76 and CN− = 0.755.

Answer: pHactivities = 5.20. pH = 5.30. pHactivities is 1.92% less than pH. pH is 1.89% greater than pHactivities.

Hard Solve the weak-acid equilibrium problem with and without activity coefficients, then calculate the percent difference between the pH values.

18.

Calculate the percent dissociation for a 0.01 M HCN solution, neglecting activity coefficients. Ka = 6.2 × 10−10 for HCN. Answer: 0.0249% (Neglecting the change in x with respect to HCN because 0.0249% is less than 1%.)

Intermediate Calculate [CN−], divide by 0.01 M and multiply by 100.

9-4

Weak-Base Equilibria

19.

Calculate the pH of a 0.2 M ethylamine solution. Kb = 5.0 × 10−4

Answer: pH = 12.30 (If students report 12.00 then they failed to notice that neglecting the change in 0.2 M results in an error of 5% of 0.2.)

Hard Simple weak-base titration problem, and must be worked without assuming that the change in the ethylamine concentration is less than 1% of the Kb.

72 9-5

Buffers

20.

One liter of a pH 5.00 propionic acid buffer with a total concentration of 50 mM must be prepared. Calculate the mmols propionic acid and sodium propionate needed to prepare the buffer. Ka = 1.34 x 10-5 for propionic acid.

Answer: 28.6 mmols sodium propionate and 21.4 mmols propionic acid Hard Set mmol propionate equal to x and mmol propionic acid to 50-x. Plugged into the buffer equation to solve for x.

Multiple-Choice 10-1

Diprotic Acids and Bases

1.

Which of the answers below is INCORRECT for diprotic acids and bases?

A) B) C) D) E)

Solutions of H2A are treated as monoprotic to calculate [H+], [HA−] and [H2A].

The pH for HA− solutions is appro×imated with pH = ½(pK1 + pK2).

Ka1Kb1 = Kw and Ka2Kb2 = Kw Solutions of A2− is treated as monobasic to calculate [OH−], [HA−], and [A2−].

A zwitterion is a neutral molecule with both positive and negative charges.

Answer: C Intermediate 2.

Sulfurous acid is a diprotic acid with Ka1 = 1.40 × 10−2 and Ka2 = 6.73 × 10−8. Which of the statements below are FALSE? I II III IV

Kb2 = 7.14 × 10−13 Kb1 = 1.49 × 10−7 For 0.1 M H2SO3, [H+] = 7.61 × 10−2 for Ka1.

For 0.1 M NaHSO3, pH  4.515.

A) B) C) D) E)

I, II, and IV I, II, and III I and II I, II, III, and IV II, III, and IV

Answer: D Intermediate

73 10-2

Diprotic Buffers

3.

A buffer is prepared by dissolving 70 millimoles of solid NaOH in 500.0 mL of solution with a concentration of 50 mM H2A. Which of the following are NOT true for the buffer?

V

After reaction there is 30 mmol HA− and 20 mmol A2− in solution.

[A 2- ] pH is calculated using pH = pK1 + .

[HA- ] After reaction there is 50 mmol HA− and 20 mmol NaOH in solution.

[HA- ] pH is calculated using pH = pK1 + .

[H 2 A] pH = 14 – pOH where pOH is calculated from excess OH−.

A) B) C) D) E)

III, IV, and V II, III, IV, and V 1, III, and IV I, II, and V I and II

I II III IV

Answer: B Intermediate 4.

How many millimoles of NaOH or HCl must be added to 500.0 mL of 0.250 M NaO2CCO2H to prepare a pH 3.90 buffer? Assume no change in volume. pK1 = 1.250 and pK2 = 4.266.

A) B) C) D) E)

37.65 mmol NaOH 87.35 mmol HCl 124.7 mmol HCl 0.30 mmol NaOH The answer cannot be calculated from given information.

Answer: A Hard 5.

Calculate the pH of a 0.340 M NaO2CCO2H solution. pK1 = 1.250 and pK2 = 4.266.

A) B) C) D) E)

2.37 4.89 2.79 3.38 −0.029

Answer: C Intermediate

74

10-3

Polyprotic Acids and Bases

6.

For the polyprotic acid H3A, which of the equilibria below is INCORRECT?

A) B) C) D) E)

A3− ⇋ HA2− + OH− H2A− ⇋ HA2− + H+ HA2− ⇋ A3− + H+ H3A ⇋ H2A− + H+ HA2− ⇋ H2A− + OH−

Kb3 = Kw/Ka3 Ka2 = K2 Ka3 = K3 Ka1 = K1 Kb2 = Kw/Ka2

Answer: A Easy 7.

Which of the following is TRUE for an aqueous solution of the weak acid H3A?

I II III IV

H3A is acidic H2A− is acidic HA2− is amphiprotic A3− is basic

A) B) C) D) E)

II, III, and IV I, II, and IV I, II, and III I, III, and IV I, II, III, and IV

Answer: D Easy 8.

Calculate the pH of a 0.200 M H2A− solution. K1 = 1 × 10−4, K2 = 1 × 10−8 and K3 = 1 × −12 10 .

A) B) C) D) E)

4.35 6.00 8.00 8.65 10.00

Answer: E Intermediate

75 10-4

Which Is the Principal Species?

9.

Alanine is a diprotic amino acid with a pKa = 2.344 for the carboxylic group and a pKa = 9.868 for the ammonium group. At pH = 6.32, what is the principal species in solution?

A) B) C) D) E)

H2L+ L− HL H3L2+ L2−

Answer: C Easy 10.

Sodium hydroxide is added to a 0.1 M solution of phenylalanine. Phenylalanine is a diprotic acid with two pKas, 2.20 for the carboxylic acid group and 9.31 for the ammonium functional group. When the pH = 5.93, the principal species is and when the pH = 10.10, the principal species is .

A) B) C) D) E)

H2L+; H2L+ HL; L− H2L+; L− H2L+; HL HL; HL

Answer: B Easy 11.

Solid NaOH is added to 1 L of an isoleucine solution, H2L+, i.e, 100 mM. When 175 mmol of NaOH is added, the principal species is . The pKa for the carboxylic acid group is 2.318 and the pKa for the ammonium group is 9.758. Assume the isoleucine solution volume does not change appreciably.

A) B) C) D) E)

H2L+ L− HL H3L2+ L2−

Answer: B Intermediate

76 10-5

Fractional Composition Equations

12.

For a weak acid, the HA increased pH. A) B) C) D) E)

and the A-

with

increases; decreases increases; increases decreases; decreases decreases; increases is fixed; is fixed

Answer: D Intermediate 13.

The fractional composition diagram below is representative for diprotic weak acids.

Which point INCORRECTLY indicates the principal species?

A) B) C) D) E)

H2A HA− A− H2A = HA− HA− = A2−

Answer: C Easy

77 10-6

Isoelectric and Isoionic pH

14.

Which is NOT true for isoelectric and isoionic pH?

A) B) C)

Isoionic pH is the pH of the pure, neutral, polyprotic acid.

Isoelectric pH is the pH at which average charge of the polyprotic acid is +1.

The hydronium concentration at the isoionic point is calculated with the equation K1K 2 F + K1K w .

[H + ] = K1 + F The pH at the isoelectric point is calculated with the equation pH = ½(pK1 + pK2).

K1K 2 F + K1K w For the equation [H + ] = , F is the formal concentration of the weak acid.

K1 + F

D) E)

Answer: B Intermediate 15. Calculate the pH at the isoelectric point for a 0.10 M solution of the amino acid threonine. pKa1 = 2.088 and pKa2 = 9.100 for threonine.

A) B) C) D) E)

3.57 2.39 5.59 1.54 5.05

Answer: C Intermediate Calculated 10-1

Diprotic Acids and Bases

16.

Calculate the concentration of each species of oxalic acid in a 0.150 M solution of sodium oxalate, Na2C2O4. K1 = 5.62 × 10−2, K2 = 7.24 × 10−5 Answer: [A2−] = 0.150 M, [HA−] = 4.55 × 10−6 M, [H2A] = 1.78 × 10−13 Intermediate The hydrolysis of oxalate and hydrogen oxalate are calculated solving the hydrolysis of the acid oxalate, to give A2− and HA−. H2A equals the Kb2.

78 10-2

Diprotic Buffers

17.

Calculate the pH of a solution prepared by mixing 200.0 mL of 0.100 M NaOH with 150.0 mL of 0.200 M o×alic acid. K1 = 5.62 × 10−2, K2 = 7.24 × 10−5 Answer: pH = 1.55 Easy The mmol OH− added is sufficient to only convert some of the H2A to HA− which gives a buffer centered on the pK1.

10−3 Polyprotic Acids and Bases 18.

Calculate the pH for a 0.200 M sodium L−aspartic acid solution in the H2L form. pKa1 = 1.990, pKa2 = 3.900 and pKa3 = 10.002.

Answer: pH = 2.45 Hard The pH is calculated using the equation for the intermediate form, [H + ] = 10-5

K1K 2 F + K1K w .

K1 + F

Fractional Composition Equations

19.

Malonic acid is a diprotic acid with pKs of 2.847 and 5.696. Calculate the  for each malonic acid species in solution at pH =3.50.

Answer: H2A = 0.181, HA- = 0.814, A2- = 0.00518 at pH = 3.50 Intermediate Plugged the hydronium concentrations and converted pKs into the alpha equations for H2A, HA− and A2−.

10-6

Isoelectric and Isoionic pH

20.

Calculate the pH to which a malonic acid solution must be adjusted to reach the isoelectric point. pK1 = 2.847 and pK2 = 5.696.

Answer: 4.272 Easy The answer is the average of the pKs. Multiple-Choice 11-1

Titration of Strong Base with Strong Acid

1.

Fifty mL of 0.100 M HCl is titrated with 0.1500 M NaOH. The pH at the equivalence point is: A)

1.00.

79 B) C) D) E)

13.18.

1.22.

12.78.

7.00.

Answer: E Easy 2.

For the titration of a strong acid with a strong base, which statement below is FALSE?

A) B) C)

At V = 0 mL titrant added, the pH is due to initial concentration of strong acid.

At the equivalence point, the pH is due to the auto-ionization of water.

Between V = 0 mL and the equivalence point, the pH is due to diluted unreacted strong acid.

Past the equivalence point, the pH is due to diluted excess strong base.

The indicator changes color at the end point due to excess acid.

D) E)

Answer: E Easy 11-2

Titration of Weak Acid with Strong Base

3.

Which statement is FALSE regarding the titration of a weak acid with a strong base?

HA + OH− → A− + H2O

A) B) C) D) E)

At the equivalence point, the pH is due to the hydrolysis of water by the conjugate base, A−.

Past the equivalence point the pH is due to excess conjugate base, A−.

Before strong base is added, the pH is due to the ionization of the weak acid, HA. Before the equivalence point, the pH is due to a mixture of HA and A−.

The pH at the equivalence point is basic.

Answer: B Easy

80 4.

A 25.00 mL aliquot of 0.0900 M acetic acid is titrated with a 0.100 M KOH. Calculate the pH when 25.00 mL of base is added. Ka = 1.75 x 10−5 A) B) C) D) E)

2.30 5.27 5.71 8.73 11.70

Answer: E Intermediate 11-3

Titration of Weak Base with Strong Acid

5.

Which statement is false regarding the titration of a weak base with a strong acid?

B + H+ → BH+

A) B) C) D) E)

At the equivalence point, the pH is due to the weak acid dissociation of the conjugate acid, BH+.

Past the equivalence point the pH is due to excess strong acid, H+.

Before strong acid is added the pH is due to the ionization of the weak base, B. Before the equivalence point, the pH is due to a mixture of B and BH+, a buffer.

The pH at the equivalence point is acidic.

Answer: C Easy 6.

A 25.00 mL aliquot of 0.100 M NH3 is titrated with 21.00 mL of 0.0900 M HCl. Calculate the pH. Kb = 1.76 x 10−5 A) B) C) D) E)

5.25 8.44 8.75 5.56 7.00

Answer: C Intermediate

81 11-4

Titrations in Diprotic Systems

7.

Which statement is FALSE regarding the titration of a dibasic weak base with a strong acid?

B + H+ → BH+ BH+ + H+ → BH22+ A) B) C) D) E)

For V = 0, the pH is dictated by the reaction of B with H2O. Before the first equivalence point, Veq1 the buffer created by B and BH+ determines the pH. At Veq1 the pH is due to the reaction of BH+ with H2O. After the first equivalence point, Veq1 and before the second equivalence point, Veq2, the buffer created by BH+ and BH22+ determines the pH. At Veq2 the pH is due to the reaction of BH22+ with H2O.

Answer: C Easy 8.

A 40.00 mL sample of 0.1000 M diprotic malonic acid is titrated with 0.0900 M KOH. What volume KOH must be added to give a pH of 5.697? Ka1 = 1.42 × 10−3 and Ka2 = 2.01 × 10−6.

A) B) C) D) E)

66.67 mL 44.44 mL 22.22 mL 88.89 mL 111.11 mL

Answer: A Intermediate 11-5

Finding the End Point with a pH Electrode

9.

The end point for a titration can be determined graphically. The end point volume for a first derivative plot corresponds to: A) B) C) D) E)

the volume base added where the first derivative curve crosses the x-axis. the volume base added where the first derivative curve has maximum slope. the volume base added where the first derivative curve crosses the y-axis. the volume base added where the first derivative curve has minimum slope. the volume base added where the first derivative curve has a maximum negative slope.

Answer: B Difficulty: intermediate

82 10.

The end point for a titration can be determined graphically. The end point volume for a second derivative plot corresponds to: .

A) the volume titrant added where the second derivative curve crosses the x-axis.

B) the volume titrant added where the second derivative curve has maximum slope.

C) the volume titrant added where the second derivative curve crosses the y-axis.

D) the volume titrant added where the second derivative curve has minimum slope.

E) the volume titrant added where the second derivative curve has a maximum negative slope.

Answer: B Intermediate 11.

Which statements are TRUE for Gran plots?

I II III IV

Gran plots are more accurate around end point than first and second derivative plots.

The end point is determined using concentration/pH data prior to the end point.

For a constant slope, where the Gran plot crosses the x-axis is the end point volume.

Gran plots are used with polyprotic titration curves as well as monoprotic titration curves.

A) B) C) D) E)

I, II, and IV II, III, and IV I, II, and III I, II, III, and IV I, III, and IV

Answer: D Intermediate 11-6

Finding the End Point with Indicators

12.

Which statement is FALSE for indicators?

A) B) C) D) E)

Indicators are acids or bases whose various protonated forms have different colors.

The pH range over which an indicator changes color is the transition range Indicators are chosen to change color as close to the end point as possible.

The volume difference between the equivalence point and the end point is indicator error.

Indicators can change color more than once.

Answer: C Easy

83 11-7

Practical Notes

13.

Strong base solutions require standardization before use because strong bases:

A) B)

precipitate as insoluble metal hydroxides, decreasing the hydroxide concentration. react with atmospheric carbon dioxide to form carbonates, decreasing the hydroxide concentration. react with atmospheric nitrogen to form nitrogen oxides, decreasing the hydroxide concentration. adhere to the inside surface of the solution container, reducing the hydroxide concentration. react with solution impurities to form volatile compounds, increasing the hydroxide concentration.

C) D) E)

Answer: B Easy 11-8

Kjeldahl Nitrogen Analysis

14.

I II III IV

Arrange the steps below in the order required to perform a Kjeldahl Nitrogen Analysis.

Distillation of ammonia into standard hydrochloric acid.

Neutralize ammonium to release ammonia.

Digest organic sample with boiling sulfuric acid to convert nitrogen to ammonium.

Titrate unreacted standard hydrochloric acid with sodium hydroxide.

A) B) C) D) E)

III, II, I, IV II, I, IV, III I, IV, III, II IV, III, II, I III, I, IV, II

Answer: A Easy

84 11-9

The Leveling Effect

15.

Which statement is TRUE for the leveling effect?

I II III IV

Acids stronger than water react with water to form hydronium.

Hydronium is the strongest acid in water.

In water, strong acids have the same strength.

The differences in the strength of strong acids are observed in nonaqueous solvents such as acetic acid.

A) B) C) D) E)

I, III, and IV II and IV II and III II, III, and IV I and III

Answer: D Easy Calculated 11-1

Titration of Strong Base with Strong Acid

16.

Calculate the pH when 50.00 mL of 0.100 M HNO3 is titrated with 40.00 mL of 0.0750 M NaOH. Answer: pH = 1.653 Easy

11-2

Titration of Weak Acid with Strong Base

17.

A 50.00 mL aliquot of a 0.1000 M propenoic acid solution, H2CCHCO2H, is titrated with 0.1250 M NaOH. Calculate the pH when 25.00 mL of NaOH is added. Ka = 5.52 × 10−5 Answer: pH = 4.48 Intermediate 25 mL NaOH gives a buffer, with 3.125 mmol A− and 1.875 mmol HA.

85 11-3

Titration of Weak Base with Strong Acid

18.

Calculate the pH for the titration of 50.00 mL of a 0.1500 M methylamine with 35.00 mL of 0.1000 M HCl. Kb = 4.42 × 10−4 Answer: pH = 10.70 Intermediate Buffer with 3.5 mmol BH+ and 4.0 mmol B. pOH subtracted from 14 to give pH. 11-4

Titrations in Diprotic Systems

19.

A 0.100 M oxalic acid, HO2CCO2H, solution is titrated with 0.100 M KOH. Calculate the pH when 25.00 mL of oxalic acid solution is titrated with 35.00 mL of NaOH. Ka1 = 5.4 × 10−2 and Ka2 = 5.42 × 10−5 for oxalic acid.

Answer: pH = 4.090 Hard The diprotic acid has two buffers. The added base exceeds first equivalence point, but is less than needed for second equivalence point. The result is a buffer calculated using the pKa2 with 1.5 mmol HA− and 1 mmol A2−.

11-8

Kjeldahl Nitrogen Analysis

20.

A 1.000 gram organic sample containing nitrogen is subjected to Kjeldahl analysis. The ammonia gas generated when ammonium is neutralized is distilled into 50.00 mL of 0.1000 M HCl to reform ammonium. If 17.29 mL of 0.1000 NaOH is required to titrate the unreacted HCl, calculate the percent by mass nitrogen in the original sample.

Answer: 4.58% N Hard Fifty mL of 0.1 M HCl reacts with the ammonium generated by the Kjeldahl treatment. The mmol base added to neutralize XS HCl is subtracted from the 5 mmol original HCl to give mmol ammonium generated, which of course is the mmol nitrogen in the 1 gram sample. Convert mmol to mols and then to grams N to calculate the percent by mass N. Multiple-Choice 12-1

Metal-Chelate Complexes

1.

Which statement is NOT true?

A) B) C) D) E)

A monodentate ligand binds to a metal atom through only one atom.

A metal is multidentate if it can bind to more than one ligand.

Metal ions are Lewis acids.

Ligands are Lewis bases.

Most transition metals bind six ligand atoms.

86 Answer: B Easy 2.

The is the ability of multidentate ligands to form more stable metal complexes than those formed by similar monodentate ligands.

A) B) C) D) E)

ligand effect multidentate effect chelate effect Lewis effect isomer effect

Answer: C Easy 12-2

EDTA

3.

EDTA is a hexadentate ligand containing four carboxylic acid groups and two amines.

Which is true for the acid-base properties of EDTA? I II III IV

EDTA is a hexaprotic system, H6Y2+, when the amines are protonated.

Neutral EDTA, H4Y, is tetraprotic when the amines are not protonated.

Deprotonated EDTA, Y4−, is the only form of EDTA that binds to metal cations.

The fraction of EDTA in any of the six EDTA species is pH dependent.

A) B) C) D) E)

III and IV I and II I, III, and IV II, III, and IV I, II, and IV

Answer: E Easy 4.

Which statement is incorrect for conditional formation constants, Kf′?

A)

Kf′ is pH dependent.

B) C)

Kf′ Kf′ allows the view of uncomplexed EDTA as being in only one form, Y4−, in equilibrium calculations.

As pH drops the value of Kf′ increases.

The values of Kf′ must be large to guarantee sharp end points during titrations.

D) E)

Answer: D Intermediate

87 12-3

EDTA Titration Curves

5.

For EDTA titrations the titration reaction is Mn+ + EDTA ⇋ MYn-4. Which statements are true for EDTA titration curves?

IV

At the equivalence point, there is exactly as much EDTA in solution as metal. [Mn+] is calculated from the dissociation of MYn-4 formed.

After the equivalence point, the concentration free EDTA equals excess EDTA and virtually all metal ion is in the form MYn-4. [Mn+] is calculated from the dissociation of MYn-4.

Before the equivalence point, free [Mn+] equals excess unreacted Mn+ after EDTA has been consumed. Dissociation of MYn-4 is negligible. pM is plotted on the y-axis and milliliters EDTA solution added on the x-axis.

A) B) C) D) E)

I, II, and III I, III, and IV III and IV I, II, III, and IV II, III, and IV

I II

III

Answer: D Intermediate

88 6.

Calculate pBa when 35.00 mL 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+. For the buffered pH of 10, = 0.30. Kf = 7.59 × 107 for BaY2−.

A) B) C) D) E)

1.39 1.75 4.56 4.82 4.70

Answer: B Intermediate 7.

Calculate pBa when 50.00 mL 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+. For the buffered pH of 10, = 0.30. Kf = 7.59 × 107 for BaY2−.

A) B) C) D) E)

1.30 7.88 4.59 7.36 4.33

Answer: E Intermediate 8.

Calculate pBa when 75.00 mL 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+. For the buffered pH of 10, = 0.30. Kf = 7.59 × 107 for BaY2−.

A) B) C) D) E)

7.06 3.53 4.33 7.58 4.59

Answer: A Intermediate

89 12-5

Auxiliary Complexing Agents

9.

Which statement is NOT true for auxiliary complexing agents?

A) B)

Auxiliary complexing agents prevent the precipitation of metal hydroxide at high pH. Auxiliary complexing agents are chosen to bind metal strong enough to prevent metal hydroxide precipitation but weak enough to be displaced by EDTA. is the effective formation constant for fixed pH and fixed concentration auxiliary complexing agent.

Low concentrations auxiliary complexing agent obliterates the end point of the titration.

K''f = aMa Y4- K f

C) D) E)

Answer: D Easy 10.

Calculate pBa when 35.00 mL of 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+ in the presence of 0.1 M nitrilotriacetate. a Ba 2 + = 1.48 × 10−4. = 0.30 for pH 10. Kf = 7.59 × 107 for BaY2−.

A) B) C) D) E)

4.82 2.64 1.75 4.55 3.53

Answer: B Intermediate

90 11.

Ammonia, 0.10 M, is used as an auxiliary complexing agent for a Co2+ titration with EDTA. There are 6 s for the complex between ammonia and Co2+. Which equation is used to calculate aCo2 + ?

A) B) C) D) E)

aCo =

1

1 + b1 + b2 + b3 + b4 + b5 + b6 1 aCo2 + = 1 + b1[L] + b2 [L] + b3[L] + b4 [L] + b5[L] + b6 [L] [H + ] aCo2 + = 1 + b1[L] + b2 [L] + b3[L] + b4 [L] + b5[L] + b6 [L] 1 aCo2 + = 2 3 1 + b1[L] + b2 [L] + b3[L] + b4 [L]4 + b5 [L]5 + b6 [L]6 1 aCo2 + = + + 2 + 3 1 + b1[H ] + b2 [H ] + b3[H ] + b4 [H + ]4 + b5 [H + ]5 + b6 [H + ]6 2+

Answer: D Easy 12-6

Metal Ion Indicators

12.

Which is NOT an end point detection method used with EDTA titrations?

A) B) C) D) E)

metal ion indicators adsorption indicators glass (pH) electrode mercury electrode ion-selective electrode

Answer: B Easy 13.

A metal that does not freely dissociate from an indicator is said to

A) B) C) D) E)

hinder block restrict impair handicap

Answer: B Easy

the indicator.

91 12-7

EDTA Titration Techniques

14.

A titration is necessary for analytes that precipitates in the absence of EDTA, that reacts slowly with EDTA or that blocks the indicator.

A) B) C) D) E)

direct displacement indirect masking back

Answer: E Easy 15.

A with EDTA. A) B) C) D) E)

is a reagent that protects a component of the analyte from reaction

hindrance agent displacement agent masking agent blocking agent reducing agent

Answer: C Easy Calculated 12-2

EDTA

16.

Calculate for a 0.2 M EDTA solution buffered at pH 10.00. For EDTA K1 = 1.0, K2 = 31.6, K3 = 100, K4 = 490, K5 = 1.35 × 106 and K6 = 2.34 × 1010.

Answer: 0.2990 Hard The value of D is calculated using K1 through K6 and the hydronium concentration. The product of K1 through K6 is divided by the value of D to give the fraction of EDTA in the form Y4−.

92 12-3

EDTA Titration Curves

17.

Calculate [Mn2+] when 50.00 mL of 0.1000 M Mn2+ is titrated with 17.00 mL of 0.2000 M EDTA. The titration is buffered to pH 11 for which a Y4- = 0.81. Kf = 7.76 × 1013 Answer: 0.02388 M Mn2+ Easy Find mmol EDTA and mmol Mn2+. Compare mmols to see which is in excess, [Mn2+]. Divide by total volume to get the concentration Mn2+ in solution.

18.

Calculate [Mn2+] when 50.00 mL of 0.1000 M Mn2+ is titrated with 25.00 mL of 0.2000 M EDTA. The titration is buffered to pH 11 for which a Y4- = 0.81. Kf = 7.76 × 1013 Answer: 3.26 × 10−8 M Mn2+ Intermediate All Mn2+ reacts with all EDTA added to form complex. Divide by total volume to get concentration complex. Calculate the conditional formation constant. Solve for concentration Mn2+ using the equilibrium expression.

12-5

Auxiliary Complexing Agents

19.

Calculate [Ba2+] when 35.00 mL of 0.2000 M EDTA is added to 50.00 mL of 0.1000 M Ba2+ in the presence of 0.1 M nitrilotriacetate. a Ba 2 + = 1.48 x 10−4. a Y4- = 0.30 for pH 10. Kf = 7.59 × 107 for BaY2−.

Answer: 7.424 × 10−4 M Mn2+ Difficulty: Intermediate Solve for mmols of EDTA and barium cation to see which is in excess or if the mmols are equal (equivalence point). There are more mmol EDTA than barium so essentially all barium is bound up in the complex. Since an auxiliary complexing agent is used, the conditional formation constant must also include the alpha barium. Write the K expression, plug in the concentrations complex and EDTA, and solve for barium. The assumption is made that K’’f is large enough that the change in complex concentration is negligible.

12-7

EDTA Titration Techniques

20.

The reaction between Al3+ and EDTA is too slow for direct titration. To a 25.00 mL Al3+ sample 50.00 mL of 0.2000 M EDTA is added and allowed to react for 30 minutes. The excess EDTA was titrated with 21.23 mL of 0.1000 M Zn2+. What is the concentration of Al3+ in the sample?

Answer: 0.3151 M Al3+ Intermediate

93 Calculate mmol EDTA added and mmol EDTA in excess from mL Zn2+ required. The difference in mmol equals mmol Al3+ in the 25 mL sample. Divide mmol Al3+ by 25.00 mL to give the concentration Al3+.

Multiple-Choice 13-1

General Approach to Acid-Base Systems

1.

A student is tasked to determine the pH of a solution prepared by dissolving 20.0 mmol sodium acetate, 10.0 mmol sodium oxalate (NaO2CCO2H), and 15.0 mmol potassium chloride in enough water to prepare 1.00 L of solution. Which of the following will simplify the algebra used to solve the problem?

I.

III IV

Write fractional composition equations for each acid and base in the charge balance equation.

Substitute the fractional composition equations into the charge balance equation and enter the known values of sodium, potassium, and chloride.

Assume the weakest acid will not contribute significantly to the pH of the solution.

Use a spreadsheet to perform the calculations.

A) B) C) D) E)

IV only I, II, and III I, II, III, and IV II, III, and IV I, II, and IV

II

Answer: E Easy 2.

Which of the fractional composition equations is incorrectly written for the diprotic acid H2A and the monoprotic acid HA? A)

[HA] = a HA FHA =

B)

[A2- ] = aA2- FH2A =

C) D) E)

[H + ]FHA [H + ] + K a K1K 2 FH2A

[H + ]2 + [H + ]K1 + K1K 2 [H + ]2 FH2A [H 2 A] = a H2A FH2A = + 2 [H ] + [H + ]K1 + K1K 2 K a FHA [A- ] = a A- FA- = [H + ] + K a [H + ]K1FH2A [HA- ] = a HA- FH2A = + 2 [H ] + [H + ]K1 + K1K 2

Answer: D Intermediate

94

95 3.

The pH of a solution composed of NaCH3CO2, NaO2CCO2H, and KCl must be determined. To get a more complete picture of the equilibrium, the ion-pairing equilibria are included in the calculations. Which of the following is NOT valid for ion pairing?

A)

K+ + CH3CO2− ⇋ KCH3CO2

B)

Na+ + -O2CCO2− ⇋ NaO2CCO2−

C)

Na+ + Cl− ⇋ NaCl

D)

Na+ + CH3CO2− ⇋ NaCH3CO2

E)

CH3CO2− + Cl− ⇋ CH3CO2Cl2−

[KCH3CO2 ] [K + ][CH3CO-2 ] [NaO2CCO-2 ] K NaO CCO- = 2 2 [Na + ][ - O2CCO-2 ] [NaCl] K NaCl = [Na + ][Cl- ] [NaCH3CO2 ] K NaCH3CO2 = [Na + ][CH3CO-2 ] [CH3CO2Cl2- ] K CH CO Cl2- = 3 2 [Cl- ][CH3CO-2 ] K KCH3CO2 =

Answer: E Intermediate 4.

The pH of a solution composed of NaO2CCO2H and NaCl must be determined. To get a more complete picture of the equilibrium, the mass balance equation for total oxalate is written.

Which is the ONLY valid mass balance equation?

A) B) C) D) E)

Facid = [HO2CCO2 H]+[HO2CCO-2 ]+[-O2CCO-2 ]+[NaO2CCO-2 ]+[NaO2CCO2 Na] Facid = [HO2CCO-2 ]+[-O2CCO-2 ]+[NaO2CCO-2 ]+[NaO2CCO2 Na] Facid = [HO2CCO2 H]+[HO2CCO-2 ]+[-O2CCO-2 ] Facid = [NaO2CCO-2 ]+[NaO2CCO2 Na] Facid = [HO2CCO2 H]+[HO2CCO-2 ]+[HO2CCO2 Na]+[-O2CCO-2 ]+[NaO2CCO-2 ]+[NaO2CCO2 Na]

Answer: E Easy

96 13-2

Activity Coefficients

5.

Many ions do not have tabulated activity coefficients nor can the activity coefficient be estimated, as the ion size is now known. The Davies equation is used to estimate the activity coefficient under these conditions. Which statement(s) is(are) NOT true for the Davies equation?

I II III IV

A) B) C) D) E)

The calculated activity coefficient is independent of the ion’s size.

The ion’s charge, the solutions ionic strength and the ion size are required to calculate the activity coefficient.

The Davies equation can be used up to ~ 0.5 M ionic strength.

The calculated activity coefficient is no more accurate than guessing the ion size to determine the activity coefficient.

II and IV II I III and IV III

Answer: B Intermediate 6.

The rearrangement of the equilibrium constant expression to incorporate activity coefficients at the given ionic strength is the: A) B) C) D) E)

conditional equilibrium constant, K’. temporary equilibrium constant, K’. combined equilibrium constant, K’. effective equilibrium constant, K’. activity equilibrium constant, K’.

Answer: D Easy

97 7.

Which is the correct effective equilibrium constant expression for the reaction of NaO2CCO2H with water?

-

A)

B)

C)

D)

E)

O2CCO2 H ⇋ - O2CCO-2 + H+

æg g ö [- O2CCO-2 ][H + ] O2CCO-2 H + ÷ ç K = K1 = ç g÷ [ O2CCO2 H] è O2CCO2H ø æg ö -g + [ - O CCO-2 ][H + ] K '2 = K 2 ç O2CCO2 H ÷ = - 2 ç g÷ [ O2CCO2 H] è O2CCO2H ø æ gö [ - O CCO-2 ][H + ] K '2 = K 2 ç O2CCO2H ÷ = - 2 çg ÷ [ O2CCO2 H] è O2CCO-2g H + ø æ gö [ - O CCO-2 ][H + ] K '2 = K 2 ç O2CCO2H ÷ = - 2 çg ÷ [ O2CCO2 H] è O2CCO-2g H + ø æg ö -g + [ - O CCO-2 ][H + ] K '2 = K1 ç O2CCO2 H ÷ = - 2 ç g÷ [ O2CCO2 H] è O2CCO2H ø ' 1

Answer: D Intermediate 8.

Calculate the pH for a buffer that is 0.10 M in NaO2CCO2Na and 0.50 M in NaO2CCO2H. K2 = 5.42 × 10−5, g - O CCO- = 0.333, g HO CCO- = 0.76 2

A) B) C) D) E)

4.209 4.567 3.965 4.323 3.908

Answer: A Hard

2

2

2

98 9.

Calculate the activity coefficient for phosphate in 0.0075 M KCl.

æ m ö logg = -0.51z 2 çç - 0.3m ÷÷ è 1+ m ø A) B) C) D) E)

0.441 0.140 0.804 0.913 0.116

Answer: A Intermediate 13-3

Dependence of Solubility on pH

10.

In addition to the equilibrium CaCO3 ⇋ Ca2+ + CO32-, additional equilibria can be written to give a more complete picture of calcium carbonate’s solubility. Which of the following is NOT valid?

A)

CO32- + H2O ⇋ HCO-3 + OH-

B)

HCO-3 + H2O ⇋ H2CO3 + OH-

C)

Ca2+ + H2O ⇋ CaOH+ + H+

D)

Ca2+ + CO32- ⇋ CaCO3

E)

Ca2+ + H+ ⇋ CaH3+

Answer: E Intermediate

[HCO-3 ][OH- ] [CO32- ] [H 2CO3 ][OH - ] K b2 = [HCO-3 ] [CaOH + ][H+ ] Ka = [Ca 2+ ] [CaCO3 ] K ip = [Ca 2+ ][CO32- ] [CaH 3+ ] K ip = [Ca 2+ ][H + ]

K b1 =

99 11.

Which is the correct effective solubility constant expression for PbCl2?

PbCl2 ⇋ Pb2+ + 2 Cl−

A) B) C) D) E)

K'sp = g Pb2+g Cl2 - Ksp = [Pb2+ ][Cl- ]2 g K [Pb2+ ][Cl- ]2 K'sp = PbCl2 2 sp = g Pb2+g Cl[PbCl2 ] K sp K'sp = = [Pb 2+ ][Cl- ]2 2

g Pb g Cl K sp K'sp = = [Pb 2+ ][Cl- ]2 g Pb g Cl Ksp K'sp = = [Pb2+ ][2 Cl- ] g Pb 2g Cl 2+

-

2+

-

2+

-

Answer: C Intermediate 12.

The charge balance equation for the solubility of cobalt sulfide is [H+] + 2 [Co2+] = 2 [S2−] + [OH−]

To determine a more accurate solubility for cobalt sulfide, additional equilibria must be written for cobalt(II) and sulfide in solution. Which of the charge balance equations below is the only valid charge balance equation?

A) B) C) D) E)

[H+] + 2 [Co2+] + [CoOH+] + 3 [CoH3+] = 2 [S2−] + [HS−] + [OH−] [H+] + 2 [Co2+] + [CoOH+] = 2 [S2−] + [HS−] + [OH−] + 3 [SOH3−] [H+] + 2 [Co2+] + [CoOH+] = 2 [S2−] + [HS−] + [OH−] [H+] + 2 [Co2+] + 3 [CoH3+] = 2 [S2−] + [HS−] + [OH−] + 3 [SOH3−] [H+] + 2 [Co2+] + [CoOH+] + 3 [CoH3+] = 2 [S2−] + [HS−] + [OH−] + 3 [SOH3−]

Answer: C Intermediate

100 13.

The mass balance equation for the solubility of copper(II) carbonate is [Cu2+] = [ CO32- ]

To determine a more accurate solubility for copper(II) carbonate, additional equilibria must be written for copper(II) and carbonate in solution. Which of the mass balance equations below is the only valid charge balance equation?

A)

[Cu2+] + [CuOH+] + [CuCO3] = [ CO32- ] + [ HCO-3 ]

B)

[Cu2+] + [CuCO3] = [ CO32- ] + [ HCO-3 ] + [ H 2CO3 ]

C)

[Cu2+] + [CuOH+] = [ CO32- ] + [ HCO-3 ] + [ H 2CO3 ] + [CuCO3]

D)

[Cu2+] + [CuOH+] = [ CO32- ] + [ HCO-3 ]+ [ H 2CO3 ]

E)

[Cu2+] = [ CO32- ] + [ HCO-3 ]

Answer: D Intermediate 13-4

Analyzing Acid-Base Titrations with Difference Plots

The mean fraction of protons, n H , bound to H3A ranges from . n H (theoretical) =

14.

A)

0 to 3; 3aH2A + 2aH A- + aHA- + aA2-

B)

1 to 3; 3a H2A + 2a H A- + a HA-

C)

0 to 3; aH2A + aH A- + aHA- + aA2-

2

2

2

D)

1 to 2; 2a H A- + a HA2

E)

0 to 2; a H A- + a HA- + a A22

Answer: B Intermediate

and

101 15.

The equilibrium constants for a diprotic acid are determined from a Bjerrum plot by:

A)

fitting the theoretical curve to the experimental curve by adjusting the pH values to minimize the square of the residuals. determining the slope for the upper curve which corresponds to K1, and the slope for the lower curve which corresponds to K2. varying the pH to determine the range over which the slope of the theoretical curve is zero. Lower end of range is K1 and upper end of range is K2. fitting the theoretical curve to the experimental curve by the method of least squares to find the values of K1 and K2 that minimize the square of the residuals. taking the first derivative of the experimental data to determine minimums in slope. Each minimum is an equilibrium constant.

B) C) D) E)

Answer: D Hard Calculated 13-1

General Approach to Acid-Base Systems

16.

Calculate the concentration of each fumaric acid, HO2CCHCHCO2H, species in a 0.175 M solution at pH = 5.000. Ka1 = 9.5 × 10−4 and Ka2 = 3.3 × 10−5.

Answer: [H2A] = 0.000431 M, [HA−] = 0.0410 M and [A2−] = 0.1337 M. Intermediate Calculate three fractions from the pH and Ka data. Multiply total concentration by fraction to give concentration of each species.

17.

A solution is 0.120 M sodium oxalate (NaO2CCO2H), 0.05 M chloroacetic acid (ClCH2CO2H) and 0.01 M KOH. Express the charge balance equation for the solution in terms of [H+].

0.240K1[H+ ] + 0.120K1K 2 K K + + 1 + w+ + 2 + [H ] + [H ] + K1K 2 [H ] + K1 [H ] K or [H+ ] + 0.130 = 0.240a - O CCO H + 0.120a - O CCO- + aClCH CO- + w+ 2 2 2 2 2 2 [H ] Intermediate Solve the problem in terms of Ks and [H+] or incorporate the Ks and [H+] into s. [Na+] and [K+] have their molarities inserted.

Answer: [H + ] + 0.130 =

102 13-2

Activity Coefficients

18.

A solution is 0.200 M in acetic acid and 0.100 M KCl. Calculate the concentration acetic acid and acetate in solution for a pH 4.500. g CH CO- = 0.775, g CH CO H = 1, g H+ = 0.83 and Ka = 1.75 −5

3

2

3

2

× 10 .

Answer: [CH2CO2H] = 0.1074 M, [CHCO2−] = 0.0924 M Intermediate Calculate the fraction HA and fraction A− and multiply by 0.200 M to get fraction of acetate and acetic acid. The activity coefficients are incorporated into the Ks to give effective equilibrium constants.

19.

Calculate the pH of a sulfurous acid buffer prepared from 0.010 M NaHSO3 and 0.030 M Na2SO3. Ka1 = 1.39 × 10−2 and Ka2 = 6.73 × 10−8 for sulfurous acid.

Ionic strength (, M) Ion 0.001 0.005 0.01 0.05 0.1 Charge =  1 Activity coefficient () + 0.964 0.928 0.902 0.82 0.775 Na , HSO3 Charge =  2 Activity coefficient () 20.867 0.742 0.665 0.455 0.37 SO3 Answer: 7.33 (7.328) with activities or 7.65 (7.649) without activities Hard Determine the ionic strength from the concentrations of the acid and its salt (both ionic) and then the activity coefficient for the acid and conjugate base, plug into Henderson-Hasselbalch equation with concentrations to solve for pH. Assumption made: Ks do not change with changes in ionic strength.

103 13-3

Dependence of Solubility on pH

20.

Express [Ca2+] in terms [H+] for the solubility of CaSO4 using the equations below.

Assume all activity coefficients are unity.

CaSO4 ⇋ Ca2+ + SO24 Ca2+ + H2O ⇋ CaOH+ + H+ Ca2+ + SO2⇋ CaSO4 4 + HSO 4 ⇋ H + SO24

Ksp = 2.40 × 10−5 Ka = 2.00 × 10−13 Kip = 229 Ka2 = 1.03 × 10−2

æ [H + ] ö K sp ç1+ ÷ è K a2 ø 2+ Answer: [Ca ] = æ Ka ö ç1 + ÷ [H + ] ø è Hard (and time consuming) Use the mass balance equation to solve for the various species in terms of Ca2+. The concentration ion pairing cancels out in the mass balance equation leaving [Ca2+] + [CaOH+] = [HSO4−] + [SO42−].

Multiple-Choice 14-1

Basic Concepts

1.

For the reaction below, the oxidizing agent is electrons, and the reducing agent is and it 8 H+ + MnO-4 + 5 Cu+ ⇋ Mn2+ + 5 Cu2+ + 4 H2O A) B) C) D) E)

MnO-4 ; gains 5; Cu+; loses 1 H+; loses 1; Cu+; gains 1 H+; gains 1; MnO-4 ; loses 5 Cu+; loses 1; MnO-4 ; gains 1 Cu+; loses 1; H+; gains 1

Answer: A Easy 2.

Which relationship is NOT correctly defined?

A) B)

Work = E  q G = -n · F · E E I= R P=EI

C) D)

and it electrons.

104 E)

q = n N F

Answer: B Intermediate 14-2

Galvanic Cells

3.

Which of the following is NOT true for galvanic cells?

A) B) C) D) E)

Galvanic cells are spontaneous.

Oxidation occurs at the anode and reduction occurs at the cathode.

Electrons move toward more negative electrical potential.

Galvanic cells are composed of two half-cells connected by salt bridge.

The salt bridge maintains electroneutrality throughout the cell.

Answer: C Intermediate

105 4.

For the line notation below, which is NOT true?

Fe(s)| Fe2+ (aq, A = 1) || Cu2+ (aq, A = 1) | Cu (s)

A) B) C) D) E)

Fe is the anode. || represents the salt bridge.

Cu is the anode.

1 M Cu2+ is the electrolyte concentration and species for the copper electrode.

1 M Fe2+ is the electrolyte concentration and species for the iron electrode.

Answer: C Intermediate 14-3

Standard Potentials

5.

All of the following are true for the measurement of standard reduction potentials, except:

I

IV

standard conditions for reduction potentials is 25 oC and A = 1 for all aqueous and gaseous species. reduction potentials are measured relative to the standard hydrogen electrode (S.H.E.). standard reduction potentials are measured while connected to the negative electrode of the potentiometer. the E° for the S.H.E. is defined as 0 volts.

A) B) C) D) E)

I and IV I and III I only III only II and IV

II III

Answer: D Intermediate

106 6.

Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer?

Ag+ + 1 e- → Ag (s) E° = 0.7993 V V2+ + 2 e- → V (s) E° = −1.125 V I II III IV V

Ag+ is reduced.

V is oxidized.

Eocell = 1.924 V V2+ is reduced.

Ag is oxidized.

A) B) C) D) E)

I and II III, IV, and V I, II, and III III only IV and V

Answer: E Intermediate 14-4

Nernst Equation

7.

For the silver half-reaction, Ag+ + 1 e- → Ag (s), when the concentration of silver cation is increased, the reduction potential: A) B) C) D) E)

becomes more positive. becomes more negative. remains constant. increases or decreases depending on the voltage of the other half-reaction. increases or decreases depending on the temperature.

Answer: A Hard

107 8.

The half-cells below are connected via a salt-bridge and potentiometer. What is the voltage read by the potentiometer when AAg+ = 2.00 and AV2+ = 1.50?

Ag+ + 1 e− → Ag (s) Eo = 0.7993 V V2+ + 2 e− → V (s) Eo = −1.125 V A) B) C) D) E)

1.912 V −0.326 V −0.313 V 1.937 V 1.924 V

Answer: D Intermediate 14-5

Eo and the Equilibrium Constant

9.

A galvanic cell with a large equilibrium constant has a

A) B) C) D) E)

negative constant zero voltage positive variable

cell potential.

Answer: D Easy 10.

Calculate the equilibrium constant for the reaction between Sn metal and Zn2+ solution.

Sn + Zn2+ ⇋ Sn2+ + Zn

Sn2+ + 2 e− → Sn (s) E° = −0.141 V Zn2+ + 2 e− → Zn (s) E° = −0.762 V A) B) C) D) E)

9.86 × 1020 1.21 × 10−21 3.37 × 1030 3.18 × 10−11 1.84 × 1015

Answer: B Intermediate

108 11.

A galvanic cell is at equilibrium when:

A) B) C) D) E)

Ecell is a constant value. cell resistance is very low.

Ecell = 0 V. Ecell fluctuates between  E°cell. cell resistance is infinitely large.

Answer: C Easy 14-6

Cells as Chemical Probes

12.

Which is true when electrochemical cells are used as chemical probes?

I. II III IV V

All concentrations in the half-cells must be known except for the analyte of interest.

The analyte concentration is calculated from the equilibrium cell potential.

Equilibrium between the half-cells is established.

Equilibrium within each half-cell is established and is assumed to remain at equilibrium.

Half-cells other than the S.H.E. can be used to determine pH.

A) B) C) D) E)

II, III, and IV I, IV, and V I and II III, IV, and V I, II, and V

Answer: B Intermediate 13.

A cell is constructed according to the cell diagram below. If the measured potential is 1.00 V, what is the pH of the dichromate cell?

Fe|Fe2+ (aq, 1M)|| Cr2O2(aq 1 M), Cr2+ (aq, 1 M), H+ (aq, x M)|Pt 7 + 14 H+ + 6 e− ⇋ 2 Cr3+ + 7 H2O Cr2O27 Fe2+ + 2 e− ⇋ Fe (s) A) B) C) D) E)

1.931 5.797 12.162 6.081 4.054

Answer: B Hard

E° = 1.36 V E° = −0.44 V

109 14-7

Biochemists Use Eo

14.

Eo is defined as:

A) B) C) D) E)

the formal half-cell potential for compounds that are neither acids nor bases. the formal half-cell potential at 20 C. the formal half-cell potential at pH 7. the inverse of E°.

None of these answers is the definition of E°.

Answer: C Intermediate 15.

Calculate E° for the half-cell below.

O2 + 4 H+ + 4 e− ⇋ 2 H2O

A) B) C) D) E)

E° = 1.229 V

1.643 V 0.815 V 0.307 V 1.229 V 0.711 V

Answer: B Intermediate Calculated 14-2

Galvanic Cells

16.

For the cell diagram below write out and balance the reaction that occurs in the cell.

Pt|HOCl (aq, 1 M), HClO2 (aq, 1 M), H+ (aq, 1 M)|| MnO-4 (aq, 1 M), H+ (aq, 1 M) | MnO2 (s)|Pt Answer: 2 MnO-4 + 2 H+ + 3 HOCl ⇋ 2 MnO2 + 3 HClO2 + H2O Hard

110 14-3

Standard Potentials

17.

Calculate the standard potential when the two half-cells below are connected via a salt bridge and potentiometer. Which species are the reducing agent and oxidizing agent?

IO-4 + 2 H+ + 2 e- ⇋ IO-3 + H2O Mg(OH)2 (s) + 2 e- ⇋ Mg (s) + 2 OH-

E° = 1.589 V E° = -2.690 V

Answer: 4.279 V, IO-4 is the oxidizing agent and Mg (s) is the reducing agent.

Easy 14-4

Nernst Equation

18.

Calculate the potential for the half-cell below for [ SO32- ] = 0.100 M, [ S2O32- ] = 0.500 M and the pH = 10.50.

2 SO32- + 3 H2O + 4 e− ⇋ S2O32- + 6 OH-

E° = −0.566 V

Answer: -0.280 V Intermediate Use the Nernst equation, plugging the concentration values and the moles electrons. Calculate the OH− from the pH. 14-5

E° and the Equilibrium Constant

19.

Calculate Eo and K for the cell composed of a bromate/bromide half-cell and a chlorate/chlorine dioxide half-cell.

ClO-3 + 2 H+ + e− ⇋ ClO2 + H2O BrO-3 + 3 H2O + 6 e− ⇋ Br− + 6 OH−

E° = 1.130 V E° = 0.613 V

Answer: E° = 0.517 V and K = 2.72 x 1052 Intermediate Flip the sign on the bromate/bromide half-cell and add the two half-cell potentials together. Plug o E° and n = 6 into the K = 10nE 0.05916 .

111 14-6

Cells as Chemical Probes

20.

A bromide probe is constructed from a saturated silver bromide half-cell and a Ni2+/Ni half-cell. If the measured potential is 0.952 V and [Ni2+] = 1 M, what is the bromide concentration? − AgBr (s) + e− ⇋ Ag(s) + Br E° = 0.071 V Ni2+ + 2 e− ⇋ Ni (s) E° = −0.714 V Answer: [Br−] = 1.52 × 10−3 M Intermediate Determine E°. Plug E° and [Ni2+] into the Nernst equation to solve for [Br−].

Multiple-Choice 15-1

Reference Electrodes

1.

An electrode with a fixed potential is the:

A) B) C) D) E)

cathode. anode. indicator electrode. reference electrode. working electrode.

Answer: D Easy 2.

The main advantage to using a saturated KCl solution in reference electrodes is:

A) B) C) D) E)

the low cost of potassium chloride. the ease of reaction with analyte. the chloride concentration does not change as liquid evaporates from the cell. the ease of cell construction. the ease at which chloride passes through membranes.

Answer: C Easy 3.

An electrode has a potential of 1.201 V with respect to a saturated silver-silver chloride electrode. What would the electrodes potential be relative to a saturated calomel electrode (S.C.E.)? E = 0.197 V for the saturated silver-silver chloride electrode and E = 0.241 V for the S.C.E. A) B) C) D)

1.442 1.245 V 1.004 V 1.398 V

112 E)

1.157 V

Answer: E Hard

113 15-2

Indicator Electrodes

4.

For indicator electrodes, which statements are true?

I II III IV

Indicator electrodes are made from relatively inert metals.

Indicator electrodes work best when the electrode surface is large and clean.

Platinum is a common indicator electrode.

Platinum electrodes are more inert than gold electrodes.

A) B) C) D) E)

I, III, and IV II and III I, II, and IV I, II, and III I, II, III, and IV

Answer: D Intermediate 5.

The potential of a silver electrode is measured relative to an Ag-AgCl electrode for the titration of 100.0 mL of 0.100 M Cl with 0.100 M Ag+. What is the potential after 75.00 mL of titrant is added. Eo = 0.799 V for Ag+, E = 0.197 V for the Ag-AgCl electrode and Ksp = 1.8 × 10−10.

A) B) C) D) E)

0.493 V 1.070 V 0.521 V 1.267 V 0.134 V

Answer: E Intermediate 15-3

What Is a Junction Potential?

6.

The are in contact.

A) B) C) D) E)

ion potential electrolyte potential cation potential junction potential anion potential

Answer: D Easy

is the voltage difference when two dissimilar electrolyte solutions

114 7.

A 0.100 M KCl solution is placed in contact with a 0.100 M LiCl solution. Which of the following is true for junction between the two solutions?

I II III IV

A) B) C) D) E)

The KCl side of the junction will be positive as Cl− diffuses faster across the junction than K+.

The LiCl side of the junction will be negative as Li+ diffuses faster across the junction than K+.

The KCl side of the junction will be positive as K+ diffuses slower across the junction than Li+ The LiCl side of the junction will be more negative as Li+ diffuses faster across the junction than Cl−.

I and III II and IV I and II III and IV II and III

Answer: E Intermediate

115 15-4

How Ion-Selective Electrodes Work

8.

All of the following are true for ion-selective electrodes specific to aqueous ions, except:

I

two reference electrodes measure the voltage difference across the ion-selective electrodes organic polymer membrane. the organic polymer membrane separates the inner solution from the outer analyte solution. the organic polymer membrane is hydrophobic and is impregnated with a viscous organic solution containing an ion exchanger and sometimes a ligand specific for the analyte ion. the potential difference across the organic polymer membrane occurs when the analyte ions embedded in the membrane diffuse into the inner aqueous solution creating a junction potential. the potential difference is dependent only on the activity of analyte ion in the outer solution.

II III IV

V

A) B) C) D) E)

IV I and V I and IV IV and V V

Answer: A Intermediate

116 9.

A sodium selective electrode increased in voltage by 59.16 mV when the sodium cation concentration was increased by a factor of 10. A S2− selective electrode measured a decrease of 29.58 mL when the sulfide concentration increased by a factor of 10. Why? I II III IV

A) B) C) D) E)

Cations always increase the measured potential by n × 59.16 mV where n is the charge on the cation.

Anions always decrease the measured potential by n × 59.16 mV where n is the charge on the anion.

Cations always increase the measured potential by 59.16 mV n where n is the charge on the cation.

Anions always decrease the measured potential by 59.16 mV n where n is the charge on the anion.

II and III III and IV I and II I and IV None of the statements explain the behavior.

Answer: B Hard 15-5

pH Measurement with a Glass Electrode

10.

Which is NOT an error associated with pH measurement that is incorrectly defined?

A)

junction potential drift – AgCl or Ag precipitates from solution due to dilution of the KCl or the presence of reducing agents sodium error – electrode responds to Na+ instead of H+ when [Na+] is low and [H+] is high acid error – the measured pH is higher than the actual pH because the glass is saturated with H+ and cannot be further protonated temperature – pH meter calibrated at a different temperature than the sample equilibration time – the electrode did not have sufficient time to equilibrate with the solution

B) C) D) E)

Answer: B Intermediate

117 15-6

Ion-Selective Electrodes

11.

Which is NOT a class of ion-selective electrode?

A) B) C) D) E)

glass membranes solid-state electrodes liquid-based electrodes compound electrodes biochemical electrodes

Answer: E Easy 12.

The gives the relative response of an ion-selective electrode to different species with the same charge.

A) B) C) D) E)

counter ion coefficient isoionic coefficient ion coefficient selectivity coefficient transient coefficient

Answer: D Easy 13.

Solid-state electrodes that respond to anions create a potential difference across the solidstate membrane by , which allows anions to migrate through the crystal.

A) B) C) D) E)

doping the membrane crystal to create anion vacancies drilling small channels through the crystal reacting the membrane crystal with a strong acid polishing the surface of the membrane crystal etching the surface of the membrane crystal with small grooves

Answer: A Intermediate

118 15-7

Using Ion-Selective Electrodes

14.

To compensate for a complex or unknown matrix, the be used to determine the analyte concentration.

A) B) C) D) E)

method

can

calibration curve standard addition standardization least-squares analysis dilution method

Answer: B Easy 15-8

Solid-State Chemical Sensors

15.

Solid-state chemical sensors are manufactured using semiconductors such as Si and Ge. Doping enhances the conductivity of Si. Which is NOT true for doping of silicon?

A) B) C) D) E)

Doping replaces some of the silicon atoms with atoms of different elements.

Doping silicon with phosphorous enhances silicon’s conductivity by providing an extra conduction electron that is free to move through the crystal.

Doping silicon with aluminum enhances silicon’s conductivity by creating a vacancy in the crystal called a hole, which behaves as a negative charge carrier. n-type silicon has excess conduction electrons. p-type silicon has excess holes.

Answer: C Intermediate Calculated 15-1

Reference Electrodes

16.

If the measured potential of a half-cell is 0.584 V when measured against the S.C.E, what is the potential if the measurement is made using the saturated Ag-AgCl reference electrode? E = 0.197 V for the saturated Ag-AgCl electrode and E = 0.241 V for the S.C.E. Answer: EAg-AgCl = 0.628 V Intermediate The potential is positive, so Eo for the half-cell of interest is more positive than the E for the S.C.E. 0.584 = x  0.241, x = 0.825 V. Next calculate the potential relative to Ag-AgCl by x = 0.825  0.197 = 0.625 V

119 15-2

Indicator Electrodes

17.

Calculate E for the titration of 50.00 mL of 0.100 M Ag+ when 50.00 mL of 0.150 M Cl− is added. Silver wire is the indicator electrode and the saturated Ag-AgCl electrode is the o reference electrode. E = 0.197 V for the saturated Ag-AgCl electrode, E Ag+ = 0.7993 V and Ksp = 1.8 × 10−8 for AgCl. Answer: 0.239 V Hard Calculate mmol Ag+ and Cl− after addition and determine which species is in excess (Cl−).

Calculate concentration Cl− using mmol excess Cl− and total volume. Use Ksp to determine [Ag+] and plug into the Nernst equation (E = Eo −(0.0592/n) ∙ log(Q) −Eref) and solve for E. 18.

Calculate E for the titration of 50.00 mL of 0.100 M Ag+ when 25.00 mL of 0.150 M Cl− is added. Silver wire is the indicator electrode and the saturated Ag-AgCl electrode is the o reference electrode. E = 0.197 V for the saturated Ag-AgCl electrode, E Ag+ = 0.7993 V and Ksp = 1.8 × 10−8 for AgCl. Answer: 0.497 V Hard Calculate mmol Ag+ and Cl− after addition and determine which species is in excess (Ag+).

Calculate concentration Ag+ using mmol excess Ag+ and total volume. Plug into the Nernst equation (E = Eo −(0.0592/n) ∙ log(Q) −Eref) and solve for E. 15-5

pH Measurement with a Glass Electrode

19.

A pH probe is calibrated against pH 4 and pH 7 buffers. For pH 4 buffer, E = 200 mV and for pH 7 buffer, E = 23 mV. If the potential for an unknown solution is 157 mV, calculate the pH of the solution.

Answer: pH = 4.73 Hard Calculate E/pH for the buffers and then solve for pHunknown when the potential Eunknown is measured using the relationship E/pH = (Eunknown – ES1)/(pHunknown – pHS1).

15-6

Ion-Selective Electrodes

20.

The selectivity coefficient for a fluoride ion-selective electrode is K F- , OH- = 0.15. What

Pot

will the change in the electrode potential be for a 1.4 × 10−5 M F− at pH 5.5 if the pH is raised to pH 11.0?

Answer: The potential decreases by 43.23 mV or E = −43.23 mV or −0.04323 V. Hard

120 Calculate the potential at pH 5.5 where [OH−] is negligible and the selectivity coefficient can be ignored. Then calculate the potential at pH 11.0 where [OH−] is not negligible and use the selectivity coefficient to correct for OH−. The potential difference is final – initial.

Multiple-Choice 16-1

The Shape of a Redox Titration Curve

1.

Which of the statements below are true for the titration of Fe2+ with standard Ce4+?

I The indicator electrode potential before the addition of Ce4+ solution is calculated using 3+ the Fe half-reaction after calculating the equilibrium concentrations of Fe3+ and Fe2+ in solution.

II Before the equivalence point, the indicator electrode potential is calculated using the Fe3+ half-reaction because only [Fe3+] and [Fe2+] are known.

III At the equivalence point, the indicator potential is the average of Eo for Fe3+ and Ce4+.

IV After the equivalence point, the indicator electrode potential is calculated using the Ce4+ half-reaction because only [Ce4+] and [Ce2+] are known.

A) B) C) D) E)

II and IV II, III, and IV I, III, and IV I, II, and IV I, II, III, and IV

Answer: C Intermediate 2.

Calculate the potential for a redox titration when 50.00 mL of 0.100 M Co 3+ is titrated with 35.00 mL of 0.130 M MnO2. The potential is measured against the Ag-AgCl reference 4 electrode, E = 0.197 V. Co3+ + e- ⇋ Co2+ MnO-4 + e- ⇋ MnO24 A) B) C) D) E)

0.42 V 0.62 V 1.66 V 1.86 V 1.24 V

Answer: C Hard

Eo = 1.92 V Eo = 0.56 V

121 16-2

Finding the End Point

3.

Which statements are true for redox indicators?

I.

The color change occurs when the indicator goes from its oxidized form to its reduced form.

The color for the oxidized state of the indicator occurs when the ratio of the reduced state to the oxidized state is 1:10.

The range over which the indicator functions is Eo ± 0.05916 n .

The color for the reduced state of the indicator occurs when the ratio of the reduced state to the oxidized state is is 1:10.

II III IV

A) B) C) D) E)

I, II, III, and IV III and IV I, III, and IV I, II, and IV I, II, and III

Answer: E Intermediate 4.

In addition to redox indicators and starch-iodine complex, the equivalence point for a redox titration can be determined using a Gran plot. Which is NOT true for Gran plots?

A) B) C) D) E)

Gran plots give best results using redox data well before the equivalence point.

Gran plots are constructed by plotting the volume titrant added on the x-axis and V10−nE/0.05916 on the y-axis.

The plot of V versus V10−nE/0.05916 is a straight line with the x-intercept = equivalence point volume.

Constant ionic strength over the course of the titration minimizes error in the equivalence point.

Gran plots are more error prone than first derivative and second derivative plots.

Answer: E Intermediate

122 16-3

Adjustment of Analyte Oxidation State

5.

For redox titrations, the oxidation state of the analyte may need adjusting. The analyte can be preoxidized with an oxidizing agent to quantitatively oxidize the analyte, with the excess oxidizing agent eliminated to prevent interference with titration. Which of the oxidizing agents is paired with INCORRECT elimination chemistry?

A) B) C) D) E)

Excess peroxydisulfate is eliminated by boiling, converting the peroxydisulfate to sulfate.

Excess silver(I, III) oxide in mineral acid is eliminated by boiling, converting silver(III) cation to silver(I) cation.

Excess solid sodium bismuthate is removed by filtration.

Excess hydrogen peroxide in basic solution is eliminated by boiling, converting hydrogen peroxide to water and oxygen gas.

Excess stannous chloride in hot HCl is eliminated by boiling, converting stannous cation to stannic cation.

Answer: E Hard 6.

For redox titrations, the oxidation state of the analyte may need adjusting. The analyte can be prereduced with a reducing agent to quantitatively reduce the analyte, with the excess reducing agent eliminated to prevent interference with titration. Which of the reducing agents is paired with INCORRECT elimination chemistry?

A) B) C) D) E)

Stannous chloride in hot HCl is eliminated by the addition of excess HgCl 2, converting stannous cation to stannic cation.

Chromous chloride is eliminated by boiling, converting chromium(II) cation to chromium(III).

Sulfur dioxide is expelled by boiling in acid.

Hydrogen sulfide is expelled by boiling in base.

Excess hydrogen peroxide in acidic solution is eliminated by boiling, converting hydrogen peroxide to water and oxygen gas.

Answer: D Hard

123 16-4

Oxidation with Potassium Permanganate

7.

Potassium permanganate is a strong oxidant commonly used in redox titrations. Which is(are) TRUE for permanganates titration chemistry?

I II III

A) B) C) D) E)

In strongly acidic solution the permanganate half-reaction is MnO-4 + 8 H+ + 5 e- ⇋ Mn2+ + 4 H2O. In neutral or alkaline solution the permanganate half-reaction is MnO-4 + 4 H+ + 3 e- ⇋ MnO2 (s) + 2 H2O. In strongly alkaline solution the permanganate half-reaction is .

MnO-4 + e- ⇋ MnO24 III only II and III I and III I and II I, II, and III

Answer: E Hard 8.

Potassium permanganate is NOT a primary standard. Why?

I

Solid KMnO4 contains traces of MnO2. [ MnO2] < [ MnO2] 4 actual 4 calculated Permanganate reacts with organic material in distilled water. [ MnO2] > [ MnO2] 4 actual 4 calculated Permanganate is unstable in aqueous solution, producing MnO2, O2, and OH−. [ MnO2] < [ MnO2] 4 actual 4 calculated 2+ The presence of MnO2, Mn , heat, light, acids, and bases increase the rate of reaction between permanganate and water. [ MnO2] < [ MnO2] 4 actual 4 calculated

II III IV

A) B) C) D) E)

I and II I, II, and IV II, III, and IV I, III, and IV I, II, and III

Answer: D Hard

124 16-5

Oxidation with Ce4+

9.

Cerium(IV) is a strong oxidizing agent commonly used in redox titrations. Which of the following is NOT true for cerium(IV)?

A) B) C) D) E)

Ce4+ is yellow and Ce3+ is colorless. The color change is abrupt enough that Ce4+ can serve as its own indicator.

Reduction of Ce4+ to Ce3+ proceeds cleanly in acidic solutions.

Ce4+ does not form aqueous complexes when dissolved in water.

Ce4+ is indefinitely stable in sulfuric acid solution.

Ce4+ reacts rapidly with chloride in hot hydrochloric acid solutions to form chlorine gas.

Answer: A Hard 10.

A Ce4+ standard solution is prepared with primary-standard-grade ammonium hexanitratocerate(IV), (NH4)2Ce(NO3)6. Which of the follow will NOT decrease the concentration of Ce4+ in solution over time?

A) B) C) D) E)

Ce4+ solution prepared in distilled water.

Ce4+ prepared in hot HCl solution.

Ce4+ prepared in H2SO4 solution.

Ce4+ prepared in HClO4 solution.

Ce4+ prepared in HNO3 solution.

Answer: C Hard 16-6

Oxidation with Potassium Dichromate

11.

Which is NOT true for oxidation with potassium dichromate?

A) B) C) D) E)

In acid solution, dichromate ion is orange.

In acid solution, dichromate ion is reduced to chromous ion.

Dichromate is a less powerful oxidizing agent than permanganate and cerium(IV).

In basic solution, dichromate is converted into yellow chromate ion.

Potassium dichromate is a primary standard.

Answer: B Hard

125 12.

Which are TRUE for the use of indicators with dichromate?

I

Dichromate is orange and changes color to green when reduced. The color change is abrupt enough to serve as an indicator.

The equivalence point for dichromate can be determined using Pt and calomel electrodes.

Dichromate end points are determined using the distinct color changes of diphenylamine sulfonic acid or diphenylbenzidine sulfonic acid indicators.

In basic solution, the orange dichromate solution changes color to yellow when reduced to chromate anion. The color change is abrupt enough to serve as an indicator.

II III IV

A) B) C) D) E)

II and III I and IV I, II, and IV I and III II and IV

Answer: D Hard 16-7

Methods Involving Iodine

13.

A) B) C) D) E)

is the titration of iodine produced by a chemical reaction.

Iodigravimetric titration Iodimetry Iodogravimetric titration Iodometry Iodotration

Answer: D Intermediate 14.

Starch is an indicator for iodine. For iodimetry, starch solution is added:

A) B) C) D) E)

at the beginning of the titration. at the half-way point of the titration. right before the equivalence point of the titration. after the equivalence point of the titration has been reached. to the titrant before starting the titration.

Answer: A Intermediate

126 15.

Sodium thiosulfate is used almost exclusively to titrate I-3 . Which of the following is NOT true for sodium thiosulfate titrant?

A) B)

In neutral or acidic solution, triiodide oxidizes thiosulfate to tetrathionate.

One mole of I-3 is equivalent to two moles of I2.

C)

In basic solution, I-3 disproportionates to I− and HOI, which can oxidize S2O32- to SO2.

4

Sodium thiosulfate is standardized with freshly made I-3 solution prepared from KIO3 and KI. E) Dissolved CO2 and metal ion catalyzed atmospheric oxidation of thiosulfate decreases the concentration of prepared thiosulfate solution.

D)

Answer: B Intermediate Calculated 16-1

The Shape of a Redox Titration Curve

16.

Calculate the potential for a redox titration when 50.00 mL of 0.100 M Co 3+ is titrated with 50.00 mL of 0.110 M MnO2. The potential is measured against the Ag-AgCl reference 4 electrode, E = 0.197 V. Co3+ + e- ⇋ Co2+ MnO-4 + e- ⇋ MnO24

Eo = 1.92 V Eo = 0.56 V

Answer: 0.422 V Hard Using the volume and molarity data, you will calculate that titrant is in excess, so you will calculate the potential using the Nernst equation with 0.5 mmol excess MnO42− and 5 mmol MnO4−, E = 0.619 V. Subtract the reference electrode potential to get the final answer.

127 16-4

Oxidation with Potassium Permanganate

17.

A permanganate solution is prepared by dissolving 20.0123 g KMnO4 in 500 mL of distilled water and boiled for 1 hour to remove any organic material. Following sintered-glass filtration, the solution is quantitatively transferred to a 1.0 L volumetric flask and diluted to volume with distilled water. The permanganate solution was titrated against 0.1023 M oxalic acid prepared in sulfuric acid solution. A 50.00 mL aliquot of oxalic acid solution required 16.68 mL of permanganate solution. A titration blank required 0.04 mL of permanganate. What is the permanganate molarity?

5 H2C2O4 + 2 MnO-4 + 6 H+ ⇋ 10 CO2 + 2 Mn2+ + 8 H2O Answer: 0.1230 M Hard You may be tempted to calculate the molarity and ignore the titration value. The answer if you chose this path is 0.1266 M. The correct answer is calculated by multiplying molarity oxalic acid by volume taken (50 mL) to get mmol oxalic acid, use the 2 oxalic acid:5 permanganate mole-tomole ratio to determine the mmol permanganate reacted. Subtract the titration blank from the volume observed to get true volume permanganate. Mmol permanganate/mL solution used = 0.1230 M permanganate.

16-5

Oxidation with Ce4+

18.

The percent purity of a malonic acid reagent must be calculated. A student weighs out 0.1511 grams of malonic acid, CH2(CO2H)2, and quantitatively transfers the solid to a 250 mL Erlenmeyer flask. To the flask 50.00 mL of distilled water is added to dissolve the malonic acid.

After dissolution, the student uses a 50 mL volumetric pipet to add 50 mL of 0.200 M Ce4+, which is allowed to react with the malonic acid. After complete reaction, the excess Ce4+ is back titrated with 14.58 mL of 0.100 M Fe2+ solution. Calculate the percent purity malonic acid.

CH2(CO2H)2 + 2 H2O + 6 Ce4+ → 2 CO2 + HCO2H + 6 Ce3+ + 6 H+ Ce4+ + Fe2+ ⇋ Ce3+ + Fe3+ Answer: 98.0% pure malonic acid Hard Calculate mmol cerium(IV) added and mmol cerium(IV) in excess. The difference is the mmol cerium(IV) consumed. Using the mole-to-mole ratio, calculate the mmol malonic acid reacted, convert to moles and then to grams with the molar mass. Next, the mass reacted divided by the mass taken × 100 gives the percent purity malonic acid.

128 16-6

Oxidation with Potassium Dichromate

19.

A student is given the assignment to determine the mass iron in a brand of women’s vitamins. He begins by taking four vitamin tablets, grinding the tablets into a fine powder to undergo the necessary sample preparation to free all iron from the sample matrix and to reduce all iron to the +2 state. The resulting Fe2+ solution is transferred to a 250 mL volumetric flask and diluted to volume with distilled water. A 50 mL aliquot of the sample requires 42.98 mL of 1.00 x 10−3 M Cr2O2. Calculate the mg Fe per tablet.

7 Cr2O2+ 6 Fe2+ + 14 H+ ⇋ 2 Cr 3+ + 6 Fe3+ + 7 H2O 7 Answer: 18.0 mg Fe/tablet Hard Begin by taking the volume dichromate used and concentration to calculate the mmol dichromate reacted, using the mole-to-mole ratio to obtain the mmol iron reacted. Next, convert mmol Fe to mg Fe by multiplying by the molar mass of Fe. Mg Fe/50 mL gives the mg Fe/mL for the original sample volume (250 mL). Multiply by 250 mL to find total mg Fe and divide by four tablets to get 18.0 mg Fe/tablet.

16-7

Methods Involving Iodine

20.

A 50.00 mL aliquot of a water sample is reacted with excess potassium iodide in acidic solution to generate I-3 . Carbon dioxide is bubbled through the solution to remove nitrogen monoxide generated. The water sample is transferred to a 500 mL volumetric flask and diluted to volume. A 50.00 mL aliquot is then titrated against 1.092 x 10−4 M thiosulfate, requiring 15.48 mL to reach the starch end point. What is the NO-2 concentration in ppm? Assume solution density of 1.000 g/mL. 2HNO2 + 2H+ + 3I- ⇋ 2NO + I-3 + 2H2O

I-3 + 2S2O32- ⇋ 3I- + S4O26 Answer: 15.56 ppm NO-2 Hard The acidic solution protonates the nitrite and the excess iodide reacts with nitrite to form I-3 .

Percent yield is assumed to be 100% due to excess iodide. The sample is diluted to 500 mL and titrated against thiosulfate to give the mmol’s I-3 in the 50 mL aliquot. Back calculation gives a concentration of 3.380 x 10-4 M for nitrite. Assuming 1.000 g/mL density for the water sample, the ppm nitrite is calculated, 15.56 ppm.

Multiple-Choice 17-1

Fundamentals of Electrolysis

1.

The

A) B) C)

auxillary electrode counter electrode anode

is the electrode at which the reaction of interest occurs.

129 D) E)

cathode working electrode

Answer: E Easy 2.

A current of 0.500 A flows through a cell containing Fe2+ for 10.0 minutes. Calculate the maximum moles of Fe that can be removed from solution? Assume constant current over time.

A) B) C) D) E)

1.04 mmol 51.8 mol 3.11 mmol 1.55 mmol 25.9 mol

Answer: D Intermediate 3.

During electrolysis, electrons flow from the terminal of the power supply into the of the electrolysis cell. Electrons flow from the into the terminal of the power supply to complete the circuit.

A) B) C) D) E)

negative; anode; ground; positive positive; cathode; anode; negative negative; ground; cathode; positive negative; cathode; anode; positive positive; anode; cathode; negative

Answer: D Intermediate

130 4.

Which is NOT true of overpotential?

A)

Overpotential is the voltage required to overcome the activation energy for a reaction at an electrode.

The rate of the reaction will increase as the applied overpotential increases.

Activation energy, hence overpotential, is electrode-surface independent.

Applying an increasing overpotential will sustain a higher current density.

Current density is current per unit area of electrode surface, A/m2.

B) C) D) E)

Answer: C Intermediate 5.

Which statement(s) is(are) defined INCORRECTLY?

I

IV

Ohmic potential is the voltage to overcome the electrical resistance of the solution in an electrochemical cell when a current is flowing.

Concentrational polarization occurs when the concentrations of the reactants or products at the surface of an electrode is greater than those in the bulk solution.

Overpotential is the voltage required to overcome the activation energy for a reaction at only the cathode.

E = E(cathode) – E(anode) – IR – overpotentials – concentration polarization

A) B) C) D) E)

II, III, and IV I, II, and III Only III II and IV I and II

II III

Answer: A Intermediate 6.

For a three-electrode cell the working electrode is the electrode at which the reaction of interest occurs. The is used to measure the potential of the working electrode and the is the current-supporting partner of the working electrode.

A) B) C) D) E)

auxiliary electrode; reference electrode reference electrode; auxiliary electrode reference electrode; polarizable electrode cathode; anode potential electrode; reference electrode

Answer: B Intermediate

131 17-2

Electrogravimetric Analysis

7.

Which is NOT true of electrogravimetric analysis?

A) B) C)

Analyte is quantitatively deposited on an electrode via electrolysis.

The greater the analyte mass deposited correlates with the amount of analyte in solution.

If the analyte has a color in solution; disappearance of the color indicates completion of deposition.

If additional electrode surface is exposed to the solution and further deposition does not occur, the deposition of analyte is complete.

If a qualitative test for a small sample of the analyte solution tests positive, the deposition of analyte is complete.

D) E)

Answer: E Intermediate 8.

Electrolysis of 200 mL of a solution of an unknown metal increased the cathode mass by 0.0434 grams when a current of 0.50 A passed through the cell for 5 minutes. Further research determined the charge on the metal to be +2. What is the identity of the metal?

A) B) C) D) E)

Rg Si Fe Cd Mn

Answer: C Hard

132 17-3

Coulometry

9.

Which of the following statement(s) is(are) TRUE for coulometry?

I II III

Coulometric methods count electrons used in a chemical reaction to measure analyte.

Coulometry employs either constant-current or constant-potential conditions.

An advantage of coulometric methods is high precision and a disadvantage is low sensitivity.

Coulometric titrations are constant-potential methods.

Constant-potential methods utilizing a three-electrode cell are more selective than constant-current methods.

IV V

A) B) C) D) E)

I, II, and V I, II, IV and V I, III, and IV II, IV, and V I and II

Answer: A Intermediate 17-4

Amperometry

10. measures the electric current flowing between a pair of electrodes driving an electrolysis reaction. The amount of current is proportional to the concentration of the analyte.

A) B) C) D) E)

Square wave voltametry Polarography Cyclic Voltametry Amperometry Coulometry

Answer: D Easy

133 11.

Molecules have three ways to reach the surface of an electrode: diffusion, convection, and migration. The rotating disk electrode uses which of these three methods to control the flux of analyte that reaches the electrode?

A) B) C) D) E)

convection diffusion and convection migration convection and migration diffusion and migration

Answer: B Intermediate 17-5

Voltammetry

12.

Which is NOT true for square wave voltammetry, stripping analysis, and cyclic voltammetry?

A) B) C) D) E)

Square wave voltammetry applies a square waveform superimposed on a staircase waveform of increasing potential to the working electrode.

Stripping analysis is the most sensitive of the voltammetric techniques.

Cyclic voltammetry applies a triangular shaped waveform to the working electrode.

For cathodic stripping analysis, the peak current during oxidation is proportional to analyte concentration.

For cyclic voltammetry, for a reversible reaction the peak current, Ipc, for the forward sweep of the first cycle is proportional to the concentration of analyte and the square root of the sweep rate.

Answer: D Intermediate 13. is a set of techniques that observes the relationship between current and voltage during an electrochemical process.

A) B) C) D) E)

Amperometry Coulometry Electrophoresis Voltammetry Electrogravimetric analysis

Answer: D Intermediate

134 14.

Which are true for polarography?

I II

Polarography is amperometry conducted with a dropping Hg electrode.

The diffusion current in the plateau region of the polarographic wave is proportional to the concentration of the analyte and is used for quantitative analysis.

The half-wave potential for maximum current is characteristic of a given analyte in a given medium and can be used for qualitative analysis for the analyte.

Fresh Hg drops give reproducible results.

Polarography is used to primarily to study oxidations Hg may reduce when studying reductions.

III IV V

A) B) C) D) E)

I and V II, III, and V I, III, and IV III, IV, and V II, III, and IV

Answer: E Intermediate 17-6

Karl Fischer Titration of H2O

15.

The Karl Fischer titration measures the traces of water in samples such as transformer oils, foods, polymers, and so forth. Which of the following is NOT true for the chemistry involved?

I II III IV V

The anode solution contains iodide; an alcohol, ROH; a base, B; sulfur dioxide; and possibly other organic compounds.

ROH + SO2 + B → BH+ + ROSO-2 I2 is generated from I− at the cathode.

H2O + I2 + ROSO-2 + 2 B → ROSO-3 + 2 BH+ITwo moles of e− corresponds to one mole H2O

A) B) C) D) E)

III only I and III I, III, and V II and IV V only

Answer: A Intermediate

135 Calculated 17-1

Fundamentals of Electrolysis

16.

Sodium sulfate cells generate hydrogen and oxygen gas when a current is passed through the cell. For the following data, calculate the potential required for the cell to generate partial pressures of 0.25 atm for oxygen and 0.50 atm for hydrogen. At the cathode surface the pH is 10 and at the anode surface the pH is 4. A current of 0.32 amps passes through the solution, which has a resistance of 3 . The overpotentials for the anode and cathode are 0.580 V and 0.584 V respectively.

O2 + 4 H+ + 4 e- ⇋ 2 H2O 2 H2O + 2 e- ⇋ H2 + 2 OH-

Eo = 1.2291 V Eo = -0.8280 V

Answer: -3.6901 V Hard First, determine which half reaction is the cathode and which is the anode. Since this is electrolytic, the more negative potential remains the reduction and is the cathode. The more positive electrode is the anode. Plugging in the partial pressure hydrogen gas and hydroxide into the cathode half reaction and the partial pressure hydrogen and hydronium into the anode half reaction, I solved the Nernst equation for the cell, (E = Ecathode – Eanode), which includes the concentration polarization at the electrode surfaces. Then, subtract the ohmic resistance (V = IR) and subtract the overpotentials for the anode and cathode.

17-2

Electrogravimetric Analysis

17.

When a potential is applied to a solution that is 0.10 M Ni2+ and 0.20 M Fe2+, which metal will be reduced first and at what potential? Is it possible to remove the metal from solution with 99.99% efficiency before the second metal begins to be reduced from solution?

Ni2+ + 2 e- ⇋ Ni, Eo = -0.236 V Fe2+ + 2 e- ⇋ Fe, Eo = -0.44 V Answer: −0.2656 V is required to start reducing Ni2+ from solution. −0.3839 V is required to remove 99.99% of Ni2+ from solution. A potential of −0.4607 is required for reduction of Fe2+ from solution.

Hard Comparison of the reduction potentials indicates that Ni2+ will reduce first. Use the Nernst equation to calculate the potential required when the concentration is 0.10 M Ni2+, −0.2656 V. Next calculate the potential when 99.99% of Ni was removed from solution (0.01% still in solution) −0.3839 V. Calculate the potential required to start reducing Fe2+ from solution, −0.4607 V. The Fe2+ potential is too negative to start reduction at −0.3839 V.

136 17-3

Coulometry

18.

A technician is tasked with testing the newly purchased chloride autotitrator. He prepares a solution that is 3.4238 mg NaCl/mL solution, taking a 25.00 mL aliquot for titration. To minimize any problems with the in situ generation of silver cation titrant, a fresh silver anode is installed and the instrument is set to titrate with a constant current of 500.0 mA. The instrument took 286.3 seconds to reach the end point. Calculate the percent error for the instrument.

Answer: 1.297% or 1.309% depending on how the error is calculated.

Intermediate 1.297% is calculated by taking the titrator time, amperage, and the volume of the sample taken to calculate the mg/mL for NaCl. Comparison of calculated versus known concentration gives 1.297%.

The second method calculates percent error using the theoretical time the titrator should have taken and the time reported by the titrator. The known concentration, the volume sample and amperage are used to calculate the theoretical time to end point. Result is 1.309% error.

19.

Sulfide in a water sample is titrated with electrolytically generated Fe2+. If a 10.00 mL aliquot of the water sample requires a current of 0.75 A for 5.4 minutes to reach the end point, what is the molarity sulfide in the water sample?

Answer: 0.1259 M S2− Intermediate Multiply the current by time expressed in seconds to get the total coulombs passing through the cell. Next use Faraday’s constant to calculate the moles electrons passing through the cell. The stoichiometry is 2 e-:1 mol S2−. Moles sulfide / 0.01 L = molarity of sulfide.

17-5

Voltammetry

20.

An experiment measured the cathodic current for five solutions of known Fe(CN)36 concentration using cyclic voltammetry. The collected data are plotted and the best line fit applied. If an unknown ferricyanide solution has a cathodic current of 17.38 A, calculate the concentration ferricyanide in the unknown.

137

Answer: 6.95 mM Easy Use the trendline data to calculate the concentration ferricyanide for 17.38 A cathodic current.

Multiple-Choice 18-1

Properties of Light

1.

Which of the following is NOT true for light modeled as a wave?

A) B) C) D) E)

Frequency is the number of waves that pass a given point in 1 second.

Wavelength is the distance between the crest and the trough.

Light travels slower through matter than a vacuum.

In a vacuum the speed of light is 3.0 × 108 m/s.

The product of a waves frequency and wavelength is the speed of light.

Answer: B Easy 2.

Which of the following is NOT true for light modeled as a particle?

A) B) C) D) E)

A particle of light is the photon.

The energy of a photon is proportional to the photon’s frequency.

Wavenumber is inversely proportional to photon energy.

As the wavelength increases, photon energy decreases.

The product of frequency and Planck’s constant equals photon energy.

Answer: C Easy

138 3.

Calculate the energy of a photon with a frequency of 573.8 nm.

A) B) C) D) E)

3.46 x 10−37 J 3.80 x 10−40 J 3.46 x 10−28 J 3.80 x 10−31 3.46 x 10−19 J

Answer: E Intermediate

139 18-2

Absorption of Light

4.

Which of the following is NOT true for the absorption of light?

A) B) C) D) E)

Transmittance is the fraction of incident light that is not absorbed.

Molecules that absorb light are promoted to an excited state.

A = −log(T) Molecules that emit light return to the ground state.

Irradiance is the energy per second of the light beam.

Answer: E Intermediate 5.

Which of the following are correct for Beers Law?

I II III IV

Absorbance increases as concentration increases.

Absorbance decreases as pathlength increases.

The molar absorptivity value is chromophore and wavelength specific.

The absorption spectrum is a graph showing absorbance changes with wavelength.

A) B) C) D) E)

I, II, and III 1, III, and IV II, III, and IV I and III I, II, III, and IV

Answer: B Intermediate 6.

A 3.7 × 10−4 M solution of compound X has an absorbance of 0.5342 when measured at 740 nm. Assuming a pathlength of 1 cm, calculate compound X’s molar absorptivity.

A) B) C) D) E)

1443 M−1 cm−1 9.926 × 105 M−1 cm−1 5059 M−1 cm−1 735.9 M−1 cm−1 1.44 × 105 M−1 cm−1

Answer: A Intermediate

140 7.

Beer’s law works for radiation passing through a absorbing species is in a concentration-dependent equilibrium.

A) B) C) D) E)

solution in which the

polychromatic ; dilute ; not participating monochromatic; concentrated; not participating monochromatic; concentrated; participating polychromatic ; dilute ; participating monochromatic; dilute; not participating

Answer: E Intermediate 18-3

Measuring Absorbance

8.

The basic spectrophotometer is composed of four components. Which component listed below is NOT one of the four?

A) B) C) D) E)

sample monochromator beam splitter light source light detector

Answer: C Intermediate 9.

The part of a spectrophotometer that selects the wavelength to irradiate the sample is called the: A) B) C) D) E)

polychromator. monochromator. beam splitter. light source. optode.

Answer: B Easy

141 18-4

Beer’s Law in Chemical Analysis

10.

Reagent blanks are primarily used to:

A) B) C) D) E)

align the optics to achieve maximum irradiance. compensate for any absorbance due to reagents or contamination. verify reagent purity. calibrate the spectrophotometer. test detector sensitivity.

Answer: B Intermediate 18-5

Spectrophotometric Titrations

11.

Absorbance measured during a spectrophotometric titration must be corrected due to

A) B) C) D) E)

absorbing wavelength changes over time. dilution of the analyte by titrant. contaminants in the titrant. instrument drift. decomposition of analyte.

Answer: B Intermediate 12. A 10.00 mL sample is titrated with 0.50 mL of titrant. The observed absorbance is 0.3219.

Calculate the corrected absorbance.

A) B) C) D) E)

0.3066 0.3058 0.2912 0.3380 0.3558

Answer: D Intermediate

142 18-6

What Happens When a Molecule Absorbs Light?

13.

Which of the statements below is FALSE for the excited and ground states of molecules?

A) B) C) D) E)

Electron spins are opposed in the singlet, S0, excited state.

Electron spins are parallel in the triplet, T1, excited state.

Electrons release energy when transitioning from the S1 to the S0 state.

Electrons gain energy when transitioning from the S0 to the S1 state.

Molecular geometry in the excited state may be different than the molecular geometry in the ground state.

Answer: A Intermediate 14.

Fluorescence is:

A) B) C) D) E)

a nonradiative transition between states with the same spin, S1  S0. a nonradiative transition between states with different spin, T1  S0 the emission of a photon during a transition between states with the same spin, S1  S0. the emission of a photon during a transition between states with different spin, T1  S0 None of these answers defines fluorescence.

Answer: C Intermediate 15.

Phosphorescence is:

A) B) C) D) E)

a nonradiative transition between states with the same spin, S1  S0. a nonradiative transition between states with different spin, T1  S0 the emission of a photon during a transition between states with the same spin, S1  S0. the emission of a photon during a transition between states with different spin, T1  S0 None of these answers defines phosphorescence.

Answer: D Intermediate

143 18-7

Luminescence

16.

Which of the following is TRUE for emission spectra?

A) B) C) D) E)

Excitation wavelength is fixed and the emission wavelengths are scanned.

Excitation wavelength is scanned and the emission wavelength is fixed.

Excitation wavelength is fixed and the emission wavelength is fixed.

Excitation wavelength is scanned and the emission wavelength is scanned.

None of these answers is true.

Answer: A Intermediate

16.

Which of the following is TRUE for excitation spectra?

A) B) C) D) E)

Excitation wavelength is fixed and the emission wavelengths are scanned.

Excitation wavelength is scanned and the emission wavelength is fixed.

Excitation wavelength is fixed and the emission wavelength is fixed.

Excitation wavelength is scanned and the emission wavelength is scanned.

None of these answers is true.

Answer: B Intermediate 17.

All of the following are TRUE for luminescence, except:

A) B) C) D) E)

luminescence is emission of light from an excited state of a molecule. luminescence is inherently more sensitive than absorption. at low concentration, emission intensity is proportional to analyte concentration. at high concentration, emission intensity decreases due to dilution. light emitted from a chemical reaction is called chemiluminescence.

Answer: D Intermediate Calculated 18-1

Properties of Light

18.

A red laser emits a pulse of light every 10 s. If the laser operates at a wavelength of 695 nm and 50 W, calculate the number of photons emitted during a pulse. (1 W = 1 J/s) Answer: 1.75 × 1015 photons Hard

144 Calculate the energy of one photon (2.86 × 10-19 J/photon), then the total energy of one pulse (0.0005 J), then total energy divided by energy by photon (1.75 × 1015.)

18-2

Absorption of Light

19.

A transmittance of 0.3847 is measured for a 0.00023 M solution. Assuming a pathlength of 1 cm, calculate the molar absorptivity of the solute.

Answer: 1803 M−1 cm−1 Intermediate Calculate absorbance from transmittance (0.4149), then molar absorptivity.

18-4

Beer’s Law in Chemical Analysis

20.

The iron content of a well water sample must be quantified. A 10.00 mL aliquot of well water is transferred to a 100.0-mL volumetric flask and 20 mL of 0.10 M nitric acid is added to oxidize iron(II) to iron(III). Next 30 mL of a 0.1 M sodium thiocyanate solution is added to form the deep red complex iron(III) thiocyanate. The solution is diluted to volume with distilled deionized water. A reagent blank is similarly prepared. If the absorbance of the iron thiocyanate solution is 0.5483 and the reagent blank absorbance is 0.0282, calculate the iron concentration for the water sample. The molar absorptivity for iron(III) thiocyanate is 3200 M−1 cm−1. Assume a pathlength of 1 cm and a 100% yield.

Fe3+ (aq) + SCN- (aq) ↔ FeSCN2+ (aq) Answer: 0.001625 M Hard Calculate corrected absorbance (0.5201), then the concentration iron (0.0001625 M) and then take into account the dilution factor (1:10.)

18-5

Spectrophotometric Titrations

21.

After 10.00 mL titrant has been added to 50.00 mL analyte the absorbance was measured to be 0.4639. Calculate corrected absorbance.

Answer: 0.5567 Easy 18-6

What Happens When a Molecule Absorbs Light?

22.

Quinine sulfate in 0.5 M sulfuric acid is excited at 310 nm and fluoresces at 450 nm.

How many Joules were dissipated through nonradiative transitions on a per molecule basis?

Answer: 1.99 × 10−19 J dissipated Hard Either calculate E310 and E450, subtracting E450 from E310 to get energy dissipated, Edisp. Or 1/310 −1/× = 1/450, solving for ×, 996.4 which is the wavelength for the energy difference, 1.99 × 10−19 J.

145 Multiple-Choice 19-1

Analysis of a Mixture

1.

Which is true for spectroscopy of a mixture?

A)

The absorbance for a mixture at a particular wavelength is the sum of the absorbances for the components that absorb at the particular wavelength.

Spectrophotometers can differentiate between mixture components that absorb at the same wavelength.

Each component in a mixture has the same molar absorptivity at the same wavelength.

The concentration for each component in the mixture is easily calculated with a set of simultaneous equations if there is significant overlap of individual spectra.

The concentration for each component in a mixture is easily calculated by least squares for guesses of each component when the individual spectra are well resolved.

B) C) D) E)

Answer: A Intermediate 2.

For the plot of absorbance versus wavelength for two compounds, which two wavelengths are most suited for mixture analysis?

A) B) C) D) E)

515 nm, 515 nm 445 nm, 635 nm 380 nm, 750 nm 515 nm, 635 nm 445 nm, 515 nm

Answer: B Easy

146

19-2

Measuring an Equilibrium Constant

3.

Which is NOT true for a Scatchard plot?

A)

A Scatchard plot is a graph of [PX]/[X] versus [PX].

DA = KDe P0 - KDA [X] The binding constant, K, is the absolute value for the reciprocal of the slope.

B) C) D) E)

A = ePX [PX] + eP [P] S = K(1 - S) [X]

Answer: C Hard 4.

The Scatchard plot for the reaction P + X ⇋ PX is below. What is the equilibrium constant for the reaction?

A) B) C) D) E)

3.830 M−1 1.478 × 10−4 M−1 0.2611 M−1 6765 M−1 0.5832 M−1

Answer: D Easy

147 19-3

The Method of Continuous Variation

5.

For the method of continuous variation, which is NOT true?

A)

The method of continuous variation allows for the determination of stoichiometry of a complex.

Corrected absorbance is the measured absorbance for the complex less the absorbance for unreacted ligand and central metal.

Maximum absorbance is reached at the composition corresponding to the stoichiometry of the predominant complex.

The total concentration for various ratios of P and X must be constant.

Maximum absorbance when the mole fraction X equals 0.75 is indicative of a complex with the chemical formula of PX3.

B) C) D) E)

Answer: E Intermediate 6.

The method of continuous variation was applied to the complex formed between P and X. A plot of mole fraction X versus PaXb absorbance is plotted. What is the stoichiometry for the complex?

A) B) C) D) E)

P2X3 PX4 PX5 P3X2 P4X

Answer: B Easy

148 19-4

Flow Injection Analysis and Sequential Injection

7.

The technique in which a liquid sample is injected into a continuously flowing liquid carrier containing a reagent that reacts with the sample is called: A) B) C) D) E)

flow analysis. sequential flow analysis. sequential analysis. sequential injection analysis. flow injection analysis.

Answer: E Easy 8.

The advantages of flow injection analysis include which of the following?

I II III IV

the speed in sample analysis automation of solution handling reproducibility low cost of analysis

A) B) C) D) E)

II and III I, II, III, and IV I, III, and IV II, III, and IV I, II, and III

Answer: B Intermediate 9.

Sequential injection differs from flow injection by:

A) B) C) D) E)

flow programming. flow reversal. reagent mixing. flow reversal and reagent mixing. flow programming and flow reversal.

Answer: E Intermediate

149 10.

The advantages of sequential injection over flow injection analysis include all of the following, except: I II III IV

less waste generated. smaller volumes of expensive reagents are consumed. ability to handle large number of samples. larger reaction volumes.

A) B) C) D) E)

I and II III and IV II and IV I, II, III, and IV I, II, and IV

Answer: A Intermediate 19-5

Immunoassays

11.

Which of the following is true for ELISA?

I

An antibody, antibody 1, specific to the analyte of interest is dissolved in aqueous solution.

Analyte is incubated with antibody 1, binding to form a complex. The fraction of antibody 1 sites bound to analyte is proportional to the concentration.

Unbound substances are flushed from the ELISA surface and a second antibody, antibody 2, which binds to a different region of the analyte, is applied to the antibody 1-analyte complexes.

Antibody 2 is chemically modified to include an enzyme that converts a colorless reagent to a colored reagent. The more bound analyte, the more intense the color generated.

II III

IV

A) B) C) D) E)

I, II, III, and IV II and IV I, III, and IV II, III, and IV I, II, and IV

Answer: D Intermediate

150 12. enhances the fluorescence of immunoassays by a factor of 100 by measuring the fluorescence of Eu3+ 200 s after excitation with a laser pulse.

A) B) C) D) E)

Time-resolved fluorescence immunoassays Time-dependent fluorescence immunoassays Immunoassay fluorescence over time immunoassays Weakly bound time fluorescence immunoassays Europium-bound fluorescence immunoassays

Answer: A Intermediate 19-6

Sensors Based on Luminescence Quenching

13.

When a molecule absorbs a photon, the molecule is promoted to an excited state, M*. The rate at which M* is created is proportional to the concentration of M. There are three pathways that M* may take to return to the ground state. Which statement below is INCORRECT for these three pathways.

A) The rate at which the excited state returns to the ground state is independent of the pathway to the ground state and the concentration of excited state molecules.

B) The three possible pathways to return to the ground state are emission, deactivation, and quenching.

C) Quenching occurs when a second molecule, the quencher Q, collides with M* to return M* to the ground state. M* + Q → M + Q* D) Deactivation returns M* to the ground state by colliding with other molecules and releasing energy in the form of heat. M* → M + heat E) Emission returns M* to the ground state by emitting a photon. M* → M + hv.

Answer: A Intermediate 14.

Which is true for the relationship between quencher concentration and fluorescent intensity?

A) B) C) D) E)

As quencher concentration increases, fluorescent intensity increases.

As quencher concentration decreases, fluorescent intensity decreases.

As quencher concentration increases, fluorescent intensity decreases.

As quencher concentration increases, fluorescent intensity remains constant.

As quencher concentration decreases, fluorescent concentration remains constant.

Answer: C Intermediate

151 15.

The Stern-Volmer equation relates quencher concentration to the ratio 0/Q. Which of the following is NOT true for the Stern-Volmer equation?

A) B)

The y-intercept for the Stern-Volmer equation is always 1.

The x-intercept for the Stern-Volmer equation is always 1.

æ kq ö The slope for the graph of quencher concentration versus I0/IQ is m = ç ÷. è ke + kd ø I0/IQ is proportional to 0/Q. 0/Q is greater than 1 for any concentration Q.

C) D) E)

Answer: B Intermediate Calculated 19-1

Analysis of a Mixture

16.

For the UV-Vis absorption data below, calculate the concentration for the metals M2+ and 2+ N in a river sample if the unknown solution was prepared by diluting 10.00 mL of river water with 10.00 mL of ligand solution and diluted to a final volume of 100.00 mL. Assume the absorbance was measured in a 1.00-cm cell.

Solution Wavelength Concentration Absorbance M standard 650 nm 1.82E-05 0.723 2+ N standard 650 nm 1.82E-05 0.312 2+ M standard 425 nm 1.82E-05 0.199 2+ N standard 425 nm 1.82E-05 0.811 Unknown 650 nm ?

0.753 Unknown 425 nm ?

0.536 2+

Answer: M2+ = 1.54 × 10−5 M and N2+ = 8.24 × 10−6 M Hard Calculate the molar absorptivity for each standard. Set up two simultaneous equations, one each for 650 nm and 425 nm and solve for M2+ and N2+.

152 17.

For the acid-base indicator bromothymol blue, the protonated form, HBB, is yellow and the deprotonated form, BB−, is blue. When both forms are present the indicator has a green color, the shade of green depends on the ratio of HBB:BB−. Use the information below to calculate the concentration of HBB and BB− for a green solution of bromothymol blue.

Wavelength 1.0 × 10−5 M HBB 1.0 × 10−5 M BB− Unknown Mixture 453 nm 0.1910 0.0234 0.4680 616 nm 0.00625 0.3211 0.3269 Answer: 3.50 × 10−5 M HBB and 9.50 x 10−6 M BB− Hard Calculate the molar absorptivity’s for each species at each wavelength from the known concentrations for HBB and BB− and the absorbances. Construct two equations relating the absorbance for the unknown to the molar absorptivity times the unknowns X and Y, Aunk-453 = HBB-453X + BB—453Y and a second form 616 nm. Solve the set of simultaneous equations.

153 19-3

The Method of Continuous Variation

18.

The stoichiometry for the complex MXn is investigated using the method of continuous variation. A series of 11 solutions are prepared using 0.0100 molar M and 0.0100 molar X with the volumes in the table below. Use the standard solution composition and corrected absorbance data to determine the stoichiometry for the complex. Assume that the absorbance is due only to the complex. Include a rough graph to support your answer.

Standard Solution Volume M, mL Volume X, mL Corrected Absorbance 1 0 10 0.0000 2 1 9 0.2398 3 2 8 0.4796 4 3 7 0.7194 5 4 6 0.7194 6 5 5 0.5995 7 6 4 0.4796 8 7 3 0.3597 9 8 2 0.2398 10 9 1 0.1199 11 10 0 0.0000 Answer: MX2 Hard Plot the mole fraction X versus absorbance of the complex MXn. The mole fraction at highest absorbance is 0.66, which is 2/3 or two moles X for three moles total material (two moles X and one mole M.) Graph prepared in Microsoft Excel for Mac 2011 below.

154 19-6

Sensors Based on Luminescence Quenching

19.

An oxygen sensor is constructed from a fluorescent dye that is easily quenched by oxygen. The sensor is calibrated with a deoxygenated water sample and an oxygen saturated water sample, 2.6 x 10−4 M. The sensor is deployed to a waste treatment plant to monitor dissolved oxygen in effluent. If the sensor records a fluorescent intensity of 322.928, what is the oxygen concentration?

O2, Fluorescent Molarity Intensity 0.00 537.283 -4 1.30 x 10 372.992 2.60 x 10-4 294.482

Answer: 2.06 × 10−4 M Intermediate Divide unknown fluorescent intensity into the fluorescent intensity of 0 M oxygen, and plug it into the best fit line to obtain the molarity oxygen.

155 20.

A student determines the concentration chloride in an unknown sample by pipetting 10 mL of the sample and 10 mL of 1.000 × 10−5 M quinine solution to a 100 mL flask and diluting to volume with 1 M H2SO4. A series of six chloride standards were prepared using a 0.0100 M standard chloride solution, adding an increasing volume of chloride to 100 mL volumetric flasks, in addition to 10 mL of 1.000 × 10−5 M quinine. Each solution was diluted to volume with 1 M H2SO4. Calculate the chloride concentration for the unknown using the collected fluorescence data and the Stern-Volmer graph below for the data.

Chloride Molarity 0.000000 0.001000 0.002000 0.003000 0.004000 0.005000 Unknown

Fluorescence Intensity 140.00 137.70 134.70 131.30 127.61 123.70 133.50

Answer: 0.01886 M Cl− Hard Divide the fluorescence intensity of the 0 chloride standard by the fluorescence intensity of the unknown to obtain the I0/IQ, plug this value in for y and solve for x. X is the chloride concentration in the dilute unknown. Multiplying by 10 will give the original chloride concentration of the unknown.

Multiple-Choice 1. true?

Which of the following statements regarding double beam spectrophotometers is NOT

A) B)

The reference beam allows for automatic correction for source intensity fluctuations.

Comparison of the power emerging from the sample and reference beams determines absorbance and transmittance.

The reference beam allows for automatic correction for detector drift.

The reference beam is created using a stationary mirror.

The reference beam is created using a rotating mirror.

C) D) E)

Answer: D Intermediate 20-1

Lamps and Lasers: Sources of Light

156 2.

Spectrophotometers employ various sources of light. Which of the following is NOT true for spectrophotometer light sources?

A) B) C) D) E)

Tungsten lamps emit radiation over the visible spectrum and parts of the ultraviolet and infrared spectrums.

Dueterium arc lamps use controlled electric discharges to cause D2 to dissociate and emit visible radiation.

Light emitting diodes emit narrow bands of visible and near-infrared radiation.

Lasers emit radiation of a single wavelength.

A silicon carbide globar emits infrared radiation when heated to near 1500 K.

Answer: B Intermediate 3.

Which property of laser light is NOT correctly defined/described?

A) B) C) D) E)

monochromatic: emits one wavelength of light extremely bright: high power at one wavelength collimated: parallel rays of light polarized: electric field oscillates between two perpendicular planes coherent: all waves in phase

Answer: D Easy

157 20-2

Monochromators

4.

Which of the following monochromator terms is NOT correctly defined?

A) B) C) D) E)

grating: optical element with closely spaced lines diffraction: bending of light by a grating refraction: scattering of light by a lens or prism polychromatic: multiple wavelengths monochromatic: one wavelength

Answer: C Intermediate 5.

Calculate the minimum grating length to successfully resolve 616.50 nm from 616.59 nm for first-order diffraction. The available diffraction grating is 103 grooves per cm.

A) B) C) D) E)

13.7 cm 1.00 cm 111.1 cm 0.15 cm 6.85 cm

Answer: E Intermediate 6.

Stray light is a problem for spectrophotometers that are not tightly sealed. What is the impact of stray light on the recorded absorbance?

A) B) C) D) E)

The apparent absorbance is greater than true absorbance and increases as the amount of stray light increases.

The apparent absorbance is greater than true absorbance and decreases as the amount of stray light increases.

The apparent absorbance is less than the true absorbance and increases as the amount of stray light decreases.

The apparent absorbance is less than the true absorbance and decreases as the amount of stray light decreases.

Impossible to determine effect of stray light on measured absorbance.

Answer: C Intermediate

158 7.

A disperses light into its component wavelengths and selects a narrow band of wavelengths to pass on to the sample or detector.

A) B) C) D) E)

polychromator photodiode array monochromator charge couple device photomultiplier tube

Answer: C Easy 20-3

Detectors

8.

Which statement(s) is(are) true for photomultiplier tubes?

I Photomultiplier tubes are very sensitive detectors, amplifying the photoelectron current by approximately 1 million.

II Emitted photoelectrons are accelerated toward positively charged electrodes called dynodes.

III Electrons strike the dynode with a kinetic energy less than the original emitted kinetic energy.

IV For each electron striking a dynode, more than one electron is knocked free. The additional electrons are accelerated toward the next dynode in the detector where amplification is repeated.

A) B) C) D) E)

II I and IV II, III, and IV I, II, and IV II and III

Answer: D Intermediate 9.

Which is NOT an appropriate detector for a spectrophotometer?

A) B) C) D) E)

photomultiplier tube thermal conductivity photodiode array charged coupled device phototube

Answer: B Intermediate

159 10.

Which is NOT true for photodiode arrays?

A) B) C) D) E)

Photodiode arrays respond to more than one wavelength.

Photodiode arrays require a polychromator to detect more than one wavelength.

Photodiode arrays allow for rapid acquisition of an entire light spectrum.

Photodiode arrays have higher signal-to-noise ratios.

Spectrophotometers using photodiode arrays are dispersive spectrometers.

Answer: E Intermediate 11. current when struck by analyte concentration.

A) B) C) D) E)

response is a function of incident light and produce an electrical . For a given radiant power, current is

Optical ; photons; constant for varying Detector; photons; proportional to Analyte; analyte particles; constant for varying Detector; analyte particles; inversely proportional to Optical ; analyte particles; inversely proportional to

Answer: B Easy 20-4

Optical Sensors

12.

Which statements are true for optical fibers?

I Optical fibers carry light by total internal refraction.

II Optical fibers are constructed of a high-refractive-index transparent core enclosed in a lower refractive-index transparent cladding.

III Optical fibers exploit Snells Law, n1 sin 1 = n2 sin 2, to transmit light along the length of the optical fiber.

IV If n1/n2 < 1, there is an allowed range of angles i in which essentially all light is reflected at the walls of the core, with negligible amounts of light penetrating the cladding.

A) B) C) D) E)

I and IV I, II, and IV II, III, and IV II and III II and IV

Answer: D Intermediate

160 13.

A) B) C) D) E)

are optical sensors based on an optical fiber.

Optodes Optides Optectors Optubes Pristrodes

Answer: A Easy 20-5

Fourier Transform Infrared Spectroscopy

14.

Two components of an interferometer are a moveable mirror and a beam splitter. Which of the following are NOT true statements regarding each component in creating an interferogram?

I II III IV

When the mirror travels a distance of /4, deconstructive interference occurs.

The beam splitter alternates the incident light between a reference and the sample.

When the mirror travels a distance of /2, deconstructive interference occurs.

The beam splitter splits the incident light between the stationary mirror and the moveable mirror.

A) B) C) D) E)

I and II I and IV II and III II and IV III and IV

Answer: C Intermediate 20-6

Dealing with Noise

15.

Which property of noise does signal averaging exploit to improve the signal-to-noise ratio?

A) B) C) D) E)

Noise is loud.

Noise is random.

Noise is predictable.

Noise drifts.

The amplitude of noise is a constant.

Answer: B Intermediate

161 16.

There are many types of noise. Of the noise types below, which is NOT correctly defined?

A) B) C) D) E)

line noise – noise that occurs at discrete frequencies drift noise – low frequency noise caused by flickering or drifting of light intensity Gaussian noise – noise amplitude is dependent on frequency Johnson noise – random fluctuations of electrons in electronic devices shot noise – noise that arises from random variation in the number of photons reaching the detector or small number of electrons and holes generated in a semiconductor

Answer: C Intermediate Calculated 20-2

Monochromators

17.

The monochromator for a new instrument design must successfully resolve 0.005 nm at 631 nm. Calculate the minimum number of grooves per centimeter for the diffraction grating if the grating has a length of 5 cm. Assume first order diffraction.

Answer: 25240 Intermediate Plug the wavelength, delta wavelength, and n = 1 into the monochromator equation to determine the number of lines needed, N. N/5 cm = # of grooves for cm required.

18.

The true absorbance for a 1.0 x 10−5 M solution is 0.7526. If the percentage stray light for a spectrophotometer is 0.56%, calculate the percentage by which the apparent concentration deviates from the known concentration.

Answer: 0.30% Hard Calculate the molar absorptivity from the known absorbance and concentration. Next, calculate the apparent absorbance from true absorbance and the percent stray light. Use the apparent absorbance and the molar absorptivity to calculate the apparent concentration. ((True concentration – apparent concentration)/true concentration) ∙100 = 0.30%.

162 19.

Ni emits light at 345.85 nm and 346.17 nm in an atomic emission spectrophotometer.

Can the spectrophotometer’s monochromator successfully resolve 345.85 nm from 346.17 nm?

The grating has a length of 5.0 cm in length and there are 5500 grooves per cm. Assume firstorder diffraction.

Answer: The required resolution to differentiate between the Ni emission lines is 345.85 nm/0.32 nm = 1080.8. The resolution of the monochromator is calculated by first determining  345.85 nm/  = 1(27500 grooves)  = 0.0126 nm. Resolution is 345.85 nm/0.0126 nm = 27448.4.

The actual resolution of the monochromator is 25.40 times greater than needed.

Hard 20-6

Dealing with Noise

20.

The signal-to-noise ratio for a single spectrum is 2.18. How many additional scans are required to increase the signal-to-noise ratio by a factor of 50?

Answer: A total of 2500 total scans are required, so 2499 additional scans.

Intermediate The square of the factor increase is the number of required scans to signal average.

Multiple-Choice 21-1

An Overview

1.

Which is NOT true of the flame atomic absorbance spectroscopy of iron?

A)

The iron sample is aspirated into the flame, where the solvent evaporates and the iron analyte is atomized.

The flame replaces the cuvet of conventional spectrometers and the flame pathlength is typically 10 cm.

The hollow-cathode lamp for iron emits wavelengths of light unique for iron.

The iron atoms in the flame absorb some of the light emitted from the hollow-cathode lamp.

A detector measures the amount of light generated by the flame and not absorbed by iron.

B) C) D) E)

Answer: E Intermediate 2.

Which of the following statements are true of atomic emission spectroscopy (AES)?

I II

Atoms are promoted to an excited state with a laser.

Atoms are promoted to an excited state by gaining energy from collisions with other atoms or from the high thermal energy of the flame.

The intensity of emitted light is proportional to analyte concentration.

Plasma often replaces flame in AES.

III IV

163

A) B) C) D) E)

III and IV I, III, and IV II, III, and IV II and III I and III

Answer: C Easy

164 21-2

Atomization: Flames, Furnaces, and Plasmas

3.

There is no universal flame temperature for flame-based atomic spectroscopy that gives optimal results for all elements. gives a .

A) B) C) D) E)

Excess oxidant; cooler flame Excess oxidant; hotter flame Shorter aspiration time; cooler flame Longer aspiration time; hotter flame Excess analyte; hotter flame

Answer: B Intermediate 4.

Graphite furnace employs temperature programming to produce elemental vapor. The program begins by raising the furnace temperature to initiate , where water is removed from the sample. Next, the temperature of the furnace is then raised to begin the process, where organic material is removed. The temperature of the furnace is raised again to the sample, creating the elemental vapor required for spectroscopy.

A) B) C) D) E)

baking; desiccation; atomize drying; desiccation; fragment drying; charring; atomize drying; atomize; charring baking; atomize; desiccation

Answer: C Intermediate

165 21-3

How Temperature Affects Atomic Spectroscopy

5.

Fluctuations in flame temperature affect the reproducibility of atomic emission experiments more so than atomic absorbance experiments. Which of the following statements explain this phenomenon?

I

II

III

IV

A) B) C) D) E)

As temperature increases there is an increase in emission intensity resulting from an increase in the excited state population; the apparent concentration is greater than the actual concentration.

As temperature decreases there is a decrease in emission intensity resulting from an increase in the excited state population; the apparent concentration is greater than the actual concentration.

As temperature increases there is a decrease in emission intensity resulting from a decrease in the excited state population; the apparent concentration is less than the actual concentration.

As temperature decreases there is a decrease in emission intensity resulting from a decrease in the excited state population; the apparent concentration is less than the actual concentration.

II and III I and IV I and III II and IV None of these statements explains this phenomenon.

Answer: B Intermediate

166 6.

Which of the following statements are true regarding the effect of flame temperature on atomic absorbance and atomic emission spectroscopy?

I II III IV

A) B) C) D) E)

Varying the flame temperature by 10 K hardly affects the ground-state population of the analyte and would not noticeably affect the analyte signal in atomic absorption.

Varying the flame temperature by 10 K hardly affects the excited-state population of the analyte and would not noticeably affect the analyte signal in atomic emission.

Varying the flame temperature by 10 K affects the ground-state population of the analyte and will noticeably affect the analyte signal in atomic absorption.

Varying the flame temperature by 10 K affects the excited-state population of the analyte and will noticeably affect the analyte signal in atomic emission.

I and II I and IV II and III III and IV II and IV

Answer: B Intermediate 7.

The energy difference between the ground and excited state and temperature has the following effect on the excited-state population: As the energy difference between the ground state and excited state decreases, the excited-state population .

As the temperature of the flame increases, the excited-state population .

A) B) C) D) E)

remains the same; increases decreases; decreases increases; remains the same decreases; increases decreases; remains the same

Answer: D Intermediate

167 21-4

Instrumentation

8.

Hollow-cathode lamps emit light with very sharp linewidths. Which mechanisms are responsible for broadening atomic emission line widths?

I II III IV

pressure broadening ionization effect matrix effect Doppler effect

A) B) C) D) E)

I and IV II, III, and IV I, III, and IV II and IV III and IV

Answer: A Intermediate 9.

All of the following are true for hollow-cathode tubes, except:

A) B)

hollow-cathode tubes emit narrow emission lines, too narrow for slits to isolate. the cathode of a hollow-cathode tube is made of the same element whose emission lines we want. the low pressure Ne or Ar gas is ionized when voltage is applied, which ionizes the gas and accelerates the ions toward the cathode. the kinetic energy of the ionized gas is sufficiently high to sputter metal atoms from the cathode. the gaseous metal atoms are excited by collisions with high-energy electrons and emit photons with characteristic wavelengths that are absorbed by the analyte.

C) D) E)

Answer: A Intermediate

168 10.

All of the following are background correction techniques for atomic spectroscopy, except: A) B) C) D) E)

subtraction of adjacent pixel height from signal pixel height for a charge injection detector. a rotating beam chopper periodically blocks the analyte signal, allowing for a background measurement. the absorbance and scattering of the light from a D2 lamp is measured and subtracted from the absorbance and scattering of light from a hollow-cathode tube. the analyte sample is spiked with a compound to absorb all background signal. a strong magnetic field is pulsed on and off. The sample and background is observed when field off. Only the background is observed when the field is on. The absorbance difference is the analyte signal.

Answer: D Intermediate 21-5

Interference

11. unchanged.

A) B) C) D) E)

Scattering Interference Self-absorption Diffraction Refraction

Answer: B Easy

is any effect that changes the signal while analyte concentration remains

169 12.

Which of the following statements are true for chemical interference?

I II

Releasing agents are added to the sample to reduce chemical interference.

Chemical interference occurs when a component of the sample reduces the extent of atomization of the analyte.

Higher temperature flames will reduce chemical interference.

Chemical interference occurs when a component of the sample emits light at or close to the same frequency as the analyte.

Releasing agents will preferentially react with the analyte, protecting the analyte from components of the sample that will ionize the analyte.

III IV V

A) B) C) D) E)

III, IV, and V I, III, and IV I, II, and IV I, II, and III II, III, and V

Answer: D Intermediate 13.

Which of the following is(are) NOT true for ionization interference?

I

Ionization interference can occur with alkali metals at relatively low temperature and for analyses for other elements at higher temperature.

Ionization suppressors decrease the extent of ionization of analyte.

Ionization suppressors are more easily ionized than the analyte, the large number of electrons produced prevent ionization of the analyte.

Ionization interference occurs when a component of the sample prevents ionization of the sample.

II III IV

A) B) C) D) E)

IV II I and II III III and IV

Answer: A Intermediate

170 14.

Inductively coupled argon plasma does not suffer from many of the interferences encountered with flame atomic spectroscopy. Which of the statements below are NOT advantages of plasma over flame?

A) B) C) D) E)

Plasma is twice as hot as conventional flame.

Formation of analyte oxides and hydroxides is negligible.

Atomization is more complete and signal is enhanced.

Analyte in plasma self-absorbs at a higher rate than in flame.

Analyte residence time in the plasma is about twice as long as in flame.

Answer: D Intermediate 21-6

Sampling by Laser Ablation

15.

Laser ablation is an inductively coupled plasma sampling technique. Which of the following is true for laser ablation?

I Laser ablation destroys the entire sample.

II A laser pulse is focused onto a small area, producing an explosion of atoms, electrons, and ions in the gas phase.

III Depth profiling of the sample is possible with repeated laser pulses probing deeper and deeper into the sample.

IV Nanogram amounts of material are ablated per pulse.

V Ablation products are swept to the plasma with H2 gas.

A) B) C) D) E)

I and II II, III, and IV I, II, III, and IV II, IV, and V III and IV

Answer: B Intermediate

171 21-7

Inductively Coupled Plasma-Mass Spectrometry

16. occurs for inductively coupled plasma-mass spectroscopy when the interfering species has the same mass-to-charge ratio as the analyte ion.

A) B) C) D) E)

Ionization interference Analyte interference Isobaric interference Allotrope interference Isothermal interference

Answer: C Easy 17.

The dynamic reaction cell reduces isobaric interference with thermodynamically favorable reactions that remove interfering species with the same mass-to-charge ratio. Which of the reactions does not remove interfering species for 52Cr+ or 56Fe2+?

A)

40

Ar12C+ + NH3 → NH+3 + Ar + C

B)

40

Ar16O+ + NH3 → NH+3 + Ar + O

C) D) E)

35

Cl16OH+ + NH3 → NH+3 + ClO 40 Ca16O+ + CO → Ca+ + CO2 35 17 Cl OH+ + NH3 → NH+3 + ClO

Answer: E Intermediate 21-8

X-ray Fluorescence

18.

X-rays are generated when:

A) B) C) D) E)

metals such as W, Mo, Ag, or Rh are heated to high temperatures. high energy photons strike a metal such as W, Mo, Ag, or Rh. high-energy electrons strike an anode made of metals such as W, Mo, Ag, or Rh. high-energy protons strike an anode made of metals such as W, Mo, Ag, or Rh. metals such as W, Mo, Ag, or Rh are electrically excited.

Answer: C Intermediate

172 19.

Which of the following is NOT true for X-ray fluorescence?

A) The fluorescent X-ray photon has a lower energy than the original X-ray photon.

B) The incoming X-ray knocks free an electron from the K, L, or M shells.

C) The fluorescent X-ray photon is emitted when an outer shell electron falls into a shell closer to the nucleus.

D) K X-rays original when L or M shell electrons fall into the K shell.

E) K and K X-rays originate when L shell electrons transition to the K shell.

Answer: E Hard Calculated 20.

To measure calcium in a breakfast cereal, 0.6590 grams of crushed cereal were reacted for 1 hour in warm 1 M HNO3. The reaction mixture was filtered and the filtrate transferred to a 100 mL volumetric flask. A series of eight standards are prepared by transferring 5.00 mL aliquots to 50-mL volumetric flasks to which increasing volumes of standard Ca2+ (containing 20.0 g/mL) is added, diluting to the final volume with distilled deionized H2O. The samples were then analyzed by flame atomic absorption. Use the data and graph below to find the percent by mass calcium in the cereal.

Ca2+ Stnd (mL) 0 1.00 3.00 5.00 8.00 10.00 15.00 20.00

Absorb.

0.151 0.185 0.247 0.300 0.388 0.445 0.572 0.723

Answer: 0.340% Ca by mass Intermediate Solve the trendline equation for the x-intercept, then take the absolute value to get the concentration of the unknown as diluted when preparing the 50 mL standard solutions. Back dilute to the original solution, multiply by 100 mL, and convert mass calcium to grams and solve for the percentage.

Multiple-Choice

173

22-1

What Is Mass Spectrometry?

1.

The three essential components of any mass spectrometer are:

I II III IV V

ion source. decelerator. detector. mass separator. accelerator.

A) B) C) D) E)

I, III, and IV II, III, and IV I, III, and III II, IV, and V III, IV, and IV

Answer: A Intermediate 2.

Analytes must be ionized prior to entering the mass filter in a mass spectrometer.

Electron ionization and chemical ionization are two ionization techniques. Which of the following statements are INCORRECT regarding either technique?

I Electron ionization uses an electron beam to create molecular ions, M+.

II Chemical ionization uses an electron beam to ionize reagent gas.

III Electron ionization gives very little fragmentation of the molecular ion.

IV The ionized reagent gas in chemical ionization undergoes a complex set of chemical reactions before protonating the analyte to create MH+.

V Both ionization techniques give identical mass spectra.

A) B) C D) E)

I, III, and V III and V II, III, and IV I, II, and IV I, II, and V

Answer: B Intermediate

174 3.

The resolving power for a mass spectrometer can be calculated using the equations m m Resolving power = or Dm m1/2 Which of the following statements are NOT true regarding resolving power?

I II III IV V

m is the larger value of m/z m is the difference between the two m/z peaks m1/2 is the width of the peak at half maximum height m is the smaller value of m/z m1/2 is the average mass for the two peaks

A) B) C) D) E)

I and V II, III, and IV I, II, and III I, II, and V II and IV

Answer: A Intermediate 22-2

Oh, Mass Spectrum, Speak to Me!

4.

The molecular ion peak for a compound is 180 Da. The number of nitrogen atoms in the compound can be: A) B) C) D) E)

an odd number of nitrogens. an even number of nitrogens. zero nitrogens.

A and C B and C

Answer: E Intermediate 5.

How many rings and double bonds are there for the compound C11H7NO2Cl2?

A) B) C) D) E)

c = 10; h = 7; n = 1; R&DB = 8 c = 11; h = 9; n = 1; R&DB = 8 c = 10; h = 7; n = 3; R&DB = 9 c = 11; h = 7; n = 1; R&DB = 9 c = 11; h = 9; n = 3; R&DB = 9

Answer: B Intermediate

175 6.

When identifying the molecular ion peak, M+· , all of the following must be kept in mind, except: A) M+· will be the highest m/z value of any of the ―significant‖ peaks in the spectrum that can not be attributed to isotopes or background.

B) intensities of isotopic peaks M+1, M+2 and so forth must be consistent with the proposed formula.

C) M+· will be at least 70% of the base peak.

D) the peak for the heaviest fragment ion should not correspond to an improbable mass loss from M+· .

E) if a fragment ion is known to contain X atoms of element Z, then there must be at least X atoms of element Z in the molecular ion.

Answer: C Intermediate 22-3

Types of Mass Spectrometers

7.

Which is NOT true of mass spectrometers?

A) Double-focusing mass spectrometers couple an electric sector to a magnetic sector to increase the resolution to ~105.

B) Transmission quadrupole mass spectrometers select ions by applying a constant voltage and a radio-frequency oscillating voltage to the four parallel metal rods.

C) Orbitrap mass spectrometers do not require magnetic or radio-frequency fields to measure m/z.

D) Quadrupole mass spectrometers operate at constant resolving power.

E) The time-of-flight mass spectrometer employs a reflectron to group ions with the same m/z, regardless of initial kinetic energy.

Answer: D Intermediate

176 8.

Which statements are true for the various ion selection techniques?

I

Time-of-flight MS – Ions expelled from the source all have the same kinetic energy. For ions of different masses but equivalent kinetic energies, the lighter ions travel faster than heavier ions, separating the ions.

Quadrupole MS – A constant voltage and a radio-frequency oscillating voltage is applied to the four metal rods. The electric field deflects ions in complex trajectories as the ions migrate from source to detector, allowing only one particular m/z to reach the detector.

Three-dimensional quadrupole ion-trap MS – A constant-frequency radio-frequency voltage is applied to the central ring electrode causes ions to circulate in stable, threedimensional orbits in the cavity, with the lowest m/z in the outermost orbits. Increasing the amplitude of the radio-frequency voltage destabilizes m/z ion orbits one value at a time.

Magnetic sector MS – Ions are accelerated into the magnetic sector with equivalent kinetic energies. Ions are selected by passing through a magnetic field; lighter ions are deflected the most and the heavier ions are deflected the least. The magnetic field is varied to select individual m/z.

II

III

IV

A) B) C) D) E)

I, II, and IV I, II, III, and IV II, III, and IV II and IV III and IV

Answer: B Intermediate 22-4

Chromatography-Mass Spectrometry Interfaces

9.

Electrospray ionization is one of many techniques interfacing chromatography with mass spectroscopy. Which is NOT true for electrospray ionization?

A) B) C) D) E)

The steel nebulizer is held at 0 V while the spray chamber is held at −3500 V. The strong electric field between the nebulizer and spray chamber creates a fine aerosol.

The glass capillary leading to the mass spectrometer is held at a potential of −4500 V to attract the gas phase cations.

Typically the ions that vaporize from aerosol droplets were already charged while on the chromatography column.

There is minimal fragmentation, but fragmentation can be increased by collisionally activated dissociation.

Answer: B Intermediate

177 10.

Which is NOT a chromatography-mass spectrometry interface?

A) B) C) D) E)

electrospray direct electron ionization matrix-assisted laser desorption ionization atmospheric pressure chemical ionization photoionization

Answer: C Intermediate 22-5

Chromatography-Mass Spectrometry Techniques

11.

Which is true for using mass spectroscopy as a chromatography detector?

I

Selected ion monitoring observes only one to four m/z to create a selected ion chromatogram.

Extracted ion chromatogram records the entire mass spectrum for the compounds coming off the column, picking one m/z from the collected mass spectra to create the chromatogram.

Reconstructed total ion chromatogram records the total ion current below a certain m/z to generate the chromatogram.

Selected reaction monitoring creates a chromatogram using a triple quadrupole mass spectrometer. Quadrupole 1 selects the precursor ion. In quadrupole 2, the precursor ion is fragmented through collisions with N2 or Ar gas. All fragment ions pass to quadrupole 3, which selects one ion fragment to monitor.

II

III IV

A) B) C) D) E)

I, II, III, and IV II, III, and IV I and IV IV only I, II, and IV

Answer: E Intermediate

178 12.

Which is true for mass spectrometry of proteins?

I

III IV

MALDI and electrospray are major methods for introducing proteins into the gas phase. m - 1.008 The equation n = n+1 is used to determine the number of H+ bound to a protein m n - m n+1 from the mass spectrum.

Once n is known; the molecular mass of the protein is calculated, molecular mass = nmn.

Electron-transfer dissociation is a mass spectrometry technique to sequence proteins.

A) B) C) D) E)

I, II, and IV II and IV III and IV I, III, and IV II and III

II

Answer: A Intermediate 22-6

Open-Air Sampling for Mass Spectrometry

13.

Analytes on the surface of an object can be sampled using an open-air sampling technique by vaporizing and ionizing the analyte directly from the surface under atmospheric conditions. Which of the following statements are true for these techniques?

I II III IV

A) B) C) D) E)

Direct analysis in real time generates excited He which produces protonated water clusters from distilled water. The water clusters protonate the analyte.

Low-temperature plasma generates excited-state species in plasma, which ionizes the analyte.

Desorption electrospray ionization directs micron-sized droplets of analyte-free solvent with electrospray, dissolving the analyte in the droplets.

All techniques dislodge analyte from the surface, collecting the ions for later mass spectrometer analysis.

I and IV II and III I and III II and IV I and II

Answer: B Intermediate

179 22-7

Ion Mobility Spectrometry

14.

Which is NOT true for ion mobility spectrometry?

A) B) C) D)

Ion mobility spectrometer is not a form of mass spectrometry.

Ion mobility spectrometry is gas-phase electrophoresis.

Ions are separated according to their size-to-charge ratio.

Ions are generated when sample is desorbed and passes through a 10 millilcurie 63Ni source.

All ions travel with the same kinetic energy.

E)

Answer: E Intermediate 15.

Ion mobility spectrometry

and

A) B) C) D) E)

operates at ambient pressure; is gas-phase electrophoresis operates at low pressure; is gas-phase electrophoresis operates at ambient pressure; is a form of mass spectrometry is gas-phase electrophoresis; is a form of mass spectrometry operates at ambient pressure; high pressure

.

Answer: A Intermediate Calculated 22-1

What Is Mass Spectrometry?

16.

Calculate the resolving power for a mass spectrometer that successfully separates m/z 125.135 from 125.293.

Answer: 792 Intermediate Plug the values into resolving power = m/m.

17.

Calculate the resolving power for m/z = 283.980 if the width of half height is 0.076 m/z.

Answer: 3737 Intermediate Plug the values into resolving power = m/m1/2.

180 22-2

Oh, Mass Spectrum, Speak to Me!

18.

Calculate the number of rings and double bonds for the chemical formula C25H32N2O3.

Answer: 11 Intermediate Plug the number of carbons, hydrogens, and nitrogens into c – h/2 + n/2 + 1.

19.

Calculate the number of rings and bonds for the chemical formula C23H28N2O5S2.

Answer: 11 Intermediate Plug the number of carbons, hydrogens and nitrogens into c – h/2 + n/2 + 1.

20.

The mass spectrum of an unknown compound has a molecular ion peak at 194 m/z. The proposed chemical formula for the compound is C8H10N4O2. The M+1 peak has a relative abundance of 10.1%. Does the intensity of the M+1 peak support the proposed chemical formula?

Isotopic Abundance Element Factor for M+1 H 0.012 % C 1.08 % N 0.369 % O 0.038 % Answer: The calculated intensity is 10.3%, which is within 10% of 10.1%, or 10.1% ± 1.0%.

Intermediate Multiply each isotopic abundance factor by the number of atoms for each element and take the sum.

Multiple-Choice 23-1

Solvent Extraction

1.

A) B) C) D) E)

is the physical transfer of solute from one phase to another.

Partitioning Extraction Activated transport Sieving Miscibility

Answer: B Easy

181 2.

A solute is to be extracted using 150 mL of solvent. Of the options below, which will extract the maximum amount of solute?

A) B) C) D) E)

extract with 30 mL five times extract with 37.5 mL four times extract with 75 mL twice extract once with the entire volume extract with 50 mL three times

Answer: A Intermediate 3.

The partition coefficient is modified when the pH of the aqueous phase is adjusted or when a chelating agent is added. The partition coefficient is renamed the distribution coefficient.

What is the impact on the extraction of a weak acid into an organic solvent as the pH of the aqueous phase increases?

A) As the pH increases the value of D decreases and the percentage weak acid left in the aqueous phase decreases.

B) As the pH increases the value of D increases and the percentage weak acid left in the aqueous phase increases.

C) As the pH increases, the value of D decreases and the percentage weak acid left in the aqueous phase decreases.

D) As the pH increases the value of D increases and the percentage weak acid left in the aqueous phase decreases.

E) As the pH increases the value of D decreases and the percentage weak acid left in the aqueous phase increases.

Answer: E Hard

182 23-2

What Is Chromatography?

4.

Which form of chromatography is INCORRECTLY defined?

A) adsorption chromatography – solute is adsorbed into solid particles B) partition chromatography – solute equilibrates between the stationary liquid and the mobile phase C) ion-exchange chromatography – solute ions of the opposite charge are attracted to the stationary phase D) molecular exclusion chromatography – solute is separated based on size E) affinity chromatography – solute binds to a molecule specific for the solute Answer: A Intermediate 5.

Which statements are NOT true regarding chromatography?

I II III IV V

The stationary phase is typically a liquid bonded to the inside surface of a capillary or the surface of solid packed in the column.

The mobile phase entering the column is the eluate.

Mobile phase is the solvent (gas or liquid) that moves through the column.

Eluent is the process of passing mobile phase though a column.

Columns are either open tubular or packed.

A) B) C) D) E)

I and V II and IV I, II, and III III and V I, IV, and V

Answer: B Intermediate 23-3

A Plumber’s View of Chromatography

6.

A) B) C) D) E)

is a plot of detector response versus retention time.

Spectrum Electrophoretogram Chromatogram Chromospectrum Elution response

Answer: C Easy

183 7.

Which is NOT true for solute retention?

A)

Retention time, tr, is the elapsed time between injecting the sample to the arrival of a sample component at the detector minus the dwell time between the end of the column and the detector.

Adjusted retention time is tr’ = tr – tm, where tr is the retention time and tm is the time for mobile phase or unretained solutes to pass through the column. t' The relative retention,  = r' 2 , is the ratio for the adjusted retention time for two t r1 components of the sample, where tr2’ > tr1’ so that  > 1. t t The retention factor, k = r m , is the time, tr, required to elute a component minus the tm time, tm, for mobile phase to pass through the column divided by tm.

K The relative retention, , also equals 2 , where K2 and K1 are the partition coefficients K1 for components 2 and 1.

B) C)

D)

E)

Answer: A Intermediate 8.

The relative retention between two solutes, 1 and 2, is  = 1.35. If the adjusted retention time for solute 2 is 293 s, what is adjusted retention time for solute 1?

A) B) C) D) E)

396 s 217 s 189 s 241 s 344 s

Answer: B Intermediate 9.

Which is NOT true when scaling up preparative chromatography?

A) Keep column length constant.

B) Cross-sectional area of a column is inversely proportional to the mass of analyte.

C) Maintain constant linear velocity.

D) Sample volume applied to column is proportional to mass of analyte.

E) If you change column length, mass of sample can be increased in proportion to total length.

Answer: B Intermediate

184 23-4

Efficiency of Separation

10.

The resolution between two peaks is defined as resolution =

t r wav

=

Vr wav

=

0.589t r . w1 / 2av

Which variable is INCORRECTLY defined?

A) B) C) D) E)

tr is the retention time.

Vr is the retention volume. wav is the average peak width measured at the base of each peak. w1/2av is the average peak width measured at the base of each peak, divided by 2. resolution is the separation between two adjacent peaks.

Answer: D Intermediate 11.

The longer an analyte remains on the column, the broader the peak becomes. Peaks widen with time on column due to : A) B) C) D) E)

effusion. diffusion. adsorption. absorption. viscosity.

Answer: B Easy 12.

Which is true for plate height?

I The smaller the plate height, the narrower the peaks.

II The larger the plate height, the better separation between peaks.

III Plate height can be calculated from the column length, L, and the number of theoretical plates, N. IV For fixed number of theoretical plates, N, the longer a solute is on column, the wider the peak will become.

A) B) C) D) E)

I, II, and IV II and IV I, III, and IV II only IV only

Answer: C Intermediate

185 23-5

Why Brands Spread

13.

The van Deemter equation describes plate height in terms of constants A, B, C, and the linear velocity, ux. Which statement is NOT true for the van Deemter equation?

B + Cu x ux A takes into account multiple pathways through the column. The value of A is column specific and independent of linear flow.

B takes into account longitudinal diffusion of the analyte in the mobile phase. The value of B is column specific and the impact of B on the plate height is inversely proportional to the linear velocity.

C takes into account equilibrium time between the stationary and mobile phase. The value of C is column specific and the impact of C on plate height is proportional to the linear velocity.

To minimize plate height, the optimal flow rate is the minima for the plot of B u x + Cu x versus flow rate.

Linear velocity is the flow rate of the mobile phase.

H  A+

A) B)

C)

D) E)

Answer: D Intermediate 14.

Which comparison of open tubular columns to packed columns is INCORRECT?

A) B) C) D) E)

Open tubular columns have higher resolutions than packed columns.

Open tubular columns have increased linear velocity or a longer column or both.

Packed columns have decreased sensitivity.

Packed columns have longer analysis times.

Packed columns have a lower sample capacity.

Answer: E Intermediate

186 15.

2 The observed variance, s obs , for a peak is the sum of the variances from all contributing

2 factors. Which of the following is not a contributing factor to s obs ?

A) B) C) D) E)

2 s column 2 s detector 2 s injection 2 s tubing 2 s source

Answer: E Intermediate Calculated 23-1

Solvent Extraction

16.

Calculate the weak acid, HA, concentration that remains in the aqueous phase when 50.0 mL of 0.184 M HA at pH = 4.75 is extracted with 50 mL of toluene three times. The partition coefficient for the extraction is 5.17, favoring toluene. The Ka for HA is 8.0 x 10−4.

Answer: 72.5% remains or 0.133 M HA remains in the aqueous phase.

Hard Calculate the distribution coefficient, D, from [H+], Ka and K, substitute into the extraction equation and solve for q3. Multiply q3 by the original concentration to calculate concentration remaining HA. 18.

Suppose that the partition coefficient for an amine, B, is K = 3.0 and the acid dissociation constant for BH+ is Ka = 1.0 x 10−9. If 50.0 mL of 0.010 M aqueous amine is extracted with 100.0 mL of solvent two times, what percentage amine remains in aqueous solution if the pH of the aqueous phase is 9.00?

Answer: 32.7% Hard Calculate the distribution coefficient, D, from [H+], Ka and K, substitute into the extraction equation and solve for q2. Multiply q2 by 100 to get percentage.

23-4

Efficiency of Separation

18.

For two separated solutes the  = 1.10 and the k2 = 4.38. How many theoretical plates are required to achieve a resolution of 1.5?

Answer: 6572 theoretical plates are required.

Intermediate

187 Plug the value of , k2, and resolution value into the Purnell equation, solving for N. 19.

Calculate the resolution for two chromatographic peaks that elute at tr1 = 1770 s and tr2 = 1800 s with base peak widths of w1 = 51 s and w2 = 57 s.

Answer: 1.11 Intermediate Calculate tr and wav and plug into equation 23-23.

20.

Calculate the number of theoretical plates and plate height for a peak with a retention time of 1295 seconds, a width at the baseline of 27 seconds and elutes from a 15 m column.

Answer: 36807 plates, 0.408 mm Intermediate Calculate the number of plates, N, from the retention time and the baseline width, equation 23-30.

Then, divide the number of plates into the length of the column to get plate height.

Multiple-Choice 24-1

The Separation Process in Gas Chromatography

1.

As the temperature of a GC column increases, the increases, the retention time and the peak shapes A) B) C) D) E)

of

the

analytes .

vapor pressure; ; increases; ; sharpen solubility; ; decreases; ; sharpen vapor pressure; ; decreases; ; sharpen equilibrium; ; increases; ; broaden temperature; ; increases; ; sharpen

Answer: C Intermediate 2.

Which of the following statements are true for GC columns?

I II

GC columns are capillaries composed of fused silica and coated with a polyimide.

Column inner diameters are typically 0.10 cm to 0.53 cm and the columns are 15 cm to 100 cm in length, with 30 cm the most common.

A wall-coated open tubular column has a liquid stationary phase coating the inside wall of the column.

A porous-layer column is filled with a highly porous stationary phase.

A packed column is a column filled with a solid stationary support coated in liquid stationary phase.

III IV V

A)

I, III, and V

188 B) C) D) E)

II and IV III, IV, and V I and II I, II, III, IV, and V

Answer: A Intermediate 3.

Which statement is NOT true for GC carrier gases?

A) B) C) D) E)

N2, He, and H2 are the most common carrier gases.

Optimal linear velocity increases in the order N2 < He < H2.

The fastest separations can be achieved with N2 as carrier gas.

N2 gives a lower detection limit than He with a flame ionization detector.

Helium is the most common carrier gas and is compatible with most detectors.

Answer: C Intermediate 4.

A) B) C) D) E)

columns are used with

chromatography.

Wall-coated open tubular; gas-solid Porous-layer open tubular; liquid-solid Wall-coated open tubular; liquid-liquid Porous-layer open tubular; liquid-liquid Wall-coated open tubular; liquid-solid

Answer: B Intermediate 5.

The temperature program for a separation starts at a temperature of 50 oC and ramps the temperature up to 270 oC at a rate of 10 oC/minute. Which statement is NOT true for this separation?

A) B) low.

C) D) E)

At 10 oC/minute, a total of 22 minutes is needed to reach 270 oC. Strongly retained solutes will remain at the head of the column while the temperature is Weakly retained solutes will separate and elute early in the separation.

The vapor pressure of strongly retained solutes will increase as temperature increases.

Strongly retained analytes will give broad peaks.

Answer: E Intermediate 6.

The first 3 to 10 meters of a capillary column serves as a guard column and retention gaps. Guard columns:

189

A) improve peak shape by separating volatile solvent from less volatile solutes prior to chromatography.

B) divert a portion of the sample to waste to improve peak shape.

C) accumulate nonvolatile substances to prevent column contamination D) begin separation by retaining less volatile solutes from volatile solutes.

E) increase the number of theoretical plates, reducing plate height.

Answer: C Intermediate

190 7.

The elution order for solutes from a gas chromatography column is loosely in order of:

A) B) C) D) E)

decreasing molar mass. decreasing freezing point. decreasing dielectric constant. increasing boiling points. increasing entropy change.

Answer: C Intermediate 24-2

Sample Injection

8.

Cold trapping and solvent trapping is used with peaks.

A) B) C) D) E)

injection to give

split; sharper on-column; faster eluting splitless; sharper split; faster eluting splitless; faster eluting

Answer: C Intermediate 9.

Under what conditions are split, splitless, and on-column injection used?

I

Split injection is preferred for samples where the analyte of interest constitute > 0.1% of the sample.

On-column injection is used for samples that decompose above their boiling point.

Splitless injection is preferred for introducing small volumes of sample onto the column.

Split injection is preferred for samples with high boiling point solutes in low boiling point solvents.

Splitless injection is preferred for trace analysis of analytes that are less than 0.01% of the sample.

II III IV V

A) B) C) D) E)

I, II, and III II, III, and IV II, IV, and V II, III, and V I, II, and V

Answer: D Intermediate 10.

Which is NOT true for split injection?

191

A) The sample is completely vaporized and mixed in the mixing chamber by a combination of high temperature and brisk carrier flow.

B) At the split point, a small fraction of vapor enters the column, but most passes through the waste vent.

C) The proportion of sample that does not reach the column is the split ratio.

D) Split ratios typically range from 50:1 to 500:1.

E) After injection, the waste vent is closed and flow rate increased.

Answer: E Intermediate 24-3

Detectors

11.

For quantitative analysis with gas chromatography:

A)

the retention time reported by the instrument for a compound is proportional to the quantity of that compound. the area of a peak reported by the instrument for a compound is proportional to the quantity of that compound. the calculated response factor value for a compound is proportional to the quantity of that compound. the peak height reported by the instrument for a compound is inversely proportional to the quantity of that compound. the peak shape reported by the instrument for a compound is proportional to the quantity of that compound.

B) C) D) E)

Answer: B Intermediate 12.

Which statement is NOT true for gas chromatography detectors?

A) Thermal conductivity detectors respond to the thermal conductivity difference between analyte and carrier gas.

B) Flame ionization detectors are used only with inorganic compounds.

C) Thermal conductivity detectors respond to all analytes.

D) Flame ionization detectors generate signal by burning the analyte, generating electrons.

E) For thermal conductivity and flame ionization detectors, detector response is proportional to analyte concentration.

Answer: B Intermediate

192 13.

Which of the following is INCORRECT for quantitative or qualitative analysis with gas chromatography?

A) B) C)

D) E)

The confirmation ion collected by a mass spectrometer is used to quantify a peak.

Selected ion monitoring identifies a peak by monitoring the m/z characteristic for the compound.

Internal standards quantify compounds by adding a known amount of compound to the sample and calculating the concentration from peak areas for the compound and internal standard, and the response factor for the two compounds.

Spiking identifies a peak by a known compound to the unknown. If the peak area increases, the peak has been identified.

Linear response for concentration and detector response is required for quantitative analysis.

Answer: A Intermediate 14.

The thermal conductivity detector functions by measuring the for a hot filament that is the result of a(n) to compensate for the of the filament by the mixture of analyte and carrier gas that elutes from the column.

A) B) C) D) E)

voltage increase; decreased resistance; ; heating resistance change; increased voltage; ; cooling voltage decrease; increased resistance; ; heating resistance change; increased voltage; ; heating voltage increase; increased resistance; ; cooling

Answer: E Intermediate 15.

Which statement is NOT true for flame ionization detectors?

A)

The detector responds to hydrocarbon analytes, which are burned in a mixture of H2 and air after eluting from the column.

During combustion the sample carbon atoms produce CH radicals, which are thought to produce CHO+ ions and electrons in flames.

In the absence of analyte, a current of ~10−14 A flows between the flame tip and the collector. Analytes produce a current of ~10−12 A. Flame ionization detectors respond to non-hydrocarbon compounds, but to a lesser degree than hydrocarbons.

The detector responds proportionally to analyte mass over seven orders of magnitude.

B) C) D) E)

Answer: D Intermediate

193 24-4

Sample Preparation

16.

Which is NOT a sample preparation technique for use with gas chromatography?

A) B) C) D) E)

purge and trap solid-phase microextraction thermal desorption stir-bar sorptive extraction matrix matching

Answer: E Intermediate 17.

To maximize solid-phase microextraction analyte recovery, keep all of the following in mind, except: A) B) C) D) E)

―like dissolves like‖ in selecting fiber. agitate or stir the sample. select thick or porous film for volatiles. add 25% NaCl and adjust pH, reducing pH for acids and increasing pH for bases. purchase solid-phase microextractors from reputable suppliers.

Answer: E Intermediate 24-5

Method Development in Gas Chromatography

18.

For method development, multiple decisions must be made. Order the development decisions below in the proper order.

I II III IV V

detector goal of analysis injection sample preparation column

A) B) C) D) E)

II, III, I, IV, V II, I, V, III, IV II, V, III, I, IV II, IV, I, V, III II, IV, III, I, V

Answer: D Intermediate

194 Calculated 24-3

Detectors

19.

The response factor between an alcohol and an alkane is 2.54. An internal standard with a concentration of 25.9 nM has a peak area of 92832. The peak area for the alcohol is 23422. What is the concentration of alcohol in the sample?

Answer: 2.57 nM Hard Realize that the response factor is multiplied by the ratio of area to concentration of the internal standard. Plug the values into the equation and solve for [X].

20.

To quantify the amount of ethanol in an organic sample, a technician prepares a standard solution containing 0.1473 grams ethanol and 0.1847 grams decane in 1.203 grams dichlormethane. GC analysis returns peak areas of 9,930 for ethanol and 7,893 for decane. Next, 0.9831 grams unknown and 0.2932 grams decane are added to 1.2119 grams dichloromethane.

GC analysis returns peak areas of 6,382 for ethanol and 11,283 for decane. What is the percent by mass ethanol in the sample?

Answer: 10.69% Hard Use the mass ethanol and decane in the known sample as well as the peak areas to calculate the response factor F. Next, use the peak area for ethanol and decane as well as the mass decane added to the solution to calculate the mass ethanol in the unknown. Divide the mass ethanol by the mass unknown and multiply by 100 to get the percent by mass ethanol.

Multiple-Choice 25-1

The Chromatographic Process

1.

As particle size decreases for HPLC, the required solvent pressure the plate number , and the resolution A) B) C) D) E)

increases; increases; decreases decreases; decreases; increases increases; increases; increases increases; decreases; decreases decreases; increases; increases

Answer: C Intermediate 2.

Which of the following are characteristics of reverse-phase chromatography?

, ?

195

I II III IV V

The stationary phase is polar.

The mobile phase is more polar than the stationary phase.

Less polar mobile phase has a lower eluent strength.

More polar mobile phase has a higher eluent strength.

The stationary phase is nonpolar.

A) B) C) D) E)

I and IV II only I and III II and IV II and V

Answer: E Intermediate 3.

Gradient elution is used in HPLC. How is the gradient created for a reverse-phase HPLC experiment?

A) B) C) D) E)

Increase the pH of the mobile phase over time by adding strong base.

Decrease the pH of the mobile phase over time by adding strong acid.

Increase the vapor pressure of the analytes over time by increasing the temperature.

Increase the polarity of the mobile phase over time by mixing with a more polar solvent.

Decrease the polarity of the mobile phase over time by mixing with a less polar solvent.

Answer: E Intermediate

4.

The pressure required to push solvent through the column is dependent on several variables such as column length, particle diameter, etc. For column pressure, which relationship is INCORRECT? A) B) C) D) E)

Pressure is inversely proportional to column length.

Pressure is directly proportional to solvent viscosity.

Pressure is inversely proportional to particle diameter squared.

Pressure is inversely proportional to column radius squared.

Pressure is directly proportional to volume flow rate.

Answer: A Intermediate 5.

Which statement(s) is(are) NOT true for partition and adsorption chromatography?

I

For adsorption chromatography – the lower the eluent strength of the solvent, the more easily it displaces the solute.

196 II III IV

A) B) C) D) E)

For partition chromatography – the more time the solute spends in the mobile phase, the easier it is to elute.

For partition chromatography – solute equilibrates between the mobile phase and the liquid phase bonded to the solid support.

For adsorption chromatography – solvent molecules compete with solute molecules for binding sites on the stationary phase.

IV I III and IV II I and II

Answer: B Intermediate 6.

Elution with a constant composition mobile phase is:

A) B) C) D) E)

constant polarity elution. isopolar elution. isocratic elution. homogeneous elution. true elution.

Answer: C Easy

197 7.

Normal-phase chromatography uses a stationary phase and a mobile phase. The more the mobile phase becomes, the stronger the eluent strength.

A) B) C) D) E)

polar; polar; polar nonpolar; nonpolar; polar polar; nonpolar; nonpolar polar; nonpolar; polar nonpolar; polar; nonpolar

Answer: D Intermediate 8.

HPLC operation requires special solvents, HPLC-grade. Additionally, HPLC-grade solvents require pretreatment before use. Which of the following is NOT a characteristic of HPLC-grade solvents or pretreatment required before operation?

A) B) C) D) E)

HPLC-grade solvent prevents degradation of costly columns with impurities.

HPLC-grade solvent must be sparged prior to use if the instrument is not equipped with a degrasser.

HPLC-grade solvent minimizes background contamination signals in the detector.

HPLC-grade solvent must be filtered to remove particulate matter.

HPLC-grade solvent must be stored in special cabinetry to prevent solvent degradation.

Answer: E Intermediate 25-2

Injection and Detection in HPLC

9.

The creates gradients from up to four solvents by proportioning liquids through a four-way valve before the piston and then pumped into the column.

A) B) C) D) E)

high-pressure mixing pump multiple source mixing pump low-pressure mixing pump gradient mixing pump variable composition mixing pump

Answer: C Intermediate

198 10.

Which of the following is NOT true for HPLC sample injection.

A)

Reproducibly load injector loop with < 0.5 or > 3 loop volumes depending on injection technique.

The injection loops must be removed and cleaned after each injection.

Sample solvent strength should be less than or equal to eluent solvent strength.

Pass samples through a 0.5-m filter to remove particulate matter prior to injection.

Use a blunt HPLC syringe needle, not pointed, to prevent damage to the injection port.

B) C) D) E)

Answer: B Intermediate 11.

Which of the following is NOT a detector used with HPLC?

A) B) C) D) E)

electrochemical detector ultraviolet detector refractive index detector evaporative light-scattering detector thermal conductivity detector

Answer: E Intermediate 12.

The the mobile phase.

A) B) C) D) E)

responds to analytes that are significantly less volatile than

electrochemical detector ultraviolet detector refractive index detector evaporative light-scattering detector thermal conductivity detector

Answer: D Intermediate 13.

The is a sensitive, almost universal detector that responds to all nonvolatile compounds, regardless of their structure.

A) B) C) D) E)

electrochemical detector charged aerosol detector refractive index detector evaporative light-scattering detector thermal conductivity detector

Answer: B Intermediate

199

25-3

Method Development for Isocratic Reversed-Phase Separations

14.

The criteria for an adequate separation for isocratic reverse-phase HPLC include which of the following?

IV

The retention factor for all peaks are in the range of 0.5 – 20.

All peaks should be symmetric with an asymmetry factor B/A in the range of 0.9 – 15.

For quantification, the minimum resolution between the two closest peaks should be 1.5, though 2.0 is best to compensate for small changes in separation conditions.

The pressure must be less than or equal to 15 MPa to prolong instrument life.

A) B) C) D) E)

I, II, III, and IV I, II, and IV I, II, and III II, III, and IV III and IV

I II III

Answer: A Intermediate 15.

During method optimization, the relative retention is adjusted by altering all of the following, except: A) B) C) D) E)

column temperature. type of organic solvent. solvent composition. adjusting sample injection volume. column type.

Answer: D Intermediate

200 25-4

Gradient Separations

16.

When deciding between an isocratic elution or a gradient elution, a scouting gradient should be done. Which of the criteria below is INCORRECT? If t/tG < 0.25, use isocratic elution.

If t/tG > 0.40, use gradient elution.

For 0.25 < t/tG < 0.40 use either an isocratic or gradient elution, the decision depending on factors such as available equipment or complexity of the sample.

D) For t/tG < 0.25, a good starting isocratic elution solvent composition is the composition halfway through the t of the scouting gradient.

E) For 0.25 < t/tG < 0.40, a good starting isocratic elution solvent composition is the composition halfway through the tG of the scouting gradient.

A) B) C)

Answer: E Hard 17.

A is a gradient developed to retain the resolution of a single gradient, but reduces the analysis time.

A) B) C) D) E)

broken gradient segmented gradient choppy gradient step gradient parcel gradient

Answer: B Easy 18. is the volume between the point at which the solvents are mixed and the beginning of the column. Which is NOT true for dwell volume and dwell time?

A) B) C) D) E)

If the dwell time is 5 minutes, a solvent change initiated at 9 minutes does not reach the column until 14 minutes.

A difference in dwell volumes is the main reason why gradient separations do not always transfer well between instruments.

Dwell volume is the volume between where solvents are mixed and the beginning of the column.

The product of the dwell volume and the flow rate is the dwell time.

It is helpful to indicate the dwell volume when reporting a gradient separation.

Answer: D Intermediate Calculated

201 25-1

The Chromatographic Process

19.

What particle diameter is needed to give 15 000 plates over a 10-cm HPLC column?

Answer: 2 m Intermediate Simple plug-and-calculate problem, using equation 25-1.

20.

What is the plate height for a 10.0-cm HPLC column packed with 3.5 m solid phase?

Answer: 1.17 x 10−3 cm or 11.7 m Hard Combine equation 25-1 with H = N/L to calculate the plate height. H = 3000 L2/dp.

Multiple-Choice 26-1

Ion-Exchange Chromatography

1.

What impact does the pH of the mobile phase have on the ability of a strongly acidic exchanger to interact with ions?

A) B) C) D) E)

decreases the ability to interact with ions as pH decreases increases the ability to interact with ions as pH increases has no impact on the ability to interact with ions decreases the ability to interact with ions as pH increases increases the ability to interact with ions as pH decreases

Answer: C Intermediate 2.

X−, Y−, and Z− are separated with an anion exchange column with a mobile phase containing A−. The strength, which the anions bind to the column, is Y− > X− > Z− > A−. To elute the anions, the concentration of A− is , with eluting first, eluting second, and eluting last.

A) B) C) D) E)

decreased; Y−; X−; Z− increased; Z−; X−; Y− decreased; Z−; X−; Y− increased; Y−; X−; Z− increased; X−; Y−; Z−

Answer: B Intermediate 26-2

Ion Chromatography

202

3.

A sample of NaNO3 and K2SO4 is separated on a suppressed-ion anion chromatography column. is retained and separated on the separator column. is used to elute the analyte from the separator column. is removed in the suppressor to reduce and allow for detection of analyte.

A) B) C) D) E)

Na+ and K+ HCl NO3− and SO42− HCl + + Na and K KOH NO3− and SO42− KOH None of the above is correct.

NO3− and SO42− Na+ and K+ NO3− and SO42− Na+ and K+

Background absorbance Background conductivity Background conductivity Background absorbance

Answer: E Intermediate 4.

The commonly used detector with ion chromatography is the:

A) B) C) D) E)

conductivity detector. fluorescence detector. ion detector. gravimetric detector. oxidation detector.

Answer: A Easy 26-3

Molecular Exclusion Chromatography

5.

Which statements are true for molecular exclusion chromatography?

I II III

Molecular exclusion chromatography separates on the basis of size.

Small molecules that freely pass through the stationary phase, Kav = 0.

Small molecules can fit in the pores of the stationary phase and effectively pass through a larger volume, and elute last.

The pores in the stationary phase are too small for large molecules to pass through, so large molecules elute first.

Molecular exclusion chromatography with a hydrophilic stationary phase and an aqueous solvent is called gel permeation chromatography.

IV V

A) B) C) D) E)

I, II, III, and V II and V I, II, and IV I, III, and IV III, IV, and V

Answer: C

203 Intermediate

204 26-4

Affinity Chromatography

6. is used to isolate a single compound or class of compounds from a complex mixture by passing the sample through a column where only one solute is bound. The remaining solutes pass through the column. The bound solute is eluted from the column by changing a condition such as to weaken its binding.

A) B) C) D) E)

Adhesion chromatography; ratio of polar and nonpolar solvents Binding chromatography; pH or ionic strength Affinity chromatography; the polarity of the solvent Adhesion chromatography; the temperature of the column Affinity chromatography; pH or ionic strength

Answer: E Intermediate 26-5

Hydrophobic Interaction Chromatography

7.

Which statement is NOT true for hydrophobic interaction chromatography?

A)

Hydrophobic interaction chromatography has a hydrophobic stationary phase, which repels water and is not wetted by water. The technique is used primarily for protein purification.

Proteins are salted out of solution with ammonium, sodium, and potassium salts of phosphate and sulfate. Thiocyanate, iodide, and perchlorate salts increase the solubility of proteins in water.

The proteins are eluted from the column with a gradient of increasing salt concentration, increasing the solubility of proteins in water.

The most common stationary phase is an agarose gel with hydrophobic phenyl or alkyl groups attached and a pore size large enough to accommodate proteins.

A protein sample with a large concentration of ammonium sulfate is applied to the column. The ammonium sulfate causes the protein to stick to the hydrophobic surface of the stationary phase.

B)

C) D) E)

Answer: C Intermediate

205 26-6

Principles of Capillary Electrophoresis

8.

Under what conditions will anions be detected at the cathode during capillary electrophoresis?

A) B) C) D) E)

eo > ep eo < ep eo = ep eo + ep > 1 eo − ep < 1

Answer: A Intermediate 9.

The normal order for elution in capillary zone electrophoresis is:

A) B) C) D) E)

anions (highest mobility last), all neutrals (unseparated), cations (highest mobility first). anions (lowest mobility first), all neutrals (unseparated), cations (lowest mobility first). cations (highest mobility first), all neutrals (unseparated), anions (highest mobility last). cations (lowest mobility first), all neutrals (unseparated), anions (lowest mobility last). cations (highest mobility first), neutrals (lowest mobility first), anions (highest mobility first).

Answer: C Intermediate 10. occurs when an electric field is applied to a capillary that has an electric double layer at the negatively charged surface inside the capillary.

A) B) C) D) E)

Electropsmotic flow Electrophoretic flow Electrodynamic flow Electrodiffusion flow Electroeffusion flow

Answer: A Intermediate

206 11.

Which statements are true for electroosmotic velocity?

I Electroosmotic velocity, ueo, is calculated experimentally using the retention time of a neutral molecule and the length of the column from the injector to the detector.

II Electroosmotic velocity, ueo, is the product of the electroosmotic mobility, eo, and the electric field, E. III Electroosmotic velocity, ueo, is inversely proportional to electric field strength, E. IV Electroosmotic velocity is equal to the electroosmotic flow.

A) B) C) D) E)

II and IV II, III, and IV I, III, and IV I and II I, II, and IV

Answer: E Intermediate 12.

Which is NOT true for apparent, electroosmotic, and electrophoretic mobilities?

A)

app = eo + ep u L /t mapp = net = d r E V / Lt u L /t meo = neutral = d neutral E V / Lt u mep = ep = app + eo E uapp = (eo + ep)E

B) C) D) E)

Answer: D Hard 13.

The number of plates, N, for a capillary electrophoresis is defined to be N =

Which statement is NOT true?

A) B) C) D) E)

As the apparent mobility, app, increases, the number of plates increases.

As the total length of the capillary, Lt, increases, the number of plates increases.

As the diffusion coefficient, D, increases, the number of plates decreases.

As the distance to the detector, Ld, increases, the number of plates increases.

As the voltage, V, decreases, the number of plates decreases.

Answer: B Intermediate

mappV Ld .

2D Lt

207 26-7

Conducting Capillary Electrophoresis

14.

Capillary columns require conditioning prior to their first use. Typical conditioning is washing with 1 M NaOH for 1 hour, followed by 1 hour of water, then 1 hour of 6 M HCl, and finally 1 hour of run buffer. Which statements are INCORRECT? I II III IV

The NaOH wash is thought to generate the Si-OH groups on the silica surface.

The NaOH wash removes any acidic contaminants from the silica surface.

The HCl wash is thought to generate the Si-OH groups on the silica surface.

The HCl wash removes metal ions from the capillary surface.

A) B) C) D) E)

I and III I and IV II and III II and IV I, II, and III

Answer: C Intermediate 15.

Which statement regarding capillary electrophoresis sample injection is INCORRECT?

A)

Hydrodynamic injection uses a pressure differential between the two ends of the capillary to inject a volume of sample onto the capillary.

Electrokinetic injection uses an electric field to inject a volume of sample onto the capillary.

For electrokinetic injection, each analyte has a different app, so the composition of the loaded sample is different from the original sample.

For quantitative work, an internal standard is essential because the amount of sample injected is not reproducible between runs.

Stacking concentrates the sample cations and anions in the sample at the interfaces between the sample plug and run buffer.

B) C) D) E)

Answer: B Intermediate

208 16.

Which is NOT a detector used with capillary electrophoresis?

A) B) C) D) E)

ultraviolet absorption conductivity electrospray mass spectrometry laser induced fluorescence evaporative light scattering

Answer: E Intermediate 26-8

Lab-on-a-Chip: DNA Profiling

17. are instruments constructed on plastic or glass chips which use electroosmotic flow or pressure to move liquids through micrometer channels from one reaction site to another.

A) B) C) D) E)

Microosmotic devices Microcapillary devices Microelectrophoric devices Microfluidic devices Microscale devices

Answer: C Easy Calculated 26-6

Principles of Capillary Electrophoresis

18.

The retention time for an analyte separated via capillary electrophoresis is 115 s. The separation occurred with a 50-cm column, with the detector at 45 cm, and a potential of 30 kV. A neutral molecule took 200 s to reach the detector. What is the electrophoretic mobility for the analyte?

Answer: ep = 2.77 × 10−8 m2/Vs Hard

u L /t Calculate uapp and divide by E, or calculate app with mapp = net = d r . Next, calculate eo with E V / Lt u L /t meo = neutral = d neutral . ep = app − eo.

E V / Lt

209 19.

Sodium cation was separated from a mixture of cations via capillary electrophoresis, with negative detection. The instrument is equipped with a capillary 50 cm in length, with 45 cm to the detector. The separation took place under 30 kV with the neutral marker reaching the detector after 200 seconds. The retention time for Na+ is 85 seconds. Calculate the number of theoretical plates for the Na+ peak. The diffusion coefficient for Na+ in water is 1.0 × 10−9 m2/s.

Answer: 1.19 × 106 plates Hard Calculate app, then plug all values into N = ((appV)/2D)(Ld/Lt) 20.

Capillary electrophoresis separates two analytes with retention times of 112 s and 113 s with a 60 cm capillary, with the detector 50 cm from the inlet. The voltage applied across the capillary is 20 kV. Assuming that the peaks eluted close enough that N is essentially constant, what is the resolution for the two peaks if N = 500 000 plates?

Answer: resolution = 1.57 Hard First, determine the apparent speed for each analyte and then use the electric field to calculate the apparent mobility. Calculate the difference in mobilities, as well as the average mobility. Plug the mobility difference, average mobility, and the number of plates into the resolution equation to solve for the resolution.

Multiple-Choice 27-1

An Example f Gravimetric Analysis

1.

A 100.00 mL solution of copper(II) chloride is reacted with excess silver nitrate to precipitate 8.9260 g silver chloride. Which of the calculations below is NOT required to calculate the molarity of the copper(II) solution?

A) B) C) D) E)

calculate the molar mass of silver chloride calculate the molar mass of copper(II) chloride calculate moles silver chloride from precipitate mass calculate the moles copper(II) chloride with the mole to mole ratios silver chloride : chloride and chloride : copper(II) chloride divide moles copper(II) chloride by volume of the copper(II) chloride solution expressed in liters

Answer: B Easy 27-2

Precipitation

2.

Which of the properties below is NOT a property of the ideal solid product for gravimetric analysis?

210 A) B) C) D) E)

insoluble known composition pure easily filterable high density

Answer: E Difficulty: Easy

211 3.

Which of the following statements is(are) true for precipitate particle size?

I II III

Particle size must be large enough to not clog or pass through a filter.

Particle size is controlled by precipitation conditions.

Thermodynamics favors larger crystals because the surface atoms have higher energy than internal atoms and the fraction of atoms on the surface is smaller for larger crystals than smaller crystals.

Ostwald ripening is the process where smaller particles re-dissolve and larger crystals grow.

IV

A) B) C) D) E)

III only I, II, and IV II and IV I, II, III, and IV II, III, and IV

Answer: D Intermediate 4.

Crystallization occurs when precipitant is added to a solution, creating a temporary supersaturated solution. Crystallization occurs in two stages: nucleation and particle growth.

Which of the following is NOT true for nucleation and particle growth?

I Nucleation occurs when solutes form clusters of sufficient size which reorganize into a more ordered structure capable of growing into larger particles.

II Particle growth occurs when ions, molecules, or other nuclei condense onto a particle to form a larger crystal.

III Sources of nucleation are solution impurities or small scratches on a glass surface.

IV Particle growth occurs when amorphous particles clump together to create larger amorphous particles.

A) B) C) D) E)

I III and IV IV I, II, and IV II and IV

Answer: C Intermediate

212 5.

A) B) C) D) E)

are solid particles that lack any crystalline order.

Crystalline particles Amorphous particles Plastic particles Irregular particles Elastic particles

Answer: B Easy 6.

A) B) C) D) E)

are solid particles with long-range order.

Crystalline particles Amorphous particles Rigid particles Regular particles Inelastic particles

Answer: A Easy 7.

Which of the following are techniques to promote particle growth?

I II

Use a large volume of solution so concentration of analyte and precipitant are low.

Rapidly add the precipitant to precipitate all analyte from solution, then raise the temperature of the solution to promote crystal growth.

Slowly adding precipitant with vigorous mixing as the precipitant is added to prevent a highly supersaturated condition in solution.

Raise the temperature of the solution to increase solubility of precipitate and decrease supersaturation.

III IV

A) B) C) D) E)

III and IV II, III, and IV II and IV I, II, and III I, III, and IV

Answer: E Intermediate

213 8.

Homogeneous precipitation occurs when a precipitant is generated slowly from within an initially homogeneous solution by a chemical reaction. The main benefit from homogeneous precipitation is: A) nucleation dominates over particle grown when precipitation is slow to give smaller, purer, easily filtered crystals.

B) particle grown dominates over nucleation when precipitation is slow to give smaller, purer, easily filtered crystals.

C) nucleation dominates over particle grown when precipitation is slow to give larger, purer, easily filtered crystals.

D) particle growth dominates over nucleation when precipitation is slow to give larger, purer, easily filtered crystals.

E) None of these answers is correct.

Answer: D Intermediate 9.

Precipitates will often contain impurities and will require treatment to remove impurities.

Which of the following precipitation terms is INCORRECTLY defined?

A) B) C) D) E)

gathering – precipitate collects a trace element of the solution masking agent – compound that prevents impurities from reacting with the precipitate inclusion – pocket of impurities trapped within the precipitate digestion – precipitates remains in mother liquor after precipitation to promote slow recrystallization of the product, form larger particles and expel impurities coprecipitation – impurities are precipitated along with the desired product

Answer: C Intermediate 10.

The final precipitation product should have a known, stable chemical composition. One way to obtain a known, stable chemical composition is via , which is defined as the strong heating of a precipitate to change the chemical composition of the precipitate to a more stable form.

A) B) C) D) E)

drying ignition charring thermogravimetric analysis annealing

Answer: B Intermediate

214 27-3

Examples of Gravimetric Calculations

11.

Silver cation is precipitated from solution using sodium carbonate. Calculate the gravimetric factor to relate the mass of the product to the analyte.

A) B) C) D) E)

0.782 1.278 0.391 5.112 0.196

Answer: A Intermediate 27-4

Combustion Analysis

12.

Which of the following is NOT true for traditional combustion analysis of a hydrocarbon?

A) During combustion all carbon is converted to carbon dioxide.

B) During combustion all hydrogen is converted to water.

C) The increase in mass for the trap containing P4O10 is proportional to the mass H in the compound.

D) The increase in mass for the Ascarite II® trap is proportional to the mass C in the compound.

E) During combustion all carbon is converted to carbon monoxide.

Answer: E Intermediate 13.

Modern combustion analysis uses dynamic flash combustion. What is the main reason dynamic flash combustion is employed?

A) B) C) D) E)

to limit the time oxygen flows freely through the analyzer to leave no residue in the ceramic crucible to guarantee the complete oxidation of select reaction products because the GC requires the sample to be injected at once to guarantee the complete reduction of select reaction products

Answer: D Intermediate

215 14.

Combustion analysis of C, H, and S converts each element into nonmetal oxide for analysis. Combustion analysis of O requires a different approach. Which of the steps below is(are) not required or false for combustion analysis for O? I II III IV V

A) B) C) D) E)

Sample is thermally heated to decompose the compound in the absence of O2.

Gaseous products from pyrolysis are passed through nickelized carbon at 1 075 oC to convert oxygen into CO. Gaseous products are sent to GC for analysis.

Halogens are converted to HX during pyrolysis, dissolved in water and titrated with silver nitrate.

Gaseous products from pyrolysis are passed through nickelized carbon at 1 075 oC to convert oxygen to CO2.

III, IV, and V II and IV IV and V I, II, and III V only

Answer: C Intermediate 15.

Modern combustion analysis uses gas chromatography to measure the amounts of product gas released. Which detector is used with the GC? A) B) C) D) E)

mass spectrometry thermal conductivity flame ionization electron Ccapture UV-vis

Answer: B Intermediate Calculated 27-3

Examples of Gravimetric Calculations

16.

A 100.0 mL sample of a solution containing copper(II) cation is treated with excess sodium carbonate to precipitate 3.6055 grams of copper(II) carbonate. What is the molarity of the solution?

Answer: 0.2918 M Cu2+ Intermediate Determine the moles copper precipitated from stoichiometry and then divide by the volume of the solution to obtain molarity.

216 17.

The iron content of an iron supplement must be determined. The manufacturer states that each tablet contains 1 000.0 mg of bioavailable iron. A tablet with a mass of 4.9283 grams is digested in nitric acid to release Fe3+ into solution. After filtration, the iron in solution will be precipitate with sodium bicarbonate. What is the minimum mass sodium bicarbonate must be added to precipitate the iron in solution? Assume no sources of interfering metals are present.

Answer: 4.2775 g Fe(HCO3)3 required to precipitate 1 000.0 mg Fe from iron supplement.

Intermediate Calculate the moles iron, moles iron(III) bicarbonate, and then mass iron(III) bicarbonate from the 1 gram Fe stated in the problem.

18.

A solution is prepared from an unknown solid containing Ni and Cu by dissolving 5.2832 grams of the solid in nitric acid to release Ni2+ and Cu2+. The nitric acid is neutralized and the solution diluted to 200.0 mL. Next excess sodium carbonate is added to the solution to form Ni and Cu carbonates with a total mass of 2.5305 grams. The precipitate is then ignited to form NiO and CuO with a total mass of 1.6126 grams. What is the percent by mass Ni and Cu in the solid sample?

Answer: 11.14% Ni, 13.07% Cu Hard Set up two equations, x + y = 1.6126 g for the metal oxides and (118.9/74.89)x + (123.556/79.546)y = 2.5305. Solve for x and y to get the mass of each metal oxide. Convert metal oxide to mass metal and divide by sample mass × 100 to get the percentage.

27-4

Combustion Analysis

19.

Combustion analysis of 0.1938 grams of a hydrocarbon increased the mass of the water trap by 0.2750 grams and the mass of the carbon dioxide trap by 0.5973 grams. What is the percent by mass for C and H? Answer: 15.88% H and 84.12% C Hard Convert the masses of water and carbon dioxide to moles, then moles water to moles H, moles carbon dioxide to moles C. Next, convert moles H to grams H, and moles C to grams C. Divide mass each element by mass compound combusted to get percentage.

20.

Combustion analysis of 0.1938 grams of a hydrocarbon increased the mass of the water trap by 0.2750 grams and the mass of the carbon dioxide trap by 0.5973 grams. What is the empirical formula for the hydrocarbon?

Answer: C4H9 Hard (especially if student does not remember their general chemistry) Convert the masses of water and carbon dioxide to moles, then moles water to moles H, and moles carbon dioxide to moles C. Divide by smallest moles, then multiply the 2.25 on H and the 1 on C by 4 to get C4H9.

Multiple-Choice

217

1.

Which samplng term is incorrectly defined?

laboratory sample – smaller, homogeneous sample taken from the bulk and has the same composition as the bulk B) lot – the total material from which samples are taken C) aliquot – small portions of the bulk sample taken for individual analysis D) sample preparation – the series of steps that convert a representative bulk sample into a form suitable for analysis E) bulk sample – representative sample taken from the lot for analysis or archiving A)

Answer: C Easy 2.

Sample storage is an important aspect of sample preparation. Which is NOT true for sample storage?

A) B) C) D) E)

Glass should not be used with trace analysis of ions and proteins in solution.

Analytes may react with the container.

A stored sample may change with time.

New plastic containers do not require washing before use.

A stored sample may react with air.

Answer: D Easy 28-1

Statistics of Sampling

3.

For random errors, the overall variance, so2 , is calculated from the analytical variance, sa2 ,

and the sampling variance, ss2 . Which is the correct equation for calculating the overall variance?

A) B) C) D) E)

so2 = sa2 - ss2 so2 = ss2 - sa2 s2 so2 = s2 sa

so2 = ss2 * sa2 so2 = sa2 + ss2

Answer: E Intermediate

218 4.

A mixture contains 5% NaCl particles and 95% NaBr particles. If 103 particles are taken, what is the relative standard deviation for NaCl and NaBr? A) B) C) D) E)

13.78% for NaCl 186.2% for NaCl 95.00% for NaCl 54.39% for NaCl 6.89% for NaCl

0.73% for NaBr 9.80% for NaBr 5.00% for NaBr 2.86% for NaBr 0.36% for NaBr

Answer: A Intermediate 5.

A 4.17 g sample has a 4% sampling standard deviation. What mass sample must be taken to decrease the sampling standard deviation to 2%?

A) B) C) D) E)

8.34 g 16.68 g 10.43 g 4.17 g 15.64 g

Answer: B Intermediate 6.

To calculate the number of required replicates for analysis the equation n =

t 2 ss2 is used. e2

Which statement is NOT true for the equation and calculation?

A) B) C) D) E)

t = the Student’s t value for n − 1. ss is the sampling standard deviation. e is the difference between the sampling standard deviation and the desired sampling standard deviation.

The initial calculation assumes an infinite n value and the according Student’s t value.

The n value from a calculation dictates the Student’s t value used in the next calculation.

Answer: C Intermediate

219 28-2

Dissolving Samples for Analysis

7.

Sample preparation of a laboratory sample can involve all of the following, except:

A)

laboratory samples that do not dissolve under mild conditions may require acid digestion or fusion. organic material may be destroyed by combustion or wet ashing. solids are typically dried at 110 oC at atmospheric pressure to remove absorbed water prior to analysis. a portion of a course solid laboratory sample should be ground into a fine powder for analysis. temperature-sensitive samples may be stored in an environment to achieve a constant, reproducible moisture level.

B) C) D) E)

Answer: D Easy 8.

All of the following are used to grind solids into a fine powder, except:

A) B) C) D) E)

mortar and pestle.

Wig-L-Bug.

Shatterbox laboratory mill. ball mill. vortexer.

Answer: E Easy 9.

All of the following are true for dissolving inorganic materials in acid, except:

A) metals with positive reduction potentials should dissolve in nonoxidizing acids such as HCl, HBr, HF, H3PO4, etc.

B) volatile species formed by protonation of anions such as carbonate will be lost from hot acids in open vessels.

C) volatile metal halides such as SnCl4 and some metal oxides can be lost from hot acids in open vessels.

D) hot hydrofluoric acid dissolves silicates.

E) substances that do not dissolve in nonoxidizing acids may dissolve in oxidizing acids such as HNO3, hot concentrated H2SO4, or hot concentrated HClO4.

Answer: A Hard

220 10. inorganic A) B) C) D) E)

dissolves substances that will not dissolve in acid in a hot, molten .

Smelting; solvent Fusion; flux Smelting; flux Fusion; solvent None of these choices is correct.

Answer: B Easy 11.

Which of the following is INCORRECT for decomposition of organic substances?

A) Microwave-induced combustion is a form of dry ashing.

B) An example of wet ashing is microwave digestion with acid in a Teflon bomb.

C) Wet ashing with perchloric aicd is an extremely hazardous procedure and is best avoided.

D) Kjeldahl digestion digests organic material with sulfuric acid for nitrogen analysis.

E) The Carius method of wet ashing is performed with fuming H2SO4 in a sealed, heavywalled glass tube at 200 oC to 300 oC. Answer: E Intermediate 28-3

Sample Preparation Techniques

12.

Which of the following are NOT liquid extraction techniques?

I II III IV V

emulsive liquid-liquid microextraction dispersive liquid-liquid microextraction solid-supported liquid-liquid extraction supercritical fluid extraction solid-liquid microextraction

A) B) C) D) E)

I and II III and V I and IV IV only I and III

Answer: C Intermediate

221 13.

QuEChERS is a sample preparation procedure involving extraction and sample cleanup.

What does QuEChERS stand for?

A) B) C) D) E)

Quite Easy Cheap Effective Rugged and Safe Quiet Easy Chelating Effect Routine and Save Quite Effective Chelating Effect Running and Safe Quick Easy Cheap Effective Rugged and Safe Quiet Efficient Chopping Effect Runs and Save

Answer: D Hard 14. is the procedure in which analyte is chemically modified to make the compound easier to detect or separate.

A) B) C) D) E)

Labeling Derivatization Tagging Marking Modification

Answer: B Easy 15. uses a small volume of chromatographic stationary phase or molecularly imprinted polymer to isolate analytes from the sample matrix to simply analysis.

A) B) C) D) E)

Solid-phase extraction Polymer extraction Chromatographic extraction Antigen extraction Solvent-free extraction

Answer: A Easy Calculated

222 28-1

Statistics of Sampling

16.

A mixture contains 2.5% KBr and 97.5% inert particles. What is the relative standard deviation for KBr is 50 000 particles are taken?

Answer: 2.79% for KBr Easy Calculate the number of particles expected to be in 50 000 particles, 1 250 KBr. The standard deviation should be calculated to be 34.91, and then divided by 1250 and converted to a percentage.

17.

What mass of a solid sample containing 2.5% analyte particles and 97.5% solvent particles must be taken to give a standard deviation of 2% of the number of analyte particles, if the average particle volume is 2.54 nL and the sample has an average density of 1.548 g/mL? Answer: 0.3834 g sample Hard Begin by solving 0.02np = sqrt(npq) where p = 0.025 and q = 0.975, 0.0004pn2 – pqn = 0. The equation is a quadratic with c = 0, (y = mx +b). The value of n = 97,500. Multiply by the average volume per particle, then convert nL to L, L to mL and multiply by the density to get the mass in grams.

18.

What mass will give a sampling standard deviation of 3.5% for a sampling constant of 40.9?

Answer: 3.34 g Easy Calculate the mass, m, from R/(ss)2 where R is the sampling constant and ss is the sampling standard deviation expressed as a percentage.

19.

How many replicates must be completed for sampling deviation of 3.5% to achieve a 1% standard deviation?

Answer: 50 Hard Assume n = infinite, use Student’s t = 1.960, s = 0.035, and e = 0.01 to obtain n = 47. Then repeats the calculation using the Student’s t value of n − 1 = 46, to calculate n = 50. Repeating the calculation for 49 degrees of freedom yields the same value n.

20. Calculate the overall standard deviation if the analytical standard deviation is 6% and the sampling standard deviation is 4%.

Answer: 7.2% Intermediate Calculate the answer by using so2 = sa2 + ss2 and solving for so.