TEST BANK FOR Analytical Chemistry and Quantitative Analysis 1 EDITION. David Hage James Carr

Analytical Chemistry and Quantitative Analysis 1e David Hage James Carr (Test Bank All Chapters, 100% Original Verified, A+ Grade) Answers At The End Of Each Chapter Chapter 1: An Overview of Analytical Chemistry 1. The level of Bisphenol A (BPA; a hormone disruptor) in water stored in some plastic drinking bottles has been disputed. If you were going to design a study involving collected urine from animals to examine this dispute, how would you state the specific question to be investigated?

(A) What is the problem to be analyzed in this investigation?

(B) Polycarbonate bottles (#7 recycling code) could be part of the investigation. Describe one more aspect, besides their #7 classification, that you would want to know about the polycarbonate bottles used in an investigation?

(C) For this analysis, identify the sample (D) For this analysis, identify the analyte (E) For this analysis, identify the matrix (F) If the amounts of BPA levels were in the 10.6 ng/mL range, would that fall into the major, minor, or trace levels? (You must prove your choice by converting the level to a percentage that will classify it.)

2. The level of Bisphenol A (BPA; a hormone disruptor) in water stored in some plastic drinking bottles has been disputed.

(A) Identify one qualitative and one quantitative observation that would likely be made during the study.

(B)The following partial list provides terms that relate to common questions addressed in analytical chemistry. Which could apply to the narrative in question #1?

Chemical Identification; Structural Analysis; Property Characterization; Spatial Analysis; Time-dependent Analysis.

3. Some studies have been done to investigate the possible relationship between autism and vaccinations. Using the following paragraph as a summary, answer the following questions.

There has been some public concern about the safety of childhood vaccines. Some parents have questioned the possible links between vaccinations and symptoms of autism and other disorders. A mercury-containing compound called thimerosal was previously used as a preservative in some vaccines.

(A) Use the summary to identify a chemical problem to be analyzed regarding vaccines.

(B) Of course vaccines are important for public safety. However, side effects do exist. Suppose a certain vaccine study was tested on 7 500 people. However, after approval and use of 15 million doses, it was found that a small increase in a certain disease was detected. Was this likely an error in sample preparation or sampling? Briefly explain your answer.

(C) Currently children receive about 38 shots against 15 diseases before kindergarten. (U.S. News, Feb 09) The incidences of many of those diseases, such as polio and mumps, have dropped to very low levels, however, there are also reports of 30 000 “adverse events” or significant side effects. Of the following, which is the conclusion we could accurately report? (i) vaccines are unsafe and should never be used (ii) vaccines are not properly administered (iii) the number and frequency of vaccines should have further study (iv) vaccines are always safe (v) 30 000 out of 30 000 000 (= 1 000 ppm) vaccinations represents 10% of the children being affected

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Hage/Carr Analytical Chemistry & Quantitative Analysis C. McLaughlin

4. (A) For the investigation described in question #3 identify the Analyte, Sample, and Matrix.

(B) Using the figure shown here, indicate three areas where analytical chemistry is used in the narrative for question #3.

5. Many foods and beverages contain various levels of high fructose corn syrup (HFCS). There

is controversy about HFSC in foods for several reasons. After reading the following recent summary (Science Daily, Aug 23, 2007), briefly describe how the five steps of a traditional chemical analysis were used.

Although most carbonated soft-drinks contain HFCS, portions of eleven brands containing HFCS were studied to determine the presence of a group of reactive compounds in the sodas that are also found in the blood of persons with diabetes.

The study, testing with high-performance liquid chromatography, used solvents to extract reactive compounds from the sodas, found levels of the reactive compounds (they damage healthy tissue) five times higher in the carbonated sodas than in the blood of persons with diabetes. Cane or beet derived sugar (sucrose) did not exhibit this type of activity.

Identify the Problem: Select the Sample: Prepare the Sample: Conduct the Analysis: Analyze the Data: 6. (A) Describe one qualitative and one quantitative observation mentioned in the narrative in question #5.

(B) Of the five instrumental methods of analytical techniques listed in the following table, which are used in the techniques mentioned in question #6?

7. Trout fishing in states with many natural streams is a major recreational industry. However, in 2004 Montana had to kill around 800 000 hatchery-raised trout due to contamination found in fish. The main chemical concern turned out to be polychlorinated biphenols (PCBs). A biological technique (called ELISA) and a chemical technique (less expensive) using gas 2

Test Bank

Hage/Carr Analytical Chemistry & Quantitative Analysis C. McLaughlin

chromatography have been developed and can be applied to PCB determinations. Fat tissue from the fish was treated with a compound used to absorb PCBs for testing. The source of PCB contamination was found to be in the paint in the hatchery holding cells. Humans eating contaminated fish could encounter health problems when the contamination levels are in the 2 ppb range. Briefly provide the information for the following five components of the analysis: Identify the Problem: Select the Sample: Prepare the Sample: Conduct the Analysis: Analyze the Data:

8. (A) Are the levels of PCBs discussed above major, minor or trace components?

(B) List one qualitative and one quantitative observation about the analysis described in question #7.

(C) Describe how the analysis in question #7 could have important “time-dependent analysis” concerns.

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Hage/Carr Analytical Chemistry & Quantitative Analysis C. McLaughlin

SOLUTIONS FOR CHAPTER ONE TEST BANK: 1. (A) (i) The problem to be analyzed could be stated as: “Does BPA leach out of certain polycarbonate bottles?”

(B) Other questions about the bottles could be their history. Were the bottles all maintained at the same temperature? Were they all the same age? Other questions could include the treatment of the bottles during their use in the study.

(C) The sample being analyzed was urine.

(D)The analyte was any BPA compound.

(E)The matrix could have been any other compounds found in the urine.

(F) 10.6 ng/mL is a trace amount of an analyte. 10.6 ng x 1 g/1x109 ng = 1.06 x 10-8 g. When this is divided by 1 gram and multiplied by 100 = 1.06 x 10-6% 2. (A) Qualitative Observation: BPA was detected in the urine samples.

Quantitative Observation: 10.6 ng of BPA detected.

(B) Detection of BPA depended on some chemical properties of the compound as BPA reacted with several compounds. Time-dependent analysis could be important because time may allow the BPA to accumulate in a system.

3. (A) The key problem in this investigation was to determine if mercury levels in the preservative could be linked to disease in people receiving vaccinations.

(B) Although sample preparation could have been a problem, the more likely concern would be using a sample of 7 500 when the use was going to be so widespread. (7 500/15 000 000 = 0.05% of the total) When using a relatively small representation, careful selection of that group must be statistically valid.

(C) iii 4. (A) Analyte: thimerisol; Sample: vaccine; Matrix: compounds in the vaccine (B) Three areas used in the analysis were: Product quality control, Clinical chemistry, Pharmaceuticals.

5. Identify the Problem: Do the levels of a particular compound found in diabetics appear in beverages containing HFCS? Select the Sample: Eleven soft drink brands that contain HFCS. Prepare the Sample: Solvents extracted HFCS from the soft drinks.

Conduct the Analysis: High-performance liquid chromatography on extracts taken from beverages containing HFCS. Analyze the Data: Data revealed that the level of a compound found in diabetics was present at a level five times that in beverages sweetened with cane sugar.

6. (A) Qualitative Observation: The reactive compound was identified in soft drinks that were sweetened with HFCS. Quantitative Observation: The level of concentration in those soft drinks was five times that detected in other, cane-sweetened, soft drinks.

(B) Separation and liquid chromatography were used in this analysis.

7. Identify the Problem: What was the source of PCB compounds entering the systems of fish?

Select the Sample: The samples to be analyzed came from the fat tissue of the fish.

Prepare the Sample: A PCB- absorbing compound was used to extract the PCB for analysis.

Conduct the Analysis: ELIZA and gas chromatography were used to conduct the analysis.

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Hage/Carr Analytical Chemistry & Quantitative Analysis C. McLaughlin

Analyze the Data: It was found that the holding tank area was painted with contaminated paint.

8. (A) Qualitative Observation: PCB was being detected in some Montana fish.

Quantitative Observation: The level of PCB in some fish was found to be in the 2 ppb range.

(B) 2 ppb = 2 grams of PCB per billion grams of fish tissue. This is 2/1x109 x 100 = 2 x10-7 % This is in the trace level.

(C) The longer time the fish were held in the contaminated tanks of the fishery, the more PCB could likely show up in their tissue.

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 2: Good Laboratory Practices 1. (A) Cite one example of a standard operating procedure (S.O.P.) for operating a buret in analytical chemistry.

(B) The following NFPA warning sign was attached to a bottle of a chemical. Name two safety concerns you would have about the chemical inside the bottle. (i) (ii)

(C) What do the letters M.S.D.S. represent? What type of information about a compound is available on an M.S.D.S.?

2. (A) The National Fire Association labels universally allow safety awareness on the hazards of chemical reagents. The following label was found on a reagent bottle. What would be your assessment of the handling/storage concerns? (circle one choice)

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(i) The material is a significant high fire hazard. (ii) The material is an explosive hazard; avoid contact with water. (iii) The material is an extreme health hazard; avoid contact with water. (iv) The material is corrosive; avoid contact with water. (v) Both (i) and (iii) are true.

(B) If the top number, in the red diamond, was zero instead of “1,” how would that affect

your safety concerns?

(C) Working in a lab you glance under the exhaust hood and notice that a large bottle of liquid does not have a lid covering the opening! If the Chemical Hazard label shown here was on the opened bottle what would be your major safety concern or worry about getting the lid back on the bottle?

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3. The recommended procedures for utilizing a proper notebook in academic, research or industrial lab settings include several key instructions.

(A) A student working on a titration analysis realizes, after leaving the lab and beginning to calculate the volume of a solution delivered through a buret, that they have reversed their notations on the initial and final volume reading on the buret. How should they handle this error in their notebook?

(B) Identifying counterfeit and genuine pharmaceuticals is challenging. One method currently in use is to identify tiny pollen grains in the pills. (All pills will have some embedded pollen grains.

The region of manufacture can be identified by the type of grain.) A pharmaceutical chemist records, in their notebook, a pollen grain sample as being about 20m wide. The chemist however, needs to convert this width to millimeters. What is the width of that grain?

4. Converting measurements into convenient units is part of many scientific processes. Make the following conversions (A) Beyond the common 20 amino acids, some biochemists work with selenocysteine. Their work may involve concentrations of 2.5 x 10-5 M. Express this same concentration as M. (B) If a given vaccination injection has a volume of 3 mL, how many deciliters would have been administered? (i) 3 dL (ii) 0.3 dL (iii) 30 dL (iv) 0.03 dL (v) none of these

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

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5. (A) The carbonless paper used in many double entry lab notebooks works by having extremely small capsules of ink on one side (that are broken open when the other side is written on). If the diameter of one of the ink capsules is 6 000 nanometers, what would you report as the diameter in millimeters?

(B) Examine the following calculations. Which would yield answers with three significant digits? (*Note: it is not necessary to actually calculate the answers, just predict the resulting significant digits for the answers.) (i) (97.42 – 88.5) (ii) (42.55 x 1.66) (iii) (0.01010/0 .5000) (iv) (1.10 x 10-2 x 0.610) 6. (A) Cholesterol has been linked to heart disease, so the determination of blood cholesterol can be very important. The chemical formula of this important biological molecule is C27H46O. Use this and the atomic masses (note significant digits) of each atom to determine the molar mass of cholesterol.

(B) Round this value to four significant digits.

(C) A blood level of 240 mg/dL cholesterol (*note this is two sig dig) is considered high. That unit is based on mass of cholesterol in a volume. What would this value be if it were reported in units of g/L? 7. (A) Report the arithmetic answer, with the appropriate number of significant digits for the following operations: pH is a common way to express acid (i) subtraction: (ii) concentration. It is a logarithm Mass of an empty filtering crucible = 42.356 g function. How many significant digits Mass of crucible with dried precipitate = 43.2453 g are found when pH = 7.1010?

Mass of precipitate = ___________ 8. (A) Astronauts may someday land on Mars. If instruments are brought to Mars for some analytical chemistry, they will have to function at some extreme temperatures. If the temperature at a specific Mars location was 213K, what would be the corresponding temperature in Celcius?

(B) The average temperature of an astronaut is about 37oC. What is that value in K? 9. The graph shown here (Figure 2.10 from the textbook) shows a comparison of light absorbance and protein concentration. To what decimal place would you be able to estimate values on the Y-axis? To what decimal place would you be able to estimate values on the X-axis?

10. Round each of the following values to two significant digits: (A) The pH of human blood is approximately 7.350 (pH is the – log (activity of H+)).

(B) The pH of human urine varies, but a particular sample has a pH = 6.0250.

(C) Humans may have an average of 5.025 liters of blood.

(D) Human blood may contain 135.1 grams of hemoglobin per liter.

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis (E) There are an estimated 7.589 x 1010 cells in a human body

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C. McLaughlin

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO TEST BANK FOR CHAPTER TWO: 1. (A) (Among other answers) Always add a small amount of the solution that will be used in the analysis to be used in rinsing the buret, before using the buret in the analysis.

(B) The “3” notation for health raises the concern that the compound may be toxic or corrosive and should never contact skin. The W with the cross line indicates that water should not be used in such a way to come in direct contact with the substance.

(C) Material Safety Data Sheet. This sheet will have chemical and physical properties of the substance, fire and explosion concerns, reactivity, disposal procedures, and identification information such as formula and CAS number.

2. (A) (ii) (B) A “1” indicates a combustible material but a zero fire hazard indicates a noncombustible material.

(C) Since the Blue diamond (health) has a “3,” great care must be taken to avoid any skin contact or inhalation. (A qualified professional should be immediately notified.)

3. (A) The error should not be erased. Rather a single line could be drawn though the two entries. The correction should be made, then the student should initial and date the changes.

(B) This conversion could be done using two-dimensional analysis steps within the SI system:

1x103 mm 1m x ) = 0.02 mm ) ( 1m 1x106 m 1m 4. (A) 2.5 x 10-5 M x ( ) = 25 M 1x106 m 10dL 1L (B) 3 mL x ( ) = 0.03 dL ) x ( 3 1L 1x10 mL 20m x (

1x103 mm 1m 5. (A) 6 000 nm x ( ) x( ) = 0.006 mm 1x109 nm 1m (B) (ii); (iv) 6. (A) 27C x

12.0107g 1.00794g = 324.2889; 47 H x = 47.37318; O = 15.9994 g 1mol 1mol

324.2889 47.37318 + 15.9994 387.66148 = 387.661 g/mol (B) To four significant digits = 387.7 (C) 240 mg x

1L 1g = 0.24 g; 1dL x = 0.1L; 0.24g/0.1L = 2.4 g/L 3 10dL 1x10 mg

7. (A) 0.8893 = 0.889

(B) Since this is a logarithm value, the significant digits begin after the mantissa. This pH value has four digits beyond the mantissa, so there are four significant digits.

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

8. (A) K = oC +273; 213 = XoC +273; X = -60oC (B) K = oC +273; X = 37oC +273; X = 310K 9. The labels on the Y-axis indicate readings at the 0.1 unit. Therefore, estimates could be made between those marks at the 0.01 unit level. The labels on the X-axis indicate reading at the unit level. Therefore, readings could be estimated at the 0.1 unit level.

10. (A) The pH of human blood is approximately 7.350 (pH is the – log (activity of H+) Rounded to 7.40 (B) The pH of human urine varies, but a particular sample has a pH = 6.0250 Rounded to 6.02 (C) Humans may have an average of 5.025 liters of blood.

Rounded to 5.0 (D) Human blood may contain 135.1 grams of hemoglobin per liter Rounded to 140 (E) There are an estimated 7.589 x 1010 cells in a human body.

Rounded to 7.6 x 1010

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 3: Mass and Volume Measurements 1. (A) Accurate mass measurements are critically important to many analytical chemistry applications. Knowing the properties of the analytical balance used for those determinations is part of those determinations. Suppose a balance has a capacity of XXX and a readability of YYYY . What is the resolution of the balance?

(B) For detailed mass determinations on a balance, the buoyancy factor may need to be considered before reporting the mass of a sample. Some studies show that taking a small daily dose of aspirin (acetylsalicylic acid; ASA) offers protection from heart attacks. (This is somewhat controversial and should only be done under a doctor’s care.) If you massed 1.0810g of ASA (density of 1.40 g/cm3) what would you report as the corrected mass?

(Assume the density of air is 0.0012 g/cm3; and the density of the counterweights = 8.00g/cm3.)

2.(A) Even though we know that buoyancy effects can cause slight errors when determining masses, we often ignore the slight error. When you examine the buoyancy correction equation would the error be less or more when the density of the material being massed is closer to the assumed 8 g/mL for the calibration weights?

(B) Sometimes we calibrate the volume delivered in analytical glassware by determining the mass of water delivered by a predicted volume. Suppose the water delivered by a 25.0 mL pipet had an apparent mass (uncorrected) of 25.05 grams. If the density of air at the temperature was 0.001 2 g/mL, the density of the calibration weights was 8.0 g/mL and the density of water at the temperature was 0.9950 g/mL what would you report as the corrected mass of the sample? (report to four significant digits) (C) Using the actual mass of water delivered by the pipette, what is the volume of water delivered by the “25 mL” pipette?

3. (A) The primary standard acid potassium hydrogen phthalate (KHP) is often used to help determine the concentration of a basic solution of sodium hydroxide. If 4.826 2 grams of KHP were massed on a balance when the air had a density of 0.001 2 g/mL and the counterweights used had a density of 8.00 g/mL, what would be the true mass, accounting for possible buoyancy affects, of the KHP? The density of KHP is 1.636 g/mL. (B) Without detailed calculations, which one of the following when measured would have the greatest difference between its displayed mass (in air) and its true mass? (i) acetic acid (d = 1.05 g/mL) (ii) CCl4 (d = 1.59 g/mL) (iii) Sulfur (d = 2.07 g/mL) (iv) PbO2 (d = 9.4 g/mL) 4. Suppose a biochemist is using bovine serum albumin (BSA) in an assay. The solution on hand has a concentration of 425 mg/mL. Describe how that solution could be used to prepare 100.0 mL of a 125 mg/mL solution to use in the assay.

5. The amount of glucose in a person’s blood right now would depend on if food was eaten before working this problem or after working the problem. If the level of glucose in a blood sample (C6H12O6 molar mass = 181 g/mol) was 95 mg/100mL, what would be the Molarity of glucose in the blood?

6. During cold winters a solution of antifreeze in car cooling systems will often keep the coolant solution from freezing. If a 50.00% (by mass) solution of environmentally friendly propylene glycol, with a density of 1.04 g/mL is used, what would be the molality (m) of the solution? (molar mass of propylene glycol = 76.09 g/mol)

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

7. (A) Concentrated hydrochloric acid (HCl; molar mass = 36.46g/mol) is commonly found in analytical labs (of course, kept under the exhaust hood). The density of this dangerous solution is 1.18 g/mL. The solution is known to be 37.1%, by mass HCl. What is the molarity (M) and molality (m) of this solution?

(B) The density reported in part (A) was taken at 25oC. If the lab temperature dropped to 15oC would that make the density of the solution increase or decrease ?

(C) Which of the following best describes the effect on concentration when the temperature in the lab dropped? (i) Neither the M nor m would be affected. (ii) M would be greater; m would be unaffected. (iii) M would be smaller; m would be unaffected. (iv) M would be unaffected; m would be greater (v) M would be unaffected; m would be smaller 8. (A) Silver nitrate is often used to precipitate chloride ions in water analysis. How many grams of silver nitrate would be needed to prepare 100.0 mL of 0.100 M silver nitrate (AgNO3) solution? (1 mol = 169.9 g/mL) (B) Suppose the lab already has a solution that is 0.250 M AgNO3. Calculate and describe how to prepare, using a volumetric flask, 100.0 mL of 0.100 M AgNO3 from a stock 0.250 M AgNO3 solution. (Report the amount and brief directions for preparation.)

(C) It is a little known fact that in the USA drinking water contains, on the average, approximately 28 ppb Ag. Consuming about one liter of water in a day (recall that the density of water is 1.0g/mL) would result in ingesting how many grams of silver?

9. (A) Lactic acid can be produced by bacteria and fungi through fermentation. Lactic acid can help produce a sour taste in yogurt and sourdough bread. Suppose you are going to analyze a solution of lactic acid that is known to have a density of 1.15 g/mL and is also known to be 4.88% w/w lactic acid (C3H6O3; molar mass = 90.08 g/mol) What would you report as the molarity of the solution?

(B) Convert 4.88% w/w to ppm.

10. (A) High fructose corn syrup is made by using an enzyme reaction to convert some glucose in corn to fructose (which has a sweeter taste and is cheaper than sucrose). HFCS is a mass mixture of approximately 55.0% fructose and 45.0% glucose. What is the molarity (M) of fructose in a soft drink containing 42.0 grams of HFCS in a volume of 355 mL? (molar mass of fructose = 180.2 g/mol) (B) Molarity (M) is a very common way to express solution concentration. However, analytical concentration (C) may also be used. Which of the following is the accurate statement? (i) Analytical concentration (C ) calculations are based only on molar masses of the ions rather than the entire formula unit. (ii) C calculations express a concentration with more significant digits than M. (iii) C communicates that strong electrolyte substances are converted to other ionic species in solution. (iv) An analytical concentration of 0.44 M MgCl2 (a strong electrolyte) implies that there are 0.44 moles/L of MgCl2 in the solution.

(v) C concentrations are not temperature dependent.

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO TEST BANK FOR CHAPTER THREE: 1. (A) Resolution = capacity/readability = XXX/YYYY (B )mASA = mdisplay x

1 − (dair / dref  = 1.8010g x 1 − (.0012/8.00 = 1.8022 g

1 − (dair / dobj 

2. (A) less (B) mwater = mdisplay x

1 − (.0012/1.40

1 − (dair / dref  = 25.05g x 1 − (.0012/8.00 = 25.08 g

1 − (dair / dobj 

1 − (.0012/.9950

(C) The 25 mL pipette delivered 25.08 g x 1 mL/0.9950g = 25.2 mL 3. (A) mKHP = mdisplay x

1 − (dair / dref  = 4.8262 g x 1 − (.0012/8.00 = 4.8290 g

1 − (dair / dobj 

1 − (.0012/1.636

(B) (i) 4. d1 x V1 = d2 x V2; 425 mg/mL x V1 = 125 mg/mL x 100.0 mL; V1 = 125 x 100/425 = 29.4 mL; Dilute 29.4 mL of the more dense solution to a volume of 100.0 mL 5. 95 mg/mL x

1g

x

1mol

= 5.248 x 10-4 mol/mL; 100.0mL x

1000mg 181g -4 5.2 x 10 mol/0.1000L = 0.0052 M

1L 1000mL

=0.1000L;

6. m = mol/(kg of solvent); 1.04g/mL of solution x 0.5000 = 0.520 g propylene glycol per mL of 1mol solution. 0.520 g x = 0.00541 mol propylene glycol per mL of solution 96.05g 1kg 0.5000 x 1.04 g/mL = 0.520 g water per mL of solution. 0.520 g water x = 0.000520 Kg 1000g M = .00541mol/0.000520Kg = 10.4 m 7. (A) 1.18 g/mL of solution x 0.371 = 0.445 g HCl/mL of solution.

1000mL 1mol 0.445 g/mL x = 445 g HCl/L; 445g/L x = 12.2 M 1L 36.46g 1.18 g/mL - 0.445 g HCl = 0.735 g H2O per mL of solution 1kg 0.735 g x = 0.735 kg H2O; m = 12.2 moles/0.735kg = 16.6 m = 17 m 1000g (B) At lower temperatures the volume would be smaller, therefore, the density would increase.

(C) ii 8. (A) 0.100 mol/L x 100.0 mL x 1L/1000 mL = 0.0100 mole AgNO3 needed.

169.9 g 0.0100 mol x = 1.699 g = 1.70 g needed

1 mol

(B) M1V1 = M1V2; 0.250 M x V2 = 0.100 x 100.0 mL; V2 = 40.0 mL;

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

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Properly dilute 40.0 mL of the 0.250 M solution with enough water to make 100.0 mL of solution.

28 g Ag x (C) One liter is approximately 103 grams; ; x = 2.8 x 10-5 g Ag = 9 3 10 g solution 10 g 9. (A)M = mol/L; 1.15g/mL x 0.0488 = 0.05612 g lactic acid per mL of solution.

1mol 1000mL 0.05612 g/mL x x = 0.623 M 90.08g 1L 4.88 g Lactic acid x (B) ; X = 48 800 ppm = 106 g 102 g solution 10. (A) M = mol/L; 42.0 g x

1mol

= 0.23307 mol; 355 mL x

180.2g 0.2331mol/.355L = 0.65655 M = 0.657 M (B) iii

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1L 1000mL

= 0.355L;

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 4 : Making Decisions with Data 1. Some people prefer the taste of cane sugar (sucrose: C12H22O11; mm = 342.2 g/mol) over high fructose corn syrup in their soft drinks. A particular brand of soft drink contains 0.11 moles +/- 0.02 moles of sucrose per 0.355 L +/- 0.002 L of solution. Using the method of error propagation, report the molarity, with the absolute uncertainty, of sugar in the drink.

2. Molar masses should be reported with their uncertainty. For example, dilute solutions of hydrogen peroxide are often used as a disinfectant. Using this data from a periodic table data provided and the uncertainties reported what would you report as the molar mass and uncertainty of H2O2?

H atomic mass = 1.007 94 +/- 7 O atomic mass = 15.999 4 +/- 3 3. For each of the following lab situations involving measurements, decide if the problem is most likely a random error or systematic error (Fill in the blank with “R” (random) or “S”(systematic).

(A) Three Analytical Chemistry students use the same HCl solution to titrate identical samples of the same bleach solution, but one requires 28.45 mL to neutralize, the two others use 28.55 mL and 28.35 mL _____ (B) A bleach label states that it is a 5.0% hypochlorite solution. Fifteen students titrating, with the same HCl standard solution, report average values from 4.8% to 4.9% hypochlorite. ______ (C) Which type of error in your measurement is impossible to eliminate? ____ (D) Which type of error tends to be all too high or all too low? ____ (E) Which type of error is most likely eliminated by re-calibrating a balance? ____ 4. If an initial volume reading of 0.05 mL +/- 0.02 mL was recorded in a buret and a final reading of 15.45 +/- 0.02 mL was made after a liquid was delivered, what would be reported as the total volume including the uncertainty, delivered? *Use ALL considerations for the correct significant digits and uncertainty in the answer.

5. Chocolate contains the caffeine-like stimulant theobromine. A chemist working with a chocolate company performed five theobromine analytical tests on a chocolate sample. The results of five determinations were found as follows: Trial 1 = 1.3% Trial 2 = 1.2% Trial 3 = 1.0% Trial 4 = 0.98% Trial 5 = 1.4% (A) If the mean of the results is 1.2% what is the standard deviation?

(B) What is the relative standard deviation?

(C) What is the confidence interval (at the 95% level) that can be reported for the trials?

6. Suppose another analysis of theobromine was done on the same chocolate. However, this time a different solvent was used in the extraction procedure. This analysis was performed three times obtaining an average value of 1.4% with a standard deviation of 0.25. At the 95% confidence level, can the this mean and the 1.2% from question #5 means be considered statistically close enough to be the same?

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

7. A university Plant Science department has a project involving genetics and the amounts of amylose (unbranched starch) and amylopectin (branched starch) in peas. (This could prove to be important for diabetics.)

(A) The starch content of one genetic variety of pea was analyzed and found to be the following.

What is the standard deviation of the following data?

166 170 172 175 (this is grams of amylose per kg of pea) (mean = 171) (B) What could you claim would be true if you increased the number of peas analyzed? (i) increasing number of peas would increase the accuracy (ii) increasing number of peas likely decreases the standard deviation (iii) increasing number of peas likely increases the standard deviation (iv) increasing number of peas eliminates systematic errors (v) increasing number of peas likely increases the mean (C) Another pea analysis, using five pea samples, revealed an average amylose as 242 g/kg with a standard deviation of 12. Using the F-test, would you consider these two standard deviations to be significantly different (at the 95% confidence level)?

8. The health benefits of green tea have been touted for centuries. Modern science has also tested some of the claims. In particular, one claim is that tea will improve mental alertness. The amino acid theanine has been found in green tea and is claimed to deliver those positive mental alertness effects. Suppose a company sells theanine as a separate supplement. The company reports a mean of 125 mg of theanine per dose. A student analyzes five doses that have a mean of 119 mg with a standard deviation of 9 mg. At the 95% level, does the student’s reported mean agree with the company value?

9. Two sets of chromium analyses were performed on water samples from a stream that runs near a chromium plating plant (they make those shiny bumpers).

Before the spill After the spill mean: 0.95 ppm Cr 1.10 ppm n 5 6 s 0.05 ppm 0.08 ppm Could it be stated, at the 95% confidence level, that after a spill there was an increase in the chromium level in the stream?

10. (A) An analysis on four aspirin tablets is performed that results in an average of 79 mg ASA (acetylsalicylic acid) (with s = 1 mg). The manufacturer supplies a reference value of 81 mg per tablet. Does the experimental average value agree, at the 95% confidence level, with the reference value?

(B) Suppose the four values used in the experimental analysis to get that 79 mg average were: 76, 80, 80, 81. The 76 mg value may be suspicious since it was the first trial with this analysis technique. Apply the Q-Test to determine if the 76 mg value should be kept or eliminated.

(Justify your answer at the 90% confidence level with calculations.)

(C) If another chemist performed the aspirin analysis, they may have a slightly different result.

One comparison to make may be the F-Test. Which of the following is true about the F-Test? (i) The F-Test requires that the comparison be made between analysis plans that must have the same number of trials. (ii) The F-Test uses the “s” value of the two trials in a comparison of the accuracies of the two analysis plans. (iii) The F-Test checks to see if the precision of the two tests are statistically similar.

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

(iv) The F-Test uses a comparison of the “s” value of the two tests to see which test is more selective. (v) The F-Ttest uses the “s” values of the two tests to compare the sensitivity of the two tests.

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO TEST BANK FOR CHAPTER FOUR: 2

1. 0.11 moles/0.355L = 0.309859; relative uncertainty =

2

 .02   .002    +  = 0.18;  .11   .355 

0.18 x 0.309859 = 0.06; M = 0.31 +/- 0.06 M 2. 2 x 1.00794 = 2.01588; 2 x 15.9994 = 31.9988; Since two atoms of each element are added, the total uncertainty would be doubled for each element. Therefore, the uncertainty for H would be 0.00014. For oxygen this would be 0.0006. Next, the absolute uncertainty of adding the two atomic mass quantities would be found via: (0.00014) + (0.0006) = 0.0006; The molar mass of hydrogen peroxide would be reported as 2.01588 + 31.9988 = 32.0417 +/0.006 g/mol 2

2

3. R, R, R, S, S 4. 15.45 – 0.05 = 15.40 mL; absolute uncertainty =

(0.02)2 + (0.02) = 0.28 = 0.03 2

Therefore the volume that was delivered = 15.40 +/- 0.03 mL 5. (A)

= 0.18

(B) RSD = 0.18/1.2 = 0.15 (15%) (C) From t-table with 5-1 degrees of freedom at 95% confidence t = 2.78; Therefore C.I = 1.2 +/- (2.78 x .18) = 0.5; 1.2 +/- 0.5 6. Use “pooled” standard deviation analysis plan: spooled = Standard deviation for the pooled mean =

((5 − 1)x0.18 ) + ((3 − 1)x.25 ) = 0.205 2

2

5+3−2

0.205 = 0.1497; tcalc = |1.4-1.2|/0.1497 = 1.33 (5 x3) (5 + 3)

ttable (using 6 degrees of freedom, at 95%, = 2.45) tcalc is less than ttable, accept null hypothesis, there is no statistical difference between the two means.

7. (A)

= 3.8 (B) ii (C) Fcalc = (0.12)2/(3.8)2 = 9.97; Ftable (95%) = 9.12;

At the 95% confidence level, the two standard deviation values are not statistically similar.

8. First calculate a t value: t = |119-125|/(standard deviation of the experimental mean); Standard deviation of the mean =

9 = 4.02; tcalc = 6/4.02 = 1.49; 5

The critical student’s t value (at 95% confidence level, for 4 degrees of freedom = 2.78).

Since tcalc is less than 2.78, there is no statistical difference in the two means.

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

((5 − 1)x0.05 ) + ((6 − 1)x.08 ) = 0.068; 2

9. spooled =

C. McLaughlin

2

5+6−2

standard deviation of the means =

0.068 = 0.0411 (5 x 6) (5 + 6)

tcalc = |0.95-1.10|/0.0411 = 3.65; At 95% confidence with 9 degrees of freedom ttable = 2.26 tcalc is greater than the ttable value so the null hypothesis (no difference) is rejected. The two means are statistically different.

10. (A) tcalc = |79-81|/(standard deviation of the mean); standard deviation of the mean =

1 = 0.5; tcalc = 2/0.5 = 4; 4

ttable, at 95% confidence, with 3 degrees of freedom = 3.18 Since tcalc is larger than ttable, there is a statistical difference between the two means.

(B) Q-test =

(80 − 76) = 0.80. For n = 4 the Qtable value = 0.76. Q-Test value is larger than (81 − 76)

Qtable, therefore reject the 76 as part of the random population.

(C) iii

5

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 5: Characterization and Selection of Analytical Methods 1. New techniques are being investigated to detect small levels of compounds used in packaging of food and pharmaceuticals that may leech into the product. Suppose a new technique to investigate harmful, leachable compounds that form on the inside of microwave popcorn bags (polyfluoroalkyl phosphoric acids PAPs) is being evaluated.

One method of validation of the new technique would be to use a spiked sample. A reference sample containing 5.0 g of a PAP compound was spiked with 1.2 g of the same compound into 1.0mL of the sample. The detected response (after analysis) of the spiked sample showed 6.0 g; what is the percent recovery in the technique?

2. Athletes may be tested for excess caffeine in their systems. A known sample of urine contained 11.0 ug/L of urine and was spiked with 2.0 ug of caffeine into 10.0 mL of the urine sample. The new measured response showed a concentration of caffeine of 185 ug/L. What is the percent recovery in this assay?

3. (A) Suppose you were working a summer job (due to your qualifications learned in Analytical Chemistry) at Trees R Us. You performed an analysis on the insect repelling pouches, that are to be attached to trees. The active ingredient found, in four analysis trials on one pouch was as follows; 7.0 g 7.2 g 7.3 g 7.9 g What is the within-run precision of this work?

(B) Your supervisor used the same method on the same sample to check your work. This would be an example of which type of precision characterizing the analytical method? ______ 4. In analytical methods it is important to be aware of subtle differences in key aspects of the method. For example: (A) Limit of Detection (LOD) and Limit of Quantitation (LOQ) both deal with limits of measurement. Which uses a 10/1 signal to noise ratio? ______ (B) Dynamic range and linear range both refer to measured responses to a property of an analyte in an analysis. Which utilizes the lower and upper limits of detection? _______ (C) Sensitivity and selectivity are also related to the measured response in an analysis. Which refers to the smallest change in the amount of analyte that can be detected in an analysis? ____ 5. (A) A typical Levey-Jennings plot is shown here. In this case the X-axis shows times a standard containing the PAP (polyfluoroalkyl phosphoric acids) mentioned previously (question #1) was analyzed. The Y-axis shows the results of the level of PAP found in the reference. (The middle dotted line on the Y-axis represents the mean of all trials.) Which statement below is correct? (circle one choice) (i) The technique is not valid because the results are not equal to the mean. (ii) The technique is not valid because testing the technique 25 times is not enough trials. (iii) The technique is valid because nearly all of the trials are within 95% of the mean. (iv)The technique is valid because it reveals the true mean of the PAP level.

(B) Suppose the technique reported a level of 0.50 g of a PAP found leached into a popcorn kernel. This one value would be most directly telling you about which one of the following? (i) The selectivity of the technique 1

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

(ii) The sensitivity of the technique (iii) The linear range of the technique (iv) The precision of the technique (C) Precison and accuracy are both important when evaluating any new technique. If the number of kernels of corn analyzed by this technique were increased from 20 to 80, which would be true? (circle one) (i) s would likely decrease (iii) random errors would decrease

(ii) RSD would likely increase (iv) Confidence Interval would increase

6. Suppose a cosmetics company needs to know about a particular oil being used in a cream.

An analysis is performed using a calibration curve of response versus concentration of the oil.

The slope of this line (m) was 0.250% with a standard deviation (s) = 0.004. What are the LOD and LOQ values for this analysis?

SOLUTIONS TO TEST BANK FOR CHAPTER FIVE: 1. Percent Recovery = 100 x

(6.0 − 5.0) = 83% 1.2

2. The sample concentration originally was known to be 11.0 g/L (or 0.11 g in 10 mL.) By adding 2.0 g of caffeine to the 10 mL sample the concentration should have increased by 200 g/L. The response, with the spike in the sample, was 185 g/L. The change in concentration was 185 – 11 = 174. Therefore, the percent recovery was 174/200 x 100 = 87.0% 3. (A) s = 0.39; mean = 7.35; 100 x 0.39/7.35 = 5.3% (B) Interoperator precision 4. (A) LOQ is based on 10(sb/m) (B) Dynamic range is the response between the lower and upper limit of detection (C) Selectivity is a measurement of the smallest change that can be detected 5. (A) iii The Levey-Jennings chart allows easy visualization of points that are within +/- 2 S.D. (B) ii The others depend on more than one measurement (C) I (random errors would still be present; RSD would decrease; confidence interval would decrease) 6. LOD = 3.3 (0.004/.25) = 0.005%; LOQ = 10 (.004/0.250) = 0.160%

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 6: Chemical Activity and Chemical Equilibrium 1. (A) To keep ionic strength constant and maintain a more consistent matrix in a solution where ions could be influencing each other, another component is often added to the solution This is an ionic compound, not involved directly with the reaction, called an ionic strength adjuster (ISA) ingredient. This is an important consideration in investigations using selective electrodes. What is the ionic strength (I) of a solution that contains 0.0280 M (NH4)2SO4 as the ISA ingredient?

(NH4)2SO4 (s) → 2 NH4+ (aq) + SO42- (aq) (B) One particular use of a selective electrode is in the determination of the level of Cl- in drinking water. What would be the activity coefficient () for the chloride ion Cl- in the solution from part “A”? (Note,Table 6.3 will be necessary for this calculation.)

(C)The chemical activity (a) of an ion in solution is related to the actual concentration of the ion.

Which one, if any, of the following statements is actually true? (select one) (i) The chemical activity of an ion is closest to the actual concentration when the ion concentration is very high. (ii) Chemical activity and activity coefficient are two names that have the same meaning. (iii) A substance in its standard state will have a chemical activity (a) = zero (iv) In a solution with high ionic strength in the background, Cl- ions will have less repulsion for each other. (v) Actually, none of the above is true.

2. (A) Calcium chloride (CaCl2; very soluble) has been sold as a more effective “de-icer” than NaCl to spread on icy sidewalks. What would be the ionic strength of a solution that was 0.031 M CaCl2?

2+ CaCl2 (s) →  Ca (aq) + 2 Cl (aq) (B) Now use that value and information from Table 6.3 to determine the activity coefficient () of H+ in that solution.

3. Amino acids are the key components of proteins. When fully protonated, amino acids have similar chemical reactions as diprotic acids (such as found in H2CO3). For example, alanine undergoes the following reactions: → H2C3H5NO2+(aq) H+ (aq) + HC3H5NO2 (aq) K1 = 4.5 x 10-3  HC3H5NO2 (aq)

→ 

H+ (aq) + C3H5NO2- (aq)

H2O (l)

→ 

H+ (aq) + OH- (aq)

K2 = 1.3 x 10-10 Kw = 1.0 x 10-14

(A) What would be the [H+] in a 0.010 M solution of H2C3H5NO2+? (** Three important assumptions to use: the second K value is small enough to be ignored for this calculation. The H+ from water is also small enough to be ignored in this calculation. The calculation for H+ should be done with a graphing calculator or successive approximations.)

(B) Looking back at the three reactions, what would be the charge balance expression? __________________

= _____________________

(cation)

(anions)

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

(C) Using only the value of the first equilibrium constant value (K1) from part “A” would a solution that had the following concentrations (H2C3H5NO2+ = 0.000050M; H+ = 0.000040M; HC3H5NO2 = 0.000030 M) be at equilibrium? What is the value of Q for this system? If it is not at equilibrium, in which direction would it proceed to reach equilibrium?

4. (A) The explanation of the effects of ionic strength on activities of ions relies on the concept of the “hydration radius” that surrounds ions in solution. Would the addition of KNO3 to a solution containing CaF2 increase or decrease the solubility of this slightly soluble salt? Briefly explain.

2+ (CaF2(s) →  Ca (aq) + 2F (aq)) (B) In a certain solution the  value for Ca2+ was found to be 0.747, while the  value for F- was found to be 0.926. Using those values, and the equilibrium value of 3.3 x 10-4, what is the 2+ molarity (M) of dissolved F- in the solution? (CaF2 (s) →  Ca (aq) + 2 F (aq)) (C) Slightly soluble CaF2 salt may have a connection to music. (A recent study of the wood used in the famous Stradivari violins suggested that the unique sound quality may be due to an antifungal wood treatment containing CaF2 and borax.) Using the following reactions, write the proper mass balance equation for Ca2+.

CaF2 (s) →  Ca2+ (aq) + 2 F- (aq)

Ca2+ (aq) + H2O →  Ca(OH)+ (aq) + H+ (aq)

→ CaF- (aq)  F- (aq) + H+ (aq) →  HF (aq) Ca2+ (aq) + F- (aq)

( MASS BALANCE)

CF- = _______________________

5. Nickel (II) hydroxide is found in rechargeable batteries (NiCd batteries). The following examples show three pertinent reactions of Ni(OH)2 in water: Ni(OH)2 (s) →  [Ni2+] (aq) + 2 [OH-] (aq

Ni2+ (aq) + OH- (aq) → NiOH+ (aq) H2O (l) →  H+ (aq) + OH- (aq)

What is the mass balance equation for OH- in this system?

COH- = 6. (A) The antibiotic Penicillin G is obtained from a mold and is treated before being ready for human consumption. Penicillin G is a weak acid (Ka = 1.6 x 10-3.) Using HP to represent this weak acid, in water the following reaction represents an equilibrium that could develop. → HP (aq) H+(aq) + P- (aq)  Use the quadratic equation or a graphing calculator to solve for the equilibrium concentrations (that result from a 1.0 x 10-4 M solution) of each component.

(B) What is the value of Go for this reaction? (Assume 25.0oC.)

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

7. Barium sulfate (BaSO4) is a nearly insoluble salt used in some X-ray gastrointestinal testing.

2+ 2The solubility equation for the compound is: BaSO4 (s) →  Ba (aq) + SO4 (aq) (A) Write out the equilibrium expression for this reaction in both the concentration dependent K and the thermodynamic equilibrium expression Ko. (B) Next show the relationship between these two expressions by converting K to Ko. (C) Using Table 6.3, determine the activity coefficients for Ba2+ and SO42- in a solution that has an ionic strength of 0.02. Use these values to calculate the value of K for the reaction if Ko = 1.08 x10-10.

(D) If the system was in a NaCl solution and the ionic strength was determined to be 0.10 M, would the value of K be higher or lower than that from part C? 8. Solutions with equal concentrations of Ca(NO3)2 or Mg(NO3)2 would have equal values for .

However, the  value for Ca2+ is smaller than the  value for Mg2+. What causes the difference?

9. On some occasions it is useful to determine the activity coefficient () for a neutral compound.

Acetic acid (HC2H3O2) is a weak acid found in vinegar. If the salting coefficient (k) for acetic acid is 0.066, what would be the activity coefficient for neutral acetic acid in a solution that was known to have an ionic strength (I) of 2.0?

10. Lactic acid tends to build up during times of physical exertion. Lactic acid only dissociates slightly in water. The Go value at 25.0oC for this reaction is 22.1 kJ. What is the value of Ko for the dissociation reaction?

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO TEST BANK FOR CHAPTER SIX: 1. (A) I = ½ (cizi2); I = ½[ 2(0.0280)(1)2 +(0.0280)(2)2] = 0.0840 (B) 0.084 is between 0.05 and 0.10 ionic strength. The  value may be estimated as: − 0.51 x z 2 x I − 0.51 x (-1) 2 x 0.084 log () = = = -.114; g = 0.768 = 0.77 (a I) (300 x 0.084) 1+ 1+ 305 305 The approximate  value for Cl- may also be interpolated from the table: (.806 − .753) (.806 − X ) ; X =  = 0.77 = (0.05 − .10) (0.05 − .0840) (C) iv 2 (A) I = ½ (cizi2); I = ½[ (0.031) (2)2 + (0.031)(2) (1) 1] = 0.93 − 0.51 x (1)2 x 0.031 − 0.51 x z 2 x I (B) log  = = = -.0591;  = 0.873 = 0.87 (a I) (900 x 0.031) 1+ 1+ 305 305

3. (A) The quadratic equation solution yields: X = [H+] = − 4.5 x10 − 3  ( 4.5 x10 − 3 )2 − 4(1)( −4.5 x10 − 5 )

− B  B2 − 4 AC 2A

;=

= 4.86x10-3;

2(1) (B) [H ] + [HC3H5NO ] = [HC3H5NO-] + [HC3H5NO-]+[OH-] [H+ ][HC H NO - ] [.000040][.000030] 3 5 2 = (C) Q = = 0.000024; Q < K; reaction will proceed to [H C H NO ] [.000050] 2 3 5 2 make more products. +

+

4. (A) Adding KNO3 increases the ionic strength of the solution. This would cause the activity coefficients of Ca2+ and F- to decrease. This, in turn, would increase the solubility of CaF2.



 

2

(B) K = Ca 2 + F − = [(.747)(X)][(.926)(2X)]2 = 3.3 x 10-4; X = 0.0505 M; F- = 2X; F- = 0.101 M (C)

CaF2 (s) →  Ca2+ (aq) + 2 F- (aq)

→ CaF- (aq)  → + F (aq) + H (aq)  HF (aq) Ca2+ (aq) + H2O →  Ca(OH)+ (aq) + H+ (aq) Ca2+ (aq) + F- (aq)

CCa2+ = [Ca2+] + [CaF-] + [Ca(OH)+] 5. COH- = [OH-] + [NiOH+] 6. (A) Using the quadratic equation will produce “X” then this value can be used as follows: [H+] = X; [P-] = X; [HP] = 1.0 x 10-4 – X 4

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

− B  B2 − 4 AC

C. McLaughlin

− 1.6 x10 − 4  (1.6 x10 − 4 )2 − 4(1)( −1.6 x10 − 4 )

= 6.97x10-5 2(1) 2A [H+] = 6.97 x 10-5 M; [P-] = 6.97 x 10-5 M; [HP] = (1.0 x 10-4) – 6.97x10-5 = 3.0 x 10-3 M X=

;X=

(B) Go = -RTlnK; = - 8.3124 x 298 x ln1.6 x10-3 = -15947 J = -16 000 Joules

Ba SO  = Ba SO  ; K = ( )Ba ( 7. (A) K = 2−

2+

4

BaSO4 

(

)(

(B) K = K x  Ba 2+  SO 2− o

4

)

2+

2−

2+

4

Ba 2+

o

SO 4

2−



) SO4

2−



1

1

(C) 1.08 x 10-10 = K (0.583)(0.571); K = 3.24 x 10-10 (D) At a higher ionic strength, the activity coefficient values will be smaller, causing K to become larger.

8. The  value for Ca2+ and Mg2+ (and all ions) depends on the hydration layer. The charge (z value) is the same for both Ca2+ and Mg2, however, a calcium ion has a larger ionic radius than does a magnesium ion. Since the same charge is concentrated in a smaller ion, as is the case for Mg2+, it is able to attract more solvent toward itself. This creates a larger hydrated radius and a larger value for .

9. log () = k x I; log () = 0.066 x 2.0 = 0.132; 10132 =  = 1.36 10. Change kJ to J; change oC to K Go = -RT ln Ko; 22 100 J = - (8.314J/Kmol) (298K) lnKo; -8.920 = lnKo; Ko = 1.35 x10-4

5

22100/(-8.3124x298) = lnKo;

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 7: Chemical Solubility and Precipitation 1. While gasoline is a mixture of many hydrocarbons, most will contain some pentane. This compound has a relatively high Kow value.

(A) Does this indicate that the compound is polar or nonpolar?

(B) Which of the following would be the primary type of intermolecular attraction between pentane molecules? Ionic; Hydrogen bonding; Dipole-dipole; Dispersion (C) If a gasoline spill needed to be cleaned up, would it best be done with a water solution or an organic solvent?

(D) As pentane dissolves in octanol, would that process likely be an increase or decrease in entropy?

2. Strontium carbonate is used in some pottery glazes, helps impart the red color in fireworks and is even found in some magnets used in loud speakers. The solubility product constant (Ksp) of strontium carbonate is ~ 1.10 x 10-10. This Ksp value is the concentration based equilibrium 2+ 2constant for the reaction: SrCO3 (s) →  Sr (aq) + CO3 (aq) (A) Using this Ksp, what is the molar solubility of SrCO3?

(B) Sometimes it is necessary to use the thermodynamic equilibrium constant, in this case, Kosp.

If solid SrCO3 was placed in a solution that had an ionic strength of 0.050 M, what would you calculate as the value of Kosp for the reaction? (Note, refer to Table 6.3 for activity coefficients.)

3. (A) Calcium phosphate (Ca3(PO4)2) has many uses from cheese making to bone formation and it is found in some dietary supplements. Complete the solubility expression for this versatile salt and write the Ksp expression in terms of “X”: Ca3(PO4)2 (s) →  (B) Aluminum hydroxide (Al(OH)3) also has a wide variety of common uses. It may be found as part of some antacid products, in some water treatments, and in some fire retardants. Complete the solubility expression for this compound and write the Ksp expression in terms of “X”: Al(OH)3 (s) →  4. If 250.0 mL of 0.0010 M Sr(NO3)2 were added to 50.0 mL of .0000010 M Na2CO3 would a precipitate of SrCO3 form? (Assume that all the Sr(NO3)2 and Na2CO3 have completely dissolved.)

5. (A) The following statements all make comments or claims about solubility. However, not all are valid. Circle any and all that are true (there may be more than one!). (i) H-bonding is an intermolecular force that can occur between molecules if the molecules have hydrogen atoms that are bonded to either F, O, or N. (ii) A compound with a high Kow value will likely be non-polar. (iii) Henry’s law describes how pressure affects the solubility of ionic solids in liquids. (iv) Non-polar solvents tend to have dipole moments with low Debye values. (v) Dispersion forces are only present between polar molecules.

(B) Milk of Magnesia is actually a suspension (that’s why it should be shaken before use) of solid Mg(OH)2. If the molar solubility of Mg(OH)2 was found to be 1.12 x10-4 M, what would be the value of Ksp? 2+ Mg(OH)2 (s) →  Mg (aq) + 2 OH (aq)

1

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

(C) If solid Mg(OH)2 were placed in a solution with a [OH-] concentration of 1.0 x 10-3 M, what would be the molar solubility of Mg(OH)2 in the solution? (Note, use the Ksp value from part “B” in this calculation.)

(D) Using the Ksp value from part “B,” what is the value of Go for this reaction? (Assume 25.0oC.) 6. One method to determine the amount of dissolved Ca2+ in a solution is to add iodate ions (IO3-) that react with calcium ions to form a precipitate of solid Ca(IO3)2. However, a very small amount of the solid would still dissolve.

The equation and Ksp value is: Ca(IO3)2 (s) → Ca2+ (aq) + 2 IO3- (aq) Ksp = 7.1 x 10-7.

What is the molar solubility of Ca(IO3)2?

7. Adding 50.0 mL of 1.0 x 10-3 M Sr(NO3)2 and 50.0mL of 1.0 x10-3 M Na2CO3 would produce how many grams of SrCO3 precipitate? Ksp of SrCO3 = 1.1 x10-10 8. (A) Solubility can be expressed as moles/L or grams/L. However, when comparing the solubility of two slightly soluble salts, both stoichiometry and molar mass should be considered with the Ksp values. For example, BaSO4 has a Ksp = 1.08 x10-10. AgCl has a Ksp = 1.77 x10-10. Which has the greater molar solubility? Which has a greater value for grams/L of dissolved solid?

(B) The Ksp of MgF2 is 5.16 x10-11. Which has the greater molar solubility: BaSO4 or MgF2?

9. (A) Zinc hydroxide Zn(OH)2 has been used as an absorbent in surgical dressings. It has a fairly low Ksp value (3 x10-17). What is the molar concentration that would be expected for Zn2+ in distilled water?

(B) If a buffer maintained the solution at a pH = 9.00, what would be the expected Zn2+ concentration? (Note, excess amounts of hydroxide will dissolve precipitated Zn(OH)2.)

What effect did the higher pH (up to this value) have on the solubility of Zn2+?

10. (A) The amount of oxygen dissolved in blood depends on the partial pressure of the oxygen in the immediate atmosphere. Typically the partial pressure of oxygen at sea level is approximately 150 torr. Using the Henry’s Law constant for O2 as of 1.26 x 10-3 (mol/Lbar), what is the molar solubility of O2 in blood at that partial pressure?

(B) While flying in a pressurized jet cabin the partial pressure of O2 may drop slightly to 120 torr.

What is the molar solubility of O2 at this pressure?

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO TEST BANK CH 7 1. (A) A high value for Kow indicates the compound dissolves better in relatively nonpolar octanol, therefore, the compound would be nonpolar.

(B) Without examining the exact structure, the type of interaction that most likely facilitates dissolving in a nonpolar solvent is dispersion forces.

(C) Organic solvents are nonpolar. The evidence points to this compound also being nonpolar.

Following the “like dissolves like” concept, an organic solvent would be more effective in cleaning the spill.

(D) Typically, energy changes in nonpolar solutes dissolving in nonpolar solvents are small. As the solute dispersed through the solvent the entropy would increase.

2. (A) Ksp = [Sr2+][CO32-]; 1.10 x 10-10 = X2; 1.10 x10 −10 = X = molar solubility = 1.05 x10-5 mol/L (B) Ko = Ksp (Sr2+)(CO32-) =1.10 x 10-10 (0.464)(0.454) = 2.32 x10-11 2+ 33. (A) Ca3(PO4)2 (s) →  3 Ca (aq) + 2 PO4 (aq) 3+ (B) Al(OH)3 (s) →  Al (aq) + 3OH (aq)

Ksp = (3X)3(2X)2 = 108X5 Ksp = (X)(3X)3 = 27X4

4. Determine ion product from concentrations of Sr2+ and CO32-.

0.250 L x 0.0010 M SrNO3 = 0.00025 moles Sr2+ 0.050L x 0.0000010 M Na2CO3 = 5.0x10-8 moles CO32The molarity of ions: 2.5x10-4 moles Sr2+/0.300L = 8.3 x 10-4 M; 5.0x10-8 moles CO32-/0.300L = 1.7x10-7 M The ion product = (8.3 x 10-4)( 1.7x10-7) = 1.4 x 10-10; This is greater than the Ksp, therefore, a precipitate will form.

5. (A)i, ii, iv (B) Ksp = 4X3; Ksp = 4 (1.12 x10-4)3 = 5.63 x10-12 (C) Ksp = [Mg2+][OH-]2; 5.63 x 10-12 = X (1.0 x 10-3)2; X = 5.6 x10-6M (D) Go = -RTlnKo; Go = - 8.314J/molK (298)ln(5.63 x10-12) = 64 200 J 6. Ksp = 7.1 x 10-7 = 4X3; X = molar solubility = 0.00562 M 7. Determine molarities of the two ions: 0.050L x 1.0 x 10-3 M Sr2+ = 5.0 x10-5 moles/0.100L = 5.0 x10-4 M Sr2+; 0.050 L x 1.0 x 10-3 M CO32- = 5.0 x10-5 moles/0.100L = 5.0 x 10-4 M CO32The Ksp = for SrCO3 is 1.1x 10-10; 1.1 x 10-10 = (5.0x 10-4 –x)(5.0 x 10-4 –x ); 0 = X2 – 1.00x 10-3X + 2.49 x 10-7; X = 4.7 x10-4 4.7 x 10-4 M x 0.10L = 4.7 x 10-5 moles of SrCO3; x 147.6 g/mol = 0.0069 g precipitate 8. (A)Molar solubility of BaSO4 = Ksp = 1.04 x 10-5; Molar solubility of AgCl =

Ksp = 1.33 x10-5 AgCl has the greater molar solubility.

The amount of dissolved solid for BaSO4 = 1.04 x10-5 x The amount of dissolved solid for AgCl = 1.33 x 10-5 x

3

233.4 g

1mol 143.4 g 1mol

= 0.00243 g/L

= 0.00191 g/L

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

BaSO4 has the greater solubility expressed in g/L (B) The molar solubility of MgF2 is 3

5.16 x10 −11

= 2.35 x10-4. Even though the Ksp of MgF2 is

4 smaller than the Ksp of BaSO4, it still has a greater molar solubility.

9. (A) molar solubility of Zn2+ = 3

3 x10 −17 4

= 2 x10-6 M;

3 x10 -17 Ksp = = 3 x10-7 M 5 2 2 (1.0x10 ) (OH ) At this higher pH value, the concentration of Zn2+ (via LeChatelier’s Principle) decreased

(B) At pH = 9.00 the [OH-] = 1.0 x 10-5; Ksp = [Zn2+][OH-]2;

10. (A) 150 torr x

1bar

= 0.20 bar; 750 torr Using Henry’s Law: Csolute = kPsolute; CO2 = (1.26 10-3 mol/Lbar)(0.20 bar) = 2.5 x10-4 M

(B) 120 torr x

1bar 750 torr

= 0.16; (1.26 10-3 mol/Lbar)(0.16 bar) = 2.0 x10-4 M

4

Test BankHage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 8: Acid-Base Reactions 1. Suppose the acidity level of snow in various areas is being compared. Which of the following examples of melted snow has produced the most acidic solution? (i) [H3O+] = 1.42 x10-7; (ii) pOH = 7.55; (iii) pH = 5.89; (iv) [OH-] = 2.2 x 10-9 2. Use of the base methyl amine (CH3NH2) to replace ozone-damaging compounds is becoming more widespread. Methyl amine is now used to make pesticides and herbicides to protect potatoes and tomatoes. The Kb for CH3NH2 is 4.2 x 10-4.

(A) What would be the pH of a 0.0010 M solution of this base?

(B) While methyl amine is a weak base, NaOH is considered a very strong base. What is the pH of a solution of NaOH that has a molarity of 1.0 x 10-8? Clue: start this problem writing out the charge balance and the mass balance for what is actually in this solution.?

3. Benzoic acid is a weak monoprotic acid sometimes used to make a food preservative. If a 0.15 M solution of the acid had a pH of 2.517, what would be the value of acid dissociation constant (Ka) for the acid?

4. Lactic acid is a monoprotic acid. The equilibrium between this acid and its conjugate (lactate) can be shifted during times of physical activity and are related to muscle soreness.

The pKa of lactic acid (HLac) is 3.870. What would be the fraction of species for lactate (Lac-) lac- in a muscle fiber with a pH = 5.850?

5. (A) Citrate buffers have a wide variety of uses. For example, a citrate buffer is used to analyze the protein content of powdered milk in a process called zone electrophoresis. What would be the pH of a citrate buffer composed of 0.35 M citric acid (HCit) and 0.25 M sodium citrate (NaCit)? The Ka of citric acid used here is 7.5 x 10-4.

(B) Suppose some acidic amino acids in the proteins produced 0.030 mole of H+ per liter of solution. What would you calculate as the new pH of the citrate buffer after the buffer adjusts to the addition of this new H+?

(C) Which of the following, if any, is true? (circle choice) (i) Good buffers are prepared so that additional acid or base “attacks” cause no change in pH. (ii) The citrate buffer from part “a” would be most effective at a pH of 3.12. (iii) Buffer pH values are not affected by ionic strength of the solution. (iv) Buffers can still be considered useful as long as the pH of interest is within +/- 2 pH units of the pKa value of the acid used in the buffer. (v) None of the above statements are accurate.

6. Maleic acid is a diprotic acid found in many fruits (especially apples) that is part of the synthesis of ATP. (H2C4H2O4; Ka1 = 1.5 x 10-2; Ka2 = 8.5 10-7; abbreviated as H2Mal) (A) If the juice from an apple has a pH = 5.50, what fraction of species () of H2Mal is present?

(B) At what pH will the HMaL- have its highest fraction of species?

(C) What would be the pH of a prepared solution of 0.10 M Mal2- ?

7. Potassium hydrogen phthalate is the monopotassium salt of phthalic aicd. This compound is often used as a primary standard to determine the molarity of NaOH solutions used in lab

1

Test BankHage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

analyses. In water the salt dissociates to produce potassium ions and hydrogen phthalate ions.

What would be the pH of a 0.10 M solution of potassium hydrogen phthalate? (Phathalic acid: Ka1 = 1.12 x10-3; Ka2 = 3.90 x10-6) 8. (a) Sodium carbonate, Na2CO3, is used in processing paper, making glass, and some types of soap. This important compound is mined in large deposits in Wyoming. In water, Na2CO3 completely dissociates into sodium ions and the dibasic carbonate ion (CO32-).

(Possibly useful information: For H2CO3; Ka1 = 4.5 x10-7; HCO3- Ka2 = 4.7 x 10-11.) (i) Complete the reaction that shows CO32- acting as a base in water: CO32- + H2O →  ________________ (ii) What is the value for the Kb for that reaction? (iii) What then would be the pH of a 0.010 M solution of Na2CO3?

9. The cleaning solution for soft contact lenses may contain enzymes, salt, and a buffer. (a) Suppose a buffer for such a solution was made using boric acid (Ka = 7.2 x 10-10). What would be the pH of a buffered solution that was made from 0.010 M boric acid (HBo) and 0.050 M sodium borate (Bo-) conjugate? (b) Suppose some material on a pair of soaking lenses brings in some basic compound that, when dissolved, produces 0.0050 M OH- within the solution. When the buffer comes to equilibrium, what is the new, pH of the buffer?

10. There are now 22 known genetically encoded amino acids. Although another surprising and unusual source, amino acids have even been discovered in meteorites. Suppose you are studying one of the new amino acids and find it has the following pKa values: pKa1 value of 2.42 with a pKa2 value of 9.61.

(A) What is the value of the fraction of species for H2A+ at a pH = 3.00?

(B) Of the following statements about amino acids, which one is true? (circle choice) (i) The fraction of species for an amino acid in the form of L- will decrease as pH rises (ii) The isoelectric point for an amino acid is found when  HL =  L(iii) When pH = pKa2; the zwitterion is the principal species (iv) At the isoelectric point, the  for HL = 1. (v) The fraction of species of H2A+ will increase as pH rises.

2

Test BankHage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO TEST BANK CH 8 1. (i) pH = - log(1.42 x10-7 ) = 6.848; (ii) pH = 14.00- 7.55 =6.45; (iii) pH = 5.89; (iv) pOH = -log (2.2 x 10-9 ) = 8.66; pH = 14.00- 8.66 = 5.34

-4

2. (a) Kb = 4.2 x 10 =

X 2

0.0010 − X 

; X = [OH-] = 4.71 x10-4; pOH = - log (4.71 x10-4) = 3.270;

pH = 14.00 – 3.270 = 10.76

(b) [OH-] = [OH-]NaOH + [OH-]water = Kw/[H+] = 1.0 x10-8 + [H+]; 0 = [H+]2 + 1.0 x10-8[H+] – 1.0 x10-14 [H+] = 9.51 x 10-8; pH = 7.02 3. Convert pH to [H+]; 10-2.517 = 3.038 x10-3; this also = A-; then use the Ka set up to solve for 2 3.038 x10 −3 Ka; Ka = = 6.28 x10-5 − 3 0.15 − 3.038 x10









4. Ka = 10-3.870 = 1.38 x10-4; [H+] = 1.41 x10-6; Lac- =

K a  = 3.85x10−4  H +  + Ka  1.41x10−6  + 3.85x10−4  = 0.996

Cit  = -log (7.5 x10 ) + log .25M  = 2.98 −

5. (a) pH = -log Ka + log

HCit 

(b) pH = -log 7.5 x10-4) + log

6. (a) H2C4H2O2 =

-4

.35M 

.25M − .03 = 2.89; .35M + .03

(c) “ii”

  = H + 2 + Ka1H +  + Ka1Ka2 K a1 H +

(1.5x10−2 )3.16x10−6 = 0.82 2 − 6 − 2 − 6 − 2 − 7 3.16x10  + 1.5x10 3.16x10 + (1.5x10 x8.5x10 ) (b) pH ~ (pKa1 + pKa2)/2 = (1.82 + 6.07)/2 = 3.95 2 ( K w / H+ ) (c) K b = ; 0 =K C [H ] + K K [H ] - K C − K w / H+  B b

B

+ 2

b

w

+ 2

2 w

0 = 1.2 x10-8(0.10)(x2) + 1.2 x10-8(1.0 x 10-14)(X2) – (1.0 x10-14)2 X = [H+] = 2.9 x10-10; pH = - log 2.9 x10-10) = 9.54*

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Test BankHage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

* assumption being made here is to ignore H+ from water autoprotolysis.

7. [H+]

(

K a1 Kw + K a 2CHA − K a1 + CHA −

)

1.12 x10 − 3  1.0 x10 − 14 + 3.90 x10 − 6 (0.10 )

=



1.12 x10

−3

+ (0.10 )

pH= -log(4.32 x10-9) = 8.36 8. (i) → HCO3- + OH-; (ii) use the Ka2 for the conjugate acid HCO3-2 then Kw/Ka2 = Kb1 =2.1 x 10-4;

X  2

(iii) Kb1 =

0.10 − X 

2.1 x10-4; X = [OH-] = 0.00448; pOH = 2.35; pH = 11.65

HBo = 9.14 + log .050M  = 8.84 .10M  Bo −  .050M + .005 = 8.90 (b) pH = 9.14 + log .10M − .005 9. (a) pH = pKa + log

2  H+ 10. (a)  = = 2 + + H  + Ka1H  + Ka1Ka2 1.0x10 2 1.0x10 2 + 10 1.0x10  + 10− 2.42 10− 9.61 H2A

+

−3

−3

− 2.42

−3

(b) iv

4

= 0.21

 = 4.32 x10-9M;

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 9: Complex Formation 1. People often encounter some stannous fluoride (SnF2) when brushing their teeth. Tin ions and fluoride are known to react in the following three steps: (all Kf values are approximates) + Sn2+ (aq) + F- (aq) → Kf1 = 4.0 x 103  SnF (aq) SnF+ (aq) + F- (aq) SnF2 (aq) + F- (aq)

→ SnF 2 (aq)  → SnF 3 (aq 

Kf2 = 2.7 x 103 Kf3 = 9.1 x 101

(A) In the first reaction which species is acting as a Lewis Acid?

(B) Complete this expression for the analytical concentration of Sn2+: CSn2+ = [Sn2+] + _________________________ 2.( A) Using the reactions below, write out the chemical expression for the fraction of all species that would be [Sn2+] (note this form does not need to be solved for K values, just the compounds themselves). + Sn2+ (aq) + F- (aq) → Kf1 = 4.0 x 103  SnF (aq) SnF+ (aq) + F- (aq) SnF2 (aq) + F- (aq)

→ SnF 2 (aq)  → SnF 3 (aq 

Kf2 = 2.7 x 103 Kf3 = 9.1 x 101

 for [Sn2+] = ________________________________ (B) The following expression shows the equilibrium for the second reaction. However, the [SnF+] term may be difficult to determine. (i) So, with help from the first reaction, rewrite that Kf2 expression in terms of only Sn2+; F-; SnF2 and Kf values. (ii) Then rearrange that new form of the expression to solve for [SnF2] Kf2 =

SnF2 

SnF F  +

−

(i) New form of Kf2 =

(ii)

THEN [SnF2] =

3. The entire fraction of species expression for Sn2+ would include factors such as 1, 2, and 3.

Using information from question #2, what would be the mathematical value of 3 for the Sn2+ and F- reaction?

4. One method to determine the amount of uncomplexed Sn2+ in a solution would be to use a chelating agent such as ethylenediamminetetraacidic acid (EDTA) via the following reaction: 218 Sn2+ (aq) + EDTA4- (aq) →  [SnEDTA ] (aq) Kf = 2.0 x10 EDTA is a polyprotic acid. Therefore, the concentration of EDTA4- shown in the above reaction would depend on the fraction of EDTA4- existing at a certain pH of the solution. If this titration reaction was done at a pH = 10.00, what would be the  EDTA4-? (Use information from Table 9.8.) Then what would be the value of K’f? EDTA4- = _______; K’f = __________

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

5. (A) Examine the fraction of species graph of EDTA: What are the dominant species present at pH = 6.0?

(B) Of the following, which statements provide an accurate comment about EDTA? (There could be more than one.) (i) At higher pH values the fraction of species for EDTA4- increases because more H+ are available which helps EDTA be a stronger acid. (ii) The concentration of EDTA4- in the solution is related to all the forms of EDTA in the solution via the following equation:  [EDTA4-] = CEDTA 2+ (iii) H2EDTA is only present in a solution of EDTA if the pH is very low. (iv) In an EDTA – metal ion equilibrium, the concentration of the metal ion is zero.

6. Calcium concentrations in solutions may be determined by a complex-forming reaction with EDTA. The reaction takes place between Ca2+ and EDTA4- with a formation constant of 4.5 x1010. At a pH = 11.00 the conditional formation constant K’f = 3.9 x1010. What is the approximate fraction of species of EDTA4- at this pH? 7. (A) Cobalt ions have many uses from precipitating proteins to pottery glazes. When Co2+ and EDTA react to form a complex, it can be represented as: Co2+ + Y4- → CoY2- (Y4- represents the fully deprotonated form of EDTA). The equilibrium formation constant (Kf) for Co2+ reacting with EDTA is 2.8 x 1016. In an analysis of Co2+ with EDTA, it would be best to use the conditional equilibrium constant (K’f) that incorporates the proper fraction of species () for the Y4- form of EDTA. Write out the structure of the general expression (no numerical values) for the conditional equilibrium constant.

(B) If the minimum K’f value needed for this reaction was 1.0x1010, what is the pH value (nearest whole number) would provide the proper  to convert the formation constant (Kf) to that minimum conditional formation constant (K’f) value? (Check Table 9.8 for useful information.)

(C) Now, working with K’f of 1.0 x 1010, how much free Co2+ would be present if the CoY2concentration was 0.042M at equilibrium?

8. (A) Name three other types of ligand-analyte interactions that can lead to complex formation besides Lewis acid-base interactions.

(B) Cyclodextrins can form complexes with organic compounds. Some studies are examining ways of using cyclodextrins to complex with components of human sweat. Besides possible medical uses, this could also open up possibilities of treating fabrics. Suppose a study reveals that one particular component forms a cyclodextrin complex that has a dissociation constant (KD) of 7.8M. What would be the association constant (KA) for the complex? (Report in proper units.)

2

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO TEST BANK QUESTIONS Ch 9 1. (A) Sn2+ (B) CSn2+ = [Sn2+] + [SnF+] + [SnF2] + [SnF3-]

Sn  Sn  + SnF  + SnF  + SnF  2+

2. (A) Sn2+ = (B) (i) Kf2 =

2+

−

+

2

SnF2 

K f 1 Sn 2+ F− F− 

3

(ii) [SnF2] = Kf1Kf2[Sn2+][F-]2

;

3. 3 = Kf1 x Kf2 x Kf3 = (4.0 x10-3)(2.7 x 103)(9.1 x101) = 9.8 x 108 4. EDTA4- = 0.392 (From Table 9.8); K’f = Kf x EDTA4- = (2.0 x1018 x 0.392 = 7.8 x1017 5. (A) At pH =6.00 the dominant species are HEDTA3- and H2EDTA2- (about 50% each) (B) iii 6. K’f = Kf x EDTA4-; K’f/Kf = EDTA4-; = 3.9 x 1010/4.5 x1010 ~ 0.87

Co(Y )  ; K’ /K =  ; = 1.0 x10 /2.8 x10 = 3.6 10 ; A pH = 5 produces a Co EDTA 2−

7. (A) K’f =

4

2+

f

f

Y

4-

10

16

fraction of species close to this value (based on Table 9.8).

(C) 1.0 x1010 =

0.42 ; X = [Co2+] = 2.0 x10-6M X X 

8. (A) hydrogen bonding, dipole-dipole attractions, ionic interactions (B) KD = 1/KA; KA = 1/7.8 x10-6 = 0.128 x 106 = 1.3 x105M-1

3

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 10: Oxidation-Reduction 1. The following shows an electrochemical cell that contains a gold electrode setting in 1.00 M solution of Au3+ and a silver electrode setting in a 1.00 M solution of Ag+.

→ Au0  → Ag0 Ag+ + e

Au3+ + 3e-

(A) What is Eo for this cell?

Eo = 1.42 V Eo = 0.80 V

Eocell = _______ V

(B) The arrow points to the anode electrode, which metal is that?

(C) (i) Balance the following overall cell reaction (Use both coefficients and superscripts to properly balance the redox reaction - there are 8 __ __ blanks.)__ Ag__ + __ Au__ →  __ Ag + __ Au (ii) Now circle the oxidizing agent in the reaction (D) What is the equilibrium constant ( Ko) for this reaction?

(E) Look back at the diagram. Which direction would a NO3- ion in the salt bridge travel?

(F) Examine the gold half reaction. Does that voltage indicate that Au3+ is better or worse than H+ (as in the SHE) at obtaining electrons?

2. Shown here are the two half reactions identified as important to rusting: o O2 (g) + 4H+ + 4e- → Fe2+ +2e- →  2H2O E = + 1.23 V  Fe

Eo = - 0.41 V

(A) If the pH of a solution was 5.97 and the O2 pressure was 1bar, what would be the E value for the oxygen half reaction?

(B) If the E value for the iron half reaction in this situation was - 0.487 V, what would be the value of Enet for the combined redox reaction?

(C) (i) Would you expect rusting to be more likely at higher or lower pH values? (ii) What would be the Enet value when this system reaches equilibrium?

3. Many of the basic concepts of electrochemistry and potentiometry can be illustrated with a “battery in a beaker” set up as shown here. Suppose that one of the cells shown here consists of a strip of lead setting in a solution of Pb2+ (aq) while the other beaker gold in an Au3+ (aq) solution.

(A) The following reactions and potentials are standard for the two metal ions: Au3+ (aq) + 3e- → Au (s) Eo = +1.42 V Pb2+ (aq) + 2e- → Pb (s) Eo = - 0.13 V Label the cathode (right side) with the proper metal symbol on the metal strip in the diagram.

(B) What is the Eo value for the cell?

(C) Assume the salt bridge shown here contained dissolved KNO3. Add another arrow on the diagram, near the salt bridge to indicate the direction of K+ migration through the salt bridge.

(D) Write out the proper balanced redox equation (using only the metals and ions). _________________________

_________________________

(E) Of the following which is the reducing agent? (circle choice) (F) Write out the “line diagram” that represents this cell: 1

Pb;

Pb2+;

Al;

Al3+

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

(G) What is the value of Go for the redox reaction?

4. Suppose during the work to develop a better, longer lasting lithium ion battery a new version has a reported Ecell voltage of 3.42 compared to the SCE reference. What would the cell potential be compared to SHE? 5. While working this problem, an average person may consume approximately 1.1 moles of O2.

This will provide some of the energy needed to write the answers as the oxygen reacts with food consumed earlier. The O2 half-reaction for this is: O2 (g) + 4H+ (aq) + 4e- → 2H2O (l) Eo = 1.23 V (A) What is the Go for this reduction half reaction?

(B) If the pressure of O2 was 0.21 bar and the [H+] = 1.0x 10-7 M, without using activities, what would be the value of E? 6. During the cold and flu season, many people become concerned with their vitamin C intake.

A redox reaction can be used to measure the amount of vitamin C (ascorbic acid; C6H8O6) in a solution: C6H8O6 (aq) + I2 (aq) → C6H6O6 (aq) + 2 I- (aq) + 2 H+ (aq) (A) What is the oxidation number of one carbon atom in ascorbic acid C6H8O6 ?

(B) Is the I2 acting as an oxidizing agent or reducing agent?

7. (A) Of the following four choices, which is the best reducing agent? (circle choice) (i) Ag+ (ii) Ag (iii) Cu2+ (iv) Cu

→ Ag (s)  2+ Cu (aq) + 2e- →  Cu (s) Ag+ (aq) + e-

Eo = 0.80 V Eo = 0.34 V

(B) Cars of the future may make use of hydrogen powered fuel cells (H2 powered battery.) Two possible reactions to consider for this are: o O2 (g) + 2 H2O (l) + 4e- →  4 OH (aq) E = 0.40 V

2 H+ (aq) + 2 e-

→ H2 (g) 

Eo = 0.00 V

What would be the voltage (E) of a fuel cell under the following conditions? 25.0oC H2 gas pressure = 1 bar [H+] = 0.00100 M O2 gas pressure = 1 bar [OH-] = 0.0100 M E = _______ V 8. (A) What would be the value of Eo in the fuel cell described in question #8?

(B) What would be the Ko value for the same reaction?

9. (A) An electrochemical cell is constructed from a copper wire in a Cu2+ solution and a silver wire in a solution of Ag+. The circuit is completed with a salt bridge containing KNO3. (i) Would the copper wire act as an anode or a cathode in this cell? (ii) Write a proper line diagram for the cell: I II I (iii) Which electrode (Cu or Ag) gains mass through the operation of the battery? _____ (B) Consider that the salt bridge is made of a saturated solution of KNO3. Would K+ movement through the salt bridge proceed toward the Cu2+ solution or the Ag+ solution?

(C) What would be the value of Go for this silver-copper battery?

10. The standard reduction voltages presented in Table 10.2 assume that concentrations are 1.00 M. Permanganate (MnO4-) solutions are often used in chemical analysis in redox titrations.

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

The [H+] in those reactions may not always be 1.00 M. Assume all other concentrations remain the same, what would be the E value when the pH of a solution was 7.00?

2+ o MnO4- + 8 H+ + 5e- →  Mn + 4H2O E = 1.51 V

11. Given the following reduction potential, which statement is correct? o K+ + e- →  K E = - 2.936 V (i) K is a good reducing agent (ii) K+ is a good oxidizing agent (iii) K+ is a good reducing agent (iv) The Eo for 2K+ + 2e- →  2K is - 5.872 V

3

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO TEST BANK QUESTIONS Ch 10 1. (A) Eocell = Eocat-Eoan; = 1.42 – 80 = 0.62 V; (B) The anode is the electrode with the least + reduction voltage; Ag + 0 (C) _3 Ago + __ Au3+ →  _3 Ag + __ Au (D) Eocell = -

0.05916

n

log K; 0.62 = -

0.05916

3

log K; 10

0.62 x 3 0.05916

= K; K = 2.8 x1031

(E) Nitrate ions would flow through the salt bridge toward the anode, or silver solution.

(F) The reduction potential of Au3+ is more positive than SHE, therefore Au3+ will be reduced (gain e-) with more force than hydrogen ions.

2. (A) E = Eo -

0.05916

4

log

10

1



−5.97 4

= 0.877 = 0.88 V

(B) Enet = Ecathode – Eanode; = 0.877 – (- 0.487) V = 1.36 V (C) At lower pH values, more H+ is present; thus the reaction will be even more favored.

(D) At equilibrium E = 0 3. (A) The cathode (reduction) would be Au (B) Eonet = Eocathode – Eoanode; = 1.42 – (- 0.13) V = 1.55 V (C) K+ would move, through the salt bridge, toward the cathode (D) 2 Au3+ + 3Pb → 3Pb2+ + 2Au (E) The reducing agent is Pbo (gives up electrons) (F) Pb|Pb2+ || Au3+|Au (G) Go = - nFEo; = - (6)(96500)(1.55) = - 987450 = - 987 000 J 4. Ecell = Eind – Eref; 3.42 = Eind – 0.242; Eind = 3.42 + 0.242 = 3.66; 3.66 - SHE = 3.66 V 5. (A) Go = - nFEo ; = - 2(96500)(1.23) = 237290J; 238 000 J (B) E = 1.23 -

0.05916

1

log

(0.21)(1.0x10− 7 )4

4

= 0.806 = 0.81 V

6. (A) Assigning +1 to H and -2 to oxygen brings a subtotal of 8(+1) + 2(-2) = +4; This is distributed among the six carbon atoms for a value of 4/6 or +2/3 for each C atom.

(B) I2 becomes 2I-, therefore it gained electrons and must be an oxidizing agent.

7. (A) choice (iv); Cu (B) Ecathode = Eo Eanode = Eo -

0.05916

log

4

0.05916

2

log

− 4 [OH ]

[O2 ]

1

+ 2 [H ]

= 0.400 - 0.01479 log

= 0.00 – 0.02958 log

Ecathode - Eanode = 0.518 – (-.177) = 0.695 V 4

[0.010]

4

= 0.518 V

[1]

1 [0.00100]

2

= - 0.177 V

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

8. (A) Eo = Eocathode – Eoanode; = 0.400 – 0 = 0.400 V = 0.400 V (B) Ko = e(nFEonet/RT) = e(4(.400/0.05916) = e27.0 = 5.57 x1011 9. (A) (i) Cu wire will act as the anode (ii) Cu|Cu2+ || Ag+|Ag (iii) As Ag+ is reduced the Ag cathode will gain mass (B) K+ ions would move toward the Ag+ solution (C) Go = - nFEo = - 2(96500)(0.80 – 0.34) = 88780 = 89 000 J

Mn 2 +    10. E = Eo -.05916/5 log ; If all remains constant except for H+ changing from MnO4 − [H + ]8   1 -7 1.00 M to 1 x10 M, the calculated change would be - 0.05916/5 log

[1][1.0 x10

11. (i)

5

−7 8 ]

= - 0.663 V

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 11: Gravimetric Analysis 1. Although it involves dangerous chemicals, the use of HF provides a direct way to measure the aluminum content of various types of clay. For example, a 5.00 gram sample of kaolinite clay (used in porcelain and glossy paper production) Al2(OH)4Si2O5 was treated with excess HF and produced 2.50 grams of solid precipitate of AlF3. What percent of the clay sample was aluminum?

Al2(OH)4Si2O5 (s) + 14 HF → 9 H2O + 2 SiF4 (g) + 2 AlF3 (s) 2. (A) When a precipitate forms in gravimetric analysis, one must be concerned with contamination. Often this can happen when unwanted materials in the solution become involved with the precipitate. Adsorption and absorption are two processes that can bring this about. Of the following, which statement is correct? (i) Adsorption involves penetration beyond the surface to the inside, and absorption means attachment to the surface. (ii) Adsorption means attachment to the surface, and absorption involves penetration beyond the surface, to the inside. (iii) Adsorption involves penetration beyond the surface to the inside, and absorption refers to the interaction of light with matter. (iv) None of the above is accurate (B) Heating a precipitate, in its original solution, to a temperature just below its boiling point in attempt to produce more pure crystals is referred to as… (i) Thermogravimetric analysis (ii) Ostwald ripening (iii) peptization (iv) Hage heating (v) precipitation from homogeneous solution 3. (A) Milk of Magnesia is actually a suspension (that’s why instructions often mention shaking before use) of solid Mg(OH)2. 1.00 liter of solution of 0.055 moles of Mg2+ is held at a pH of 7.00. Would a precipitate form? If so, what fraction of the original 0.055 molar Mg2+ would precipitate at that pH? Ksp = 5.61 x10-12 2+ Mg(OH)2 (s) →  Mg (aq) + 2 OH (aq) (B) At pH = 10.00, would there be precipitation? If so, what fraction of the original 0.055 M Mg2+ precipitate?

4. The ingredients in herbicides and pesticides must be carefully monitored. Arsenic in a pesticide sample was precipitated as MgNH4AsO4 (molar mass = 181.25 g/mol). This was heated to form the more stable Mg2As2O7 (molar mass = 310.45 g/mol). Based on the following data, what is the gravimetric factor and percent As in the pesticide. A 1.134 gram pesticide sample yielded 135.8 mg of Mg2As2O7.

5. High phosphate levels in water systems may contribute to several environmental problems.

One method to gather data for this determination would be to perform a gravimetric analysis to determine the level of phosphorus (P) in an aqueous sample by converting it to P2O7. A 25.00 gram sample of unknown water was acidified, properly treated with a precipitating agent and, 1

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

when heated, showed the sample to contain 6.06% P. How many grams of P2O7 would have to be precipitated for this result?

P = 30.9738 g/mol; P2O7 = 173.94 g/mol 6. (A) Suppose an unknown sample of Ca2+(aq) was going to be analyzed by gravimetric analysis. However, two different gravimetric methods are going to be compared. The original sample is split in half. One portion is treated through a process that precipitates the calcium as CaC2O4. (molar mass = 128.1 g/mol) The other equal size sample is precipitated as CaO. (molar mass = 56.08 g/mol). Knowing that the % of Ca will be the same for both, which method would provide the greater amount of precipitate to work with?

(B) Regardless of which method used in part (A), conditions that ensure pure crystal growth would still be used. Of the following procedural statements, which describes an aid in the formation of a pure crystalline precipitate? (i) Encouraging colloid formation aids in recovering the most precipitate from its solution. (ii) Ignition aids in obtaining precipitates from sintered glass crucibles (iii) Generating a homogeneous precipitating agent increases the likelihood of pure crystals (iv)The solutions are mixed rapidly to allow the appropriate ions to make contact without interference of other materials.

7. Many breakfast cereals contain iron additives (sometimes just tiny bits of iron!). A particular brand of cereal was known to be 1.00% iron. Students performed a gravimetric analysis that produced Fe2O3 from the cereal. If they wanted to ensure that their process would produce at least 0.2000 grams of Fe2O3 precipitate, how many grams of cereal should they use to begin the analytical process? (molar mass of Fe = 55.85 g/mol; molar mass of Fe2O3 = 159.69 g/mol) 8. (A) An isolated sugar from a cereal sample is either C12H22O11 or C6H12O6. Burning 0.5576 grams of the isolated carbohydrate in a combustion analysis caused the mass of an ascarite CO2 absorber to increase by 0.8609 grams. Which compound has been isolated?

(B) Another ingredient often found in cereals is folic acid (important in maintaining healthy red blood cells; C19H19NO6). Why is this ingredient not analyzed by combustion analysis?

9. Pure calcium oxalate dehydrate may be used as a primary standard in standardizing NaOH solutions. If the CaC2O4 2H2O were pure, by what percent would the mass of a 1.566 gram sample change as it was heated in a thermogravimetric analysis from 100oC to 300oC?

2

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO CHAPTER 11 TEST BANK QUESTIONS: 1. Molar mass of Al2(OH)4Si2O5 = 258.16 g/mol; molar mass of AlF3 = 83.977 g/mol 2.50 g AlF3 x

[1mol ]

[83.977]

= 0.02977 mol AlF3 x

0.009923 mol Al2(OH)4Si2O5 x

[1mol Al (OH) Si O ] 2 4 2 5 = [3 mol AlF3 ]

[2 mol A] [1mol Al (OH) Si O ] 2

4

x

2 5

26.9815g = 0.5355 g Al [1mol A]

0.5355g/5.00 g x 100 = 10.7% 2. (A) (ii) (B) (ii) 3. (A) At pH = 7.00, the allowable [Mg2+] = Ksp/[OH-]2; = 5.61 x10-12/[1.00 x10-7]2 = 561 M; Therefore, no precipitation at this pH because 561M is a much higher concentration of Mg2+ allowed than would even be available in the 0.055 M solution.

(B) At pH = 10.00, the allowable [Mg2+] is = 5.61 x10-12/[1.00 x10-4]2 = 0.000561M; A precipitation will be observed because this is less allowed than the original 0.055. The fraction of the original that precipitated as Mg(OH)2 is: (0.055 –0.000561)/0.055 = 0.99 2 mol As

4. Gravimetric Factor =

1mol Mg2 As 2O7 1g

135.8 mg x

X

5.

1000mg

=

2 (74.92 g/mol)

= 0.48265;

310.45 g/mol

x 0.48265

1.134 g sample

x 100 = 5.780%

2 mol x 30.9738 g/mol P 173.94 g/mol P O 2 7 x 100 = 6.06 %; X = 4.254 g P2O7 precipitate 25.00 g sample

6. (A) In both samples the size of the sample and the percent of Ca2+ will be the same.

Therefore, grams of precipitate x gravimetric factor = %Ca x g sample = constant. If the gravimetric factor is larger in the comparison, the grams of precipitate will be smaller. When the other variables become constant, the smaller gravimetric factor will produce a larger mass of precipitate. In these two cases Ca/CaO = 40.078/56.08 = 0.7447 compared to Ca/CaC2O4 = 40.078/128.1= 0.3129. The calcium oxalate technique would be the choice to produce more precipitate.

(B) (iii)

3

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

2 mol Fe

7. The gravimetric factor;

1mol Fe 2O 0.2000 x 0.6994

= 3

2 (55.845 g/mol)

C. McLaughlin

= 0.6994

159.69 g/mol

x 100 = 1.00% ; X = 13.988 g of cereal

X

8. (A) 0.8609 g CO2 x 12.01 gC/44.01 gCO2 = 0.2349 g of C in the sample. In 0.5576 g of sample this would make the sample be 0.2349/0.5576 x 100 = 42.13% C. C6H12O6 is 40.00%C. C12H22O11 is 42.11% C. Therefore the sample was C12H22O11.

(B) The chemical formula of folic acid contains an atom of N. Combustion analysis consists of methods to detect carbon, hydrogen, and oxygen (by deduction) but not if N is present.

9. 1.566 g CaC2O4 2H2O x

2 mol H O 18.015 g 2 x = 0.3438 g H2O 1 mol CaC O • 2H O 1 mol H O 164.127 2 4 2 2 1mol

x

0.3438g/1.556 x 100 = 22.10%

4

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 12: Acid-Base Titration 1. Phosphoric acid is a triprotic acid (H3PO4) often found in cola drinks. It may affect the sour taste. Suppose a 25.0 mL solution of H3PO4 is neutralized by 22.5 mL of 0.125 M NaOH. What is the molarity of the acid solution?

2. An ascorbic acid (vitamin C) and sugar mixture can be used to prevent browning of sliced apples. How many grams of ascorbic acid would be found in a sugar/ascorbic mixture if 34.55 mL of 0.2378 M NaOH were required to react with all the ascorbic acid? (Assume the sugar does not react.)

(Ascorbic acid, HC6H7O6 = 176.13 g/mol; assume 1:1 stoichiometry in the NaOH titration.)

3. (A) In yet another titration, ascorbic acid was to be titrated with a NaOH solution. In order to be successful, the equilibrium constant of the titration reaction must be sufficiently large to get the acid quantitatively converted to products. What is the equilibrium constant for the reaction?

(Ka for ascorbic acid = 7.94 x10-5) Is this large enough to predict a successful titration?

(B) A 35.0 mL solution of 0.150 M ascorbic acid required 22.1 mL of that 0.2378 M NaOH to neutralize. What would be the pH at the equivalent point in the titration?

(C) If phenolphthalein was used as the indicator in the part “C” titration, what would be the ratio of pink to colorless at the equivalence point? → H+ + Ind(generic equation) Hind pK = 9.4  (colorless)

(pink)

4. Lactic acid builds up during exercise. A 1.00 gram sample of this acid required 25.88 mL of 0.4290 M NaOH to neutralize. What is the molar mass of lactic acid?

5. Muriatic acid, found in hardware stores, is an impure solution of HCl. It has several uses from cleaning swimming pools to treating concrete.

(A) Suppose you were going to titrate 20.0 mL of 10.0 M HCl solution with 4.20 M NaOH. How many mL of NaOH would be required for the titration?

(B) What would be the pH of the HCl solution after the addition of 45.0 mL of NaOH? (C) What would be the pH at the equivalence point of the titration?

6. In addition to its pharmaceutical uses, barbituric acid (HC4H3N2O3; abbreviated as HBarb; Ka = 9.8 x 10-5) has uses in textiles and plastics. Suppose a chemist was going to perform a titration analysis on 25.0 mL of 0.20 M HBarb using 0.18 M NaOH. This titration would require 28.7 mL to reach the equivalence point.

(A) What would be the pH of the solution after 10.00 mL of 0.18 M NaOH has been added?

(B) What would be the pH halfway to the equivalence point of the titration?

(C) What would be the pH at the equivalence point in the titration? (As mentioned previously, to reach the equivalence point 27.8 mL of the NaOH were required.)

(D) Suppose a new indicator (extracted from tea) has been discovered. The new indicator has a pKa of 8.42. In the Hind form it is yellow and in the Ind- form it is blue. What would be the color ratio at the end point found in part “C”?

7. The weak base propylamine (CH3(CH2)2NH2) has a Kb = 4.7 x10-4. It can be used in some pharmaceuticals and pesticides. A 25.0 mL sample of 0.185 M propylamine solution required 38.5 mL of 0.120 M HCl to neutralize. What is the pH of this solution at the equivalence point?

1

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

8. Propanoic acid (abbreviated as HPro; Ka = 1.3 x 10-5) is produced naturally by a bacterium that is also responsible for forming the holes in Swiss cheese. An extract of 45.0 mL of HPro from some cheese had a propanoic acid molarity of 0.0150 M. This was titrated with 0.034 M NaOH. The entire titration would require 20.0 mL of NaOH. (A) The initial pH of the propanoic acid solution is 3.355. What is the pH of the solution at the halway point in the titration?

(B) What is the pH after 18.0 mL of the NaOH solution has been added?

(C) What is the pH at the equivalence point?

(D) Based on your results from part “D”, which of the following indicators would be the best one to select to show the end point of this titration? (Just circle your choice, no calculation required) (i) Cresol Purple: Transition range 1.2 – 2.8 (ii) Cresol Red: Transition range 7.2 – 8.8 (iii) Alizarin Yellow: Transition range 10.1 – 12.0 9. Oxalic acid is a diprotic acid and can be used, in dilute form, as wood bleach.

(A) How many milliliters of 0.050 M NaOH would be required to neutralize 10.0 mL of 0.100 M H2C2O4?

(B) Estimate the pH at the first equivalence point of the titration.

(C) Estimate the pH at the second equivalence point of the titration.

(D) What is the pH 5.00 mL beyond the second equivalence point?

10. A solution containing a mixture of carbonate and bicarbonate is analyzed by titration with a 0.0200 M solution of HCl. What is the molarity of carbonate and bicarbonate in the 35.00 mL sample if the first end point was found at 5.00 mL and the second end point was found at 15.00 mL?

2

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO CHAPTER 12 TEST BANK: 1. H3PO4 + 3 NaOH → Na3PO4 + 3 H2O; 3VaMa = VbMb; 3 (25.0mL)(X) = (22.5)(0.125M) X = (22.5)(0.125)/(3)(25.0) = 0.0375 M 2. Vb x Mb = moles of base; at the equivalence point this also equals the moles of ascorbic acid present in the original mixture: (0.03455L)(0.2378 mol/L) = 0.008216 moles of ascorbic acid.

176.13 grams

0.008216 moles x

= 1.447 g of ascorbic acid

1 mol ascorbic acid

3. (A) K = Ka/Kw = 7.94 x 10-5/1.00 x 10-14 = 7.94 x 109; This exceeds the 104 recommendation for feasibility for the acid-base titration.

(B) This is a weak acid being titrated by a strong base: pH = 14.00 – pOH; [OH-] ~ (Kw/Ka)(Vb xMb ) /(Va + Vb ) = (1.00 x10

- 14

/7.94 x10

-5

)(0.0221L x 0.2378 mol/L/(0.0221 + 0.0350L)) = 3.239 x 10-6 M

pOH = - log(3.239 x 10 M) = 5.4896; pH = 14.000 – 5.4896 = 8.510 (C) pH = pKa + log (Ind-/Hind); 8.510 = 9.4 + log (pink/colorless); 10(8.510-9.4) = pink/colorless = 0.13/1; (approximately 12% of the indicator may have changed color -7

1.000 grams

4. MW =

= 90.07 g/mol

(0.4290 mol/L)(0.02588L)

5. (A) VaMa = VbMb; (20.0 mL)(10.0M) = X(4.20M); 200/4.2= 47.6 mL NaOH (B) [H+] =

CHA VHA  − CB VB  = 20.0(10.0) − 4.20(45.0) = 0.169 M; 20.0 + 45.0 VHA + VB 

pH = - log[H+] = - log (0.169) = 0.772 (C) For strong acids being titrated with a strong monoprotic base, the equivalence point will be a neutral solution; pH = 7.00 −  A  6.(A) pH = pKa + log ; pH = 4.01 + log [

HA

(0.0100L x 0.18M) (0.0250L x 0.200M) − (0.0100L x 0.18M)

] = 3.76

(B) At the halfway point [A-] = [HA] therefore pH = pKa; pH = 4.01 (C) Determine the [OH-] to obtain pOH, then calculate pH [OH-] = (1.00 x10 - 14 /9.8 x 10- 5 )(0.0287L x 0.18 mol/L)/(0.0287 + 0.0250L) = 3.13 x10-6M pOH = - log (3.13 x10-6M) = 5.504; pH = 14.00-pOH = 8.50 (D) pH = pKind + log

A−  ; 8.50 = 8.42 + log blue ; 10 HA

yellow

(8.50-8.42)

= 1.2/1 the indicator is mostly in

the blue form 7. Ka of the conjugate acid = Kw/Kb = 1.0 x10-14/4.7 x10-4 = 2.13 10-11; The concentration of the conjugate acid = (Mb x Vb)/(Va + Vb) = (25.0 x 0.185)/(25.0 + 38.5) = 0.0728 M The Ka expression is used to solve for [H+]; 2.13 x10-11 = X2/(0.0728 – x); x = 1.25 x10-6 + -6 -6 [H ] ~ = 1.25 x10 ; pH = - log(1.25 x10 ) = 5.90 8. (A) At the halfway point pH = pKa; pH = - log (1.3 x10-5) = 4.89 3

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

(B) pH = pKa + log

A−  ; pH = 4.89 + log HA

0.018(0.034)

(0.0450x0.015) − 0.018(0.034)

C. McLaughlin

= 5.88

(C) At the equivalence point: Kb = Kw/Ka = 1.00 x10-14/1.3 x10-5 = 7.7 x10-10; [A-] = 0.045x0.015/(0.0450+0.0200) = 0.0104 M; [OH-] = 7.7x10 −10 x 0.0104 = 2.83 x10-6 pOH = - log [2.83 x10 -6] = 5.55; pH = 14.00 – 5.55 = 8.45 (D) Cresol Red has a range that includes this pH 9. (A) 2 VaMa = VbMb; 2(10.0)(0.100) = Vb (0.0500); Vb = 40.0 mL pk

(B) pH ~

a1

+ pK 2

a2 = 1.237 + 4.187 = 2.712 2

(C) Now the H2A is essentially in the A2- form. The Kb1 value will be needed as will the [A2-] Kb1 = Kw/Ka2 = 1.00 x 10-14/6.5 x10-5 = 1.54 x10-10; [A2-] = (Vb x Mb)/(Vb + Va) = (40.0x0.0500)/(40.0+10.0) = 0.0400 M; [OH-] = 1.54x10 −10 x 0.0400 = 2.48 x10-6; pOH = 5.605; pH = 14.000-pOH = 8.395 (D) [OH-] = ((CBVB) - (2CHAVHA))/(VHA + VB) = ((0.0500 x 45.0) – ( 2 x 0.100x10.0))/(40.0 + 10.0) = 0.00500M; pOH = 2.30; pH = 14.00 – 2.30 = 11.70 10. At the first end point the HCl + CO32- reaction has converted CO32- to HCO3-. The moles of CO32- = (Va x Ma) = (0.00500L x 0.0200 M) = 1.00 x10-4 moles CO32- present. When divided by the volume of the original solution, the molarity = 1.00 x10-4/0.0350 = 2.86 x10-3 M The second end point reflects the HCl needed to neutralize the HCO3- from the original sample plus the newly formed HCO3- that was produced from the carbonate.

0.0200 mol/L(0.0150 - 0.00500) = 2.00 x10-4 moles of bicarbonate neutralized. The amount of original bicarbonate then would be 2.00 x10-4 – 1.00 x10-4 = 1.00 x10-4 moles. Dividing this by the original volume produces the molarity of HCO3-; 1.00 x 10-4/0.0350L = 2.86 x10-3M

4

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 13: Complexometric and Precipitation Titration 1. In some food aid programs, nutritious powdered (dry) milk can provide long-term storage.

The calcium (40.08 g/mol) content of the product must be monitored. This can be done via EDTA titration. A dissolved sample made from a 2.50 gram sample of powdered milk required → 25.97 mL of 0.100 M EDTA to reach the equivalent point. Ca2+ (aq) + EDTA (aq)  CaEDTA (aq)

What is the mass percent of calcium in the milk?

2. (A) In high pH enviroments, as would be used in a calcium-EDTA titration, the EDTA form that dominates is ionized and abbreviated as EDTA4-. Look at the diagram of neutral (non-ionized) EDTA. Now circle the exact hydrogen atoms that would have to be removed to create the ionic EDTA4- active form that effectively chelates with Ca2+.

(B) The complexometric titration described in part “A” requires that the EDTA molecule be in its proper form to react with Ca2+. Examine the neutral EDTA structure again. Based on what is shown there, and knowing where the complexing sites for EDTA are on the molecule, draw an arrow(s) to any chemically available site(s) on the neutral molecule that you see that would bond to Ca2+.

3. (A) The formation constant (Kf) for calcium reacting with EDTA is 4.5 x1010. The equilibrium expression that is commonly used in such a reaction is the conditional formation constant (Kf’).

A calcium-EDTA titration was done at a pH = 10.00, where the fraction of species of EDTA4- is approximately 0.39, would have what value for the conditional formation constant for the reaction?

(B) Fully protonated EDTA has six Ka values and can be represented as H6Y2+. At high pH values the complex structure of EDTA may represented as Y4-. What would be the expected charge on a metal ion (M+?) found in a neutral MH2Y compound?

4. (A) Before performing an analysis of calcium in a powdered cement sample with an EDTA titration, the EDTA must be standardized. A 50.0 mL known solution of 5.67 x10-3 M Ca2+ (aq) required 38.10 mL of an unknown EDTA solution to reach its end point.

Based on the above data, what would be the correct molarity (MEDTA) of the solution?

(B) This same EDTA solution was then used in a titration of a properly prepared solution representing the cement. Two milliliters of the cement solution (pH = 11.00) required 25.00 mL of the EDTA solution from part “A.” At the end point, a very small amount of free Ca2+ could dissociate from the complex. What is that small molarity (M) of the free Ca2+ at the end point of the titration? The pK’f for this reaction at pH = 11.00 is 10.59.

5. Of the following, which statements provide an accurate comment about EDTA? (There could be more than one.) (i) At higher pH values, the fraction of species for EDTA4- increases because more H+ are available, which helps EDTA act as a stronger acid. (ii) The concentration of EDTA4- in the solution is related to all the forms of EDTA in the solution via the following equation:  [EDTA4-] = CEDTA (iii) H2EDTA2+ is only present in a solution of EDTA if the pH is very low. (iv) At the equivalence point in an EDTA – metal ion titration, the concentration of the metal ion is zero. (v) If the  EDTA4- is small, the K’f may also become too small for a successful titration.

1

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

6. The analysis of Ca2+ and Mg2+ in urine samples could be performed in a way similar to the EDTA analysis of the calcium content in water. For example, a 25.0 mL aliquot of a urine sample (buffered to pH = 10.0) required 17.2 mL of 0.0776 M EDTA to react with all of the Ca2+ and Mg2+ ions in the sample. Then another identical 25.0 mL urine sample was analyzed, except this one was treated with NaOH to precipitate all of the Mg2+ from the solution. This second solution required 13.2 mL of the 0.0776 M EDTA to react with the Ca2+ ions. Report the moles of Ca2+ and Mg2+ in the urine.

7. Using EDTA to titrate metal ions has become a common practice. For example, the Ba2+ in solutions that may be used to produce BaSO4 for medical applications could be determined by a complexometric titration. → BaEDTA2EDTA4- (aq) + Ba2+ (aq) Kf = 7.59 x 107 (aq)  (A) If 25.0 mL of an unknown Ba2+ solution required 42.0 mL of 0.0010 M EDTA to reach the equivalence point, what was the M of the Ba2+ in the solution?

(B) Next, let’s study the entire titration. This can be done graphically by plotting the pBa2+ on the Y-axis and mL of EDTA4- added on the X-axis.

For the titration, the solution is adjusted to a pH = 10.00. (EDTA = 0.39) (i) First, calculate the pBa2+ in the 25.0 mL solution from part “A” before any titration has begun.

(0% complete) (ii) What is the pBa2+ value after the addition of 35.0 mL of the 0.0010 M EDTA4- (83.3% complete)? (iii) What is the pBa2+ value at the equivalence point in the titration? (100.0% complete) (iv) What is the pBa2+ when the equivalence point is exceeded by 10.0 mL? (C) Besides the key color change at the equivalence point, what is the most important consideration to keep in mind about the metallochrome indicator you might select for this type of titration?

(D) Use your values from parts (i), (ii), and (iii) in part “B” as “anchor points” to sketch a smooth line for the titration on the axial system shown here:

pBa2+

8 6 4 2 0

10

20 30 40 50 mL of EDTA4- added

8. (A) What is the ideal K’f Min for the titration from question #7?

(B) What is the fraction of titration at the end point for this indicator and what is the resulting titration error for the analysis?

2

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

9. (A) Proper iodide levels are critical for thyroid health. Standard silver nitrate solutions can be used in a precipitation titration to determine iodide levels. What is the molarity of an iodide solution when 20.00 mL of the solution required 15.50 mL of a 0.0045 M AgNO3 standard?

(B) The end point of the titration could be determined by allowing the precipitate to scatter light.

What is the key difference between nephelometry and turbidity measurements used for this type of titration?

10. Potassium iodide solutions have many uses. Some controversial studies suggested using them as a precaution against absorbing radioactive iodine. Suppose 20.00 mL of a 0.0100 M of KI solution is titrated with a 0.0045 M solution of AgNO3 to precipitate AgI. (A) What is the pI- in the initial solution?

(B) What is the pI- after 10.00 mL of the AgNO3 titrant?

(C) What is the pI- at the equivalence point?

(D) What is the pI- at 5.00 ml beyond the equivalence point?

SOLUTIONS TO Ch 13 TEST BANK QUESTIONS 1. CEDTA x MEDTA = moles EDTA = moles Ca2+; 0.02597Lx0.100 mol/ x40.08 g/mol = 0.104g Ca2+ 0.104 g Ca2+/2.50 g milk x 100 = 4.16% 2.

3. (A) K’f = Kf = 0.39(4.5x1010) = 1.7 x 1010 (B) Removing four H+ from H6Y2+ causes the remaining charge to be reduced by four. This results in H2Y2-. If MH2Y is neutral, the metal must have a +2 charge.

4. (A) VCa2+MCa2+ = VEDTAMEDTA; (50.0mL)(5.67 x10-3) = (38.10)(MEDTA); MEDTA = 0.00744 M

CaEDTA  ; [CaEDTA] = 0.0250L x 0.00744 mol/L = 0.000186 moles; Ca 2+ EDTA  0.00689 ; X = [Ca2+] = 0.00689 ~ 4.21 x10-7 M 0.000186 mol/0.0270 L = 0.00689 M; K’f = 10 X X  3.89x10 (B) K’f =

5. (iii) and (v) 6. Titration #1: VEDTA x MEDTA = moles EDTA = moles Ca2+ + Mg2+; (0.0172L x 0.0776 mol/L) = 0.001335 total moles.

3

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Titration #2:VEDTA x MEDTA = moles EDTA = moles Ca2+ (0.0132L)(0.0776) = 0.00102 moles Ca2+ 0.00134 total – 0.00102 moles Ca2+ = 0.00316 moles Mg2+ 7. (A) VEDTA x MEDTA = moles EDTA = moles Ba2+ (0.0420L)(0.0010 mol/L) =0.000042 mol 0.00042 mol/0.250 L = 0.00168 M = 0.0017 M (B) (i) pBa2+ = - log (0.00168) = 2.77 (ii) [Ba2+] =

(CMVM − CL VL ) (0.0017 x 25.0) − (0.0010 x 35.0) = 0.000125 M; pBa = 3.90 (VM + VL ) = (25.0 + 35.0) 2+

(iii) At pH = 10.00 K’f = EDTA(Kf) = 0.39(7.59 x107) = 2.96 x107 C V M M = K' (V + V ) f M L

[Ba2+] =

(iv) [Ba2+] =

(CMVM )

K'

(0.0017)(0.0250)

= 4.63 x10-6; pBa2+ = 5.33

7

2.96 x10 (0.0250 + 0.0420)

(0.0017 x 0.0250)

(f CL VL + CM VM ) 2.96x107 (0.0010 x 0.0520) + 0.0017 x 0.0250) = 1.5 x 10 ; pBa =

-7

2+

= 6.82 (C) The formation equilibrium constant for the indicator-metal complex must be smaller than metal-ligand formation equilibrium constant.

(D) pBa2+ 8 

6  

4  2

(line should be similar to Figure 13.7) 0

10

20 30 40 50 mL of EDTA4- added

8. (A) Ideally the conditions of the titration should be adjusted to that K’fMIn = 1/[Ba2+] at the equivalence point. = 1/4.63 x10-6 = 2.16 x105 (B) The fraction of titration when [Ba2+] = 1/K’fMIn = 2.04 x105 is: 1−

F= [M] CL

+

[M] CM K f [M]

1 + Kf[M]

1−

= [4.63 x10

-6

0.0010

]

+

[4.63 x10

-6

]

0.0017 10 -6 1.74 x10 [4.63 x10 ]

1 + 1.74 x10

10

[4.63 x10

-6

= 0.992 ]

Volume added to this point: F x VE = 0.992 x 42.0 = 41.67 Titration error: VL, end point – VL; 41.67 – 42.0 = - 0.33 mL or 0.33/42.0 x 100 = ~0.79% 9. (A) VAgNO3MAgNO3 = VI-MI-; (15.5)(.0045)/20.0 = MI- = 0.0035 M (B) Nephelometry makes use of measuring the scattering of light, at right angles to the source of light, that becomes scattered by the precipitate (AgI). Turbidity can also be measured by the diminished light that is not scattered as it passes through the sample.

4

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

10. (A) initially, pI- = -log (0.0100) = 3.000 (B) At 10.00 mL, [I-] =

CM VM − C X VX (20.00 x 0.0100) − (10.00 x 0.0045) = = 0.00517 M; pI- = 20.00 + 10.00 VM + VX

2.287 (C) At equivalence point (44.4 mL): [I-] =

K

spAgI K

(D) 5.00 mL beyond equivalence point: [I-] = 8.52 x10

- 17

(0.02000 + 0.04944)

(0.04944 x 0.00450) − (0.02000 x 0.0100)

=

8.52 x10

V + VX sp M

C V −C V X X M M

- 17

=

= 3.75 x10-15; pI- = 14.426.

5

= 9.22 x10-9M; pI- = 8.035

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 14: An Introduction to Electrochemical Analysis 1. Products that may need to be protected from rusting are often coated with zinc. This process, called galvanizing, can be done by reducing Zn2+ from zinc-containing solutions. The reaction of interest is Zn2+ (aq) + 2e- → Zn (s) (A) If a current of 35.0 amps flows for 5.00 minutes through this solution, how many moles of electrons have moved through the solution?

(B) How many grams of zinc could have been plated with this current?

2. A certain half-cell set up uses SCE as a reference. The overall cell voltage is measured at 0.542 V. However, a different lab uses the same half-cell but their reference is the Ag/AgCl electrode. What is the Ecell that should be reported in the new lab’s work?

3. (A) The following statements make comments or claims about ion selective electrodes. Which of the following statements are valid? (There could be more than one.) (i) In a pH probe, the thin glass membrane separates an inner solution of H + from a sample containing H+.

Both must be at the same concentration to develop a potential. (ii) In potentiometry measurements such as with ISE probes the current should be zero. (iii) A junction potential can form when two solutions of different ions come into contact and when two solutions of the same ions, but different concentration, come into contact. (iv) An analytical disadvantage of ISE measurements is that they typically destroy the sample analyte during the measurement. (v) ISE probes are indicator electrodes that must be paired with some type of reference electrode to produce a measurable cell voltage.

(B) Suppose you were doing work to develop a better, longer lasting lithium ion battery. Your new version has a reported Ecell voltage of 3.42 compared to the SCE reference. What would the cell potential be compared to a silver/silver chloride reference electrode?

(C) Regardless of which reference electrode you used, you would likely have problems with a junction potential. Of the following, which is the one statement that is NOT TRUE? (i) The junction potential develops due to charge separation. (ii) The junction potential in a salt bridge can be reduced somewhat if ions with similar mobilities are used inside the salt bridge. (iii) Junction potentials put a fundamental limit on the accuracy of direct potentiometric measurements. (iv)The junction potential is not part of the cathode or anode, yet we usually know its contribution Ecell. (v) The junction potential represents a balance between unequal ion mobilities and the tendency of the resulting charge imbalance to retard the movement of ions.

4. In a response to health and nutrition interests, a major soft drink company has recently announced its plans to reduce the amount of Na+ in its products by ¼ over the next five years.

One method for detecting Na+ levels could involve proper treatments and the use of a sodium ISE. A calibration curve was developed using standard Na+ solutions all treated with an ionic strength adjustment solution. The following values were collected: TRIAL [Na+] mol/L E (versus SCE) (1) 1.00 x 10-5 -70 mV -4 (2) 1.00 x 10 - 20 mV (3) 1.00 x 10- 2 + 100 mV Plot the graph for mv and -log of [Na+] to get the slope of the calibration line. The Y-intercept should equal 215 mV. Using that and the calculated slope, solve for the Na+ in an unknown extract when it produced a voltage of - 22 mV. pNa+

1

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

5. It has long been known that selenium is part of human health.

However, there are many concerns about the proper level of selenium. Suppose a selenium ISE was going to be used to monitor the level of Se2- ions. Using the following E values for known standards, make a typical ISE plot.

Cell Voltage (mv) M of Se295 1.0 x 10-7 65 1.0 x 10-6 35 1.0 x 10-5 (A) Based on the graph, the slope of the line is approximately 30. mv. (0.030 V) Does this indicate that the response is “Nernstian”? After agreeing or disagreeing, justify the basis of your choice.

(B) If an unknown solution of Se2- (aq) provided an Ecell = 75 mv, what would you report as its concentration? (Use -115 mv as the Y- intercept.)

(C) Each of the solution samples also contained some added Ionic Strength Adjustor (ISA.) What is the main purpose of adding an ISA to the solutions?

(D) The selenium electrode must be connected to a reference electrode so that the cell voltage depends directly on the selenium ion concentration in the solution. Of the following, which one is TRUE? (i) During the selenium ISE determination, Se2- ions gets oxidized. (ii) During the determination, the ion-selective membrane will dissolve, releasing Se2- from inside to create a charge separation. (iii) The reference solution on the inside of the membrane also contains Se2-, but it retains a nearly constant concentration. (iv) A liquid junction potential inside and outside of the selecting membrane must be equal to produce a measurable current.

6. (A) Of the following, which, if any, provides a true conclusion about the liquid junction potential (Ejunction) when dissimilar electrolytes come into contact? (i) Liquid junction potentials are often as large as the Ecell. (ii) Liquid junction potentials can be minimized when ions of similar mobilities are in contact. (iii) When selecting a salt for salt bridge construction, it is best to select two ions that have very different ion mobilities to provide a charge in the salt bridge that minimizes Ej. (iv)The glass membrane of a pH probe has no liquid junction potential.

(B) Ion Selective Electrodes (ISE) have a wide variety of uses. As the name implies, they must be selective. The electrode may have a glass membrane, liquid-gel membrane, combination design, enzyme on semi-permeable membrane or solid state crystal. Match each type of determination to the appropriate type of ISE: (i) CO2 (ii) glucose (iii) F- (iv) Ag+ (C) Which ONE of the following descriptions of a pH probe (ISE) is accurate? (i) H+ from the unknown solution migrates from the outside through the middle of the glass membrane to the interior side of the glass membrane and enter the interior reference solution. (ii) As H+ from the unknown solution replaces metal ions on the negative oxygen of silicate, acting as an ion exchanger, on the exterior of the glass membrane the potential changes. (iii) Most commercially available pH probes work well at very low pH values (-2 to zero), but not well at very high pH values (12 to 14) 2

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

(iv) When calibrating a pH probe (later to be used in a measurement) with standard buffers, it is best to select two standard buffer solutions that are as far apart as possible.

7. An ISE with a special membrane was developed for the detection of penicillin. (An enzyme substrate coated electrode was used for this determination.) The following data, using the ISE and constant ionic strength standards, was collected: M penicillin Potential (E) 0.0010 M 0.190 V 0.00010 M 0.135 V 0.000010 M 0.080 V (A) Sketch in a graph using –log penicillin as the X-axis and mv as the Y-axis. The slope ~ - 55mv; estimate* the Y intercept (B) Suppose a solution with an unknown penicillin concentration was measured (same temperature and same ionic strength) and found to produce a potential = 0.142 V. What would be the concentration of penicillin in the unknown?

(C) Cite two specific advantages ISE probes may have over many other analytical techniques: (1) ___________________________ and (2) _______________________

3

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO CH 14 TEST BANK: 1. (A) 35.0 amps x 5.00 min x 60s/1min = 10 500 coulombs; 10 500 coul x 1mol e-/965500 coul = 0.109 mole e(B) 0.109 mole e- x 1 mole Zn/2 mole e- x 65.39 g/1mol = 3.56 g Zn 2. Ecell = EInd – Eref; 0.542 = EInd – SHE; EInd = 0.542+ 0 = EInd = 0.542 V; Ecell = 0.542 – 0.222 = 0.320 V 3. (A) (ii) (iii) (v) (B) Ecell = EInd – Eref; 3.42 = EInd – 0.242; EInd = 3.42 + .242 = 3.66 V; Ecell = 3.66 - .222 = 3.40V (C) (iv) 4. The plot yields a slope (Y/X) of approximately – 57 mv. Substituting values for Y = mX + b yields: - 22 mv = - 57X + 215mv; X = 4.16; 10-4.16 = 6.9 x 10-5 M 5. (A) This system represents a “Nernstian” response. In theory, a 10-fold change in concentration should show a 59 mv change with an ion of +/-1. The Se2- ion represents a potential of two moles of electrons, so 59/2 ~ 30 mv still reflects a Nernstian response.

(B) Y = mX + b; Substituting the values for this system, then solving for “X”: 75 = 30X – 115; X = 6.33; 10-6.33 = 4.7 x10-7 M (C) Adding ISA keeps the ionic strength essentially constant. Otherwise, changing the Na+ concentration would change the ionic strength of the solution.

(D) (iii) 6. (A) (ii) (B) (i) – gas sensing ISE (ii) Enzyme-coated membrane (iii) solid-state ISE (iv) glass electrode (C) (ii) 7. (A) Y = mX + b; substituting the values for trial #1 allows an estimate of the Y intercept: Using mv; 190 mv = (-55)(3) + b; b ~ 350 mv (B) Using the same equation for the unknown determination: 142 mv = - 55X + 350; X = 3.78; 10-3.78 = 1.7 x10-4 M (C) ISE determinations are often faster than other methods and ISE determinations destroy only very small amounts of the analyte.

4

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 15: Redox Titrations 1. (A) Which of the following is true of a redox titration? (i) Since the redox reaction is spontaneous, the equilibrium constant does not have to be large for an effective titration. (ii) The titration effectiveness is increased when the two half-reaction potentials are far apart. (iii) Without an indicator the equivalence point cannot be detected. (iv) For an effective redox titration, the reducing agent must always be in the buret. (v) The equivalence point in a redox titration is found at the highest voltage reached during the titration.

(B) In some pretreatment situations, the analyte must be reduced prior to titration. Of the following, which does not show an analyte being converted to its reduced form(s)? (i) CrO42- → Cr2+ (ii)MnO4- → Mn2+ (iii) Fe3+ → Fe2+ (iv)Mo3+ → MoO42(v) UO22+ → U3+/U4+ (C) Which of the following, if any, provides a true statement about a redox Gran plot? (i) The redox Gran plot is useful only if a constant ionic strength is maintained (ii) The Y-axis data is obtained from VT x 10-n Ecell/0.05916V (iii)The Y-axis intercept of the line in a proper redox Gran plot shows the volume of added titrant at the equivalence point. (iv) In order to determine the volume of titrant needed to reach the equivalence point, several precise measurements near the equivalence point are required. (v) none of the above are true statements A

2. (A) Methylene blue is a common redox reaction indicator. A useful solution can be prepared by dissolving 0.10 gram in 100 mL of water. The Eo’ value for this indicator is 0.53V. (n = 2) What is the useful voltage range of this indicator?

(B) Starch indicators are useful when iodine titrations (organic oxidations) are performed. Why is it considered an advantage in these titrations to have iodine as the titrant instead of the analyte?

3. An environmental chemist is going to field test the chemical oxygen demand factor for a stream in Yellowstone Park.

(A) The dichromate that will be used for the redox titration of the stream was standardized before leaving on the hike. Based on the following data, what is the molarity of the potassium dichromate solution? 1.352 grams of K2Cr2O7 (294.19g/mol) are placed in enough water to form 250.0 mL of solution.

(B) A 25.0 mL of stream water was treated with 10.00 mL of acidified K2Cr2O7 from part “A” and warmed. After the reaction mixture was cooled, the excess K2Cr2O7 was titrated with 0.2505 M Fe2+. This back titration requires 35.67 mL. How many moles of K2Cr2O7 reacted with the original organic material in the stream? What is the COD for the stream sample? (Use 3/2 ratio for the reaction between oxygen and dichromate.)

3+ 3+ (Overall titration reaction: 6 Fe2+ + Cr2O72- + 14H+ →  6Fe + 2Cr + 7 H2O) 4. High nitrite ion (NO2-) levels in ground water can cause many health problems (nitrite can bond to hemoglobin). One method to determine nitrite levels can be titration with Ce4+.

Ce4+ (aq) + e- → Ce3+ (aq) Eo = 1.72 V NO3 (aq) + 2e → NO2 (aq) Eo = 0.940 V 1

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

3+ (A) Balance the redox equation ___ NO2- + __ Ce4+ →  ___ NO3 + ___Ce (B) How many mL of 0.00400 M Ce4+ solution would be required to titrate 100.0 mL of 0.00100 M NO2- solution to the equivalence point?

(C) At the equivalence point show the relationships indicated here by placing the proper coefficient in place of “?”: [? NO2-] = [Ce4+]; [ NO3-] = [?Ce3+]

(D) Assume that the titration is carried out with an apparatus similar to the one shown here using a S.C.E. reference electrode (+0.241V); What is the E cell after the addition of 5.00 mL of the Ce4+ solution?

(E) What is the Ecell voltage at the halfway point to the equivalence point?

(F) What is the Ecell at the equivalence point?

(G) The redox indicator Tris(5-nitro-1,10-phenanthroline)iron has been suggested to use with this particular titration. Justify, with calculations, either accepting or rejecting this choice.

Tris(5-nitro-1,10-phenanthroline)iron (pale blue → red violet) Eo = 1.25

Ce4+ Titrant

NO2-

5. Iron deficiency is one of the main world-wide nutritional problems. Studies are currently being done to add iron-containing supplements to cereal grains in an effort to increase iron in the diets of children in Africa. FeSO4 is commonly used. The iron in a bowl of cereal may be titrated with a standard solution of 0.00500 M Tl3+. (Tl = thallium) 3+ + The balanced redox titration reaction is: 2Fe2+(aq) +Tl3+(aq) →  2Fe (aq) + Tl (aq) (A) How many mL of the 0.00500 M Tl3+ titrant are required to oxidize the Fe2+ in 50.0 mL of cereal solution that is 0.00100 M Fe2+? (Essentially, what is Ve?) + Tl3+ + 2e- → Eo = + 1.25 V  Tl

2+ Fe3+ + e- → Eo = + 0.77 V  Fe (B) What would be the Ecell at the halfway point of this titration?

(C) At the equivalence point, several concentration relationships are present: For example: [Fe3+] = ?[Tl+] What is the value of the “?”?

(D)What would be the Ecell at the equivalence point?

(E) What is the Ecell when you reach 2 x Ve? (F) Which, if any, of the following would make the best indicator for this titration? (circle one choice) (i) Methylene Blue Eo = 0.53 V n = 2 (ii) Diphenylamine Eo = 0.76V n = 2 (iii) Ferroin Eo = 1.06 V n = 1 (iv) Tris(5-nitro-1, 10-phenanthroline) iron Eo = 1.25 V n =1 (v) none of the above

6. A dilute solution of hydrogen peroxide (H2O2) is commonly found as a household disinfectant.

However, it can lose some of its potency over time. You could study the decomposition of aqueous H2O2 by performing a redox titration on a solution of H2O2 using a solution of KMnO4 as the titrant. (*Note H2O2 is not a pure liquid.) o O2 (g) + 2H+(aq) + 2e- →  H2O2 (aq) E = + 0.68 V 2

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

2+ o MnO4-(aq) + 8H+(aq) + 5e- →  Mn (aq) + 4 H2O(l) E = +1.51V

2+ The overall balanced titration reaction is: 2MnO4- + 6H+ + 5 H2O2 →  2 Mn + 5 O2 + 8 H2O

In the titration of 1.00 mL of 0.884 M H2O2 it was found that 17.65 mL of 0.0200 M KMnO4 was required to reach the equivalent point. (The [H+] in the solution was maintained at 1.00 M.) (A) At the equivalence point the [Mn2+] would be equal to ___ [O2] (Fill in the blank using proper stoichiometry.)

(B) What was the cell voltage after the addition of 1.00 mL of KMnO4? ** The reference electrode used was the Ag/AgCl with a Eo = +0.222V.** (C) What was the cell voltage at the halfway point in the titration?

(D) What was the cell voltage at the equivalence point?

(E) What was the cell voltage after adding twice the KMnO4 as used at the equivalence point?

7. Some analyses are best performed via the technique of “back titration.” For example, during the analysis of vitamin C with I3- a known excess of I3-,can be added to a sample. Then the unreacted I3- excess can be back titrated with thiosulfate (S2O32-). The important analysis of glucose (C6H12O6) in nutritional and other biochemical processes can be performed in a similar I3- reaction. The stoichiometry is shown in the following balanced reaction of I3- and glucose: C6H12O6 + I3- + 3OH- → C6H11O7- + 2H2O + 3 ITo begin the analysis, 50.0 mL of 0.0526 M solution of I3- was added to a sample containing glucose. The excess (un-reacted) I3- required 23.7 mL of 0.0867 M S2O32- in a back titration.

The stoichiometry of that is shown in the following balanced reaction of I3- reacting with S2O32-.

2S2O32- + I3- → 3 I- + S4O62How many moles of glucose were in the sample?

8. Suppose a lipid (oil) is extracted from a sunflower seed and purified. This lipid sample has a mass of 12.56 g. The sample is treated with 20.00 mL of a 0.550 M Br2 solution. After excess KI is added to produce iodine, the excess iodine required 6.76 mL of 0.488 M Na2S2O3 to reach the end point.

(A) What is the iodine number of the extracted oil?

(B) Would another oil, with an iodine number of 12, have fewer or more carbon-carbon multiple bonds?

9. When hydrogen ion activity is maintained at 1.00, any [H+] components of the Nernst equation may be ignored. However, other H+ values must be utilized to determine the cell voltage. For example, suppose a permanganate titration of Fe2+ were carried out when the hydrogen ion activity was 0.010. What would be the resulting EInd value for the permanganate?

3+ 2+ Redox reaction: 5 Fe2+ + MnO4- + 8 H+ →  5 Fe + Mn + 4 H2O 10. The permanganate and dichromate titrations of unknown Fe2+ samples are both H+ dependent reactions: 3+ 2+ 5Fe2+ + MnO4- + 8 H+ →  5Fe + Mn + 4 H2O

3

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

3+ 3+ 6 Fe2+ + Cr2O72- + 14 H+ →  6 Fe + 2Cr + 7 H2O

(A) If the molarities of MnO4- and Cr2O72- were the same, which would require the greater volume to titrate identical samples of Fe2+? Explain.

(B) Which titration would experience more of a change in the Ecell at the equivalence point (assume both were using the same reference cell) if the pH of the titrations was changed by one unit? Explain.

(C) At the equivalence point the calculation for EInd will have one key difference for the dichromate titration compared to the permanganate EInd. What is that difference? Why does it arise?

4

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS FOR TEST BANK CHAPTER 15 1. (A) (ii); (B) (iv); (C) (ii) 2. (A) Ecell = EoIn +/- (0.05916/n); 0.53 + (0.05916/2) = 0.56 V; to 0.53 – (0.05916/2) = 0.50 V (B) As an analyte starch and iodine form a dark violet-purple color that is slow to dissociate.

Therefore, the titrant may have difficulty reacting with the iodine.

1.352 g x

3. (A)

1mol 294.19 g

0.0250 L

= 0.1839 M

2+

(B) 0.2505 M Fe x 0.03567 L x

1mol Cr2O7 6 mol Fe

2−

2+

= 0.001489 moles Cr2O72- excess

0.01000L x 0.1839 mol/L = 0.001839 mol Cr2O72- added; 0.001839 – 0.001489 = 0.000350 mole Cr2O72- reacted 3 mol O

0.000350 mole Cr2O72- reacted x

2

2 molCr O 2 7

31.999 g O 2

0.000525 mol O2

1 mol O 2

x

2−

= 0.000525 mol O2

1000 mg

0.0250 L

1g

= 671.9mg/L = 672 mg/L

3+ 4. (A) ___ NO2- + 2 Ce4+ →  ___ NO3 + 2 Ce

(B) nTCTVT = nACAVA; 1(0.00400)VT = 2(100.0)(0.00100); VT = (C)

[2 NO2-] = [Ce4+];

2(100.0)(0.00100)

= 50.0mL

(0.00400)

[NO3-] = [1/2 Ce3+]

 CA VA − (nT /nA )(CT VT )   (nT /nA ) CT VT  

(D) At 5.00 mL; EInd = ENO3- = Eo’NO3- - 0.5916V log  2





EInd = 0.940 – 0.02958 log  (0.00100 x 0.100) - 1/2 ( 0.00400 x 0.00500)  = 0.9117;

(1/2) 0.00400 x 0.00500





Ecell =0.9117 – 0.242 = 0.671V (E) At halfway point; NO2- = NO3- so EInd = Eo’Ind; 0.940; Ecell = 0.940 – 0.242= 0.698 V (F) At the equivalence point; EInd = EInd =

(nAEo'A + nTEo'T ) (n A + nT )

(2 x 0.940 + 1 x 1.72) = 1.20; E = 1.20 - 0.242 = 0.958 = 0.96 V cell (2 + 1)

(G) Ecell = EoInd +/- 0.05916/1 = 1.25 – 0.5916 = 1.19V – 0.242 = 0.95 V; 1.25 + 0.5916 = 1.31.242 = 1.07V; Therefore, the color-changing range is from 0.95V to 1.07 V. This range covers the 0.96 V value for the equivalence point.

5

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

5. (A) nTCTVT = nACAVA; 1(0.00500)VT = 2(50.0)(0.00100); VT =

C. McLaughlin

2(50.0)(0.00100)

= 20.0 mL

(0.00500)

(B) Halfway to equivalence point; [Fe3+] = [Fe2+]; EInd = Eo’Ind = 0.77 – 0.222 = 0.548 = 0.55 V (C) At equivalence point [Fe3+] = 2[Tl+] (D) At equivalence point; EInd =

(2 x 1.25 + 1 x 0.77) = 1.09 V; E = 1.09 - 0.222 = 0.868 = 0.87 V cell (2 + 1)

(E) At 2 x Ve the [Tl3+] = [Tl+]; EInd = Eo’T = 1.25; Ecell = 1.25 – 0.222 = 1.028 = 1.03 V (F) choice (v) none of the choices show a range that includes 0.87 6. (A) [Mn2+] = 2/5[O2]

 C V − n /n C V  (B) EInd = EH2O2 = Eo’H2O2- - 0.05916V log A A ( T A ) T T  ; 

2

(nT/nA )CTVT



 0.884x0.00100−(5/2)0.0200x0.00100   = 0.644 ; Ecell = 0.644 – 0.222 = 0.42V (5/2)0.0200x0.00100  

= 0.68 – 0.05916V log 2

(C) At the halfpoint of the titration [H2O2] = [O2]; EInd = Eo’H2O2 = 0.68; Ecell = 0.68 - 0.222 = 0.46 V (D) At the equivalence point EInd =

(2 x 0.68 + 5 x 1.25) = 1.273; E = 1.273 - 0.222 = 1.05 V cell (2 + 5)

(E) 5.00 mL + 17.65 = 22.65 mL titrant added;

     n /n C V  0.05916V  o’ T A A A  EInd = ET = E MnO4- log   5 n    CTVT − A C V   nT  A A             2/5)0.884x0.00100 ( =1.51 - 0.01183 log  =1.50V 2   (0.0200x0.02265)− 0.884x0.00100    5    

(

)

Ecell = 1.50 – 0.222 = 1.28 V 7. 0.0500 L x 0.0526 M = 0.00263 moles I3- were added.

0.0237 L x 0.0867M x

1 mol I3

−

2 mol S 2 O 3

2−

= 0.00103 mole I3- in excess.

0.00263 – 0.00103 = 0.00160 mol I3- reacted with the glucose = 0.00163 mole glucose 8. mol Br2 added = 0.0200 L x 0.550 M = 0.0110 moles; mol S2O32- used = 0.00676 L x 0.488 M = 0.00330 moles mol iodine = 0.00330/2 = 0.00165 mol mol iodine reacted: 0.0110 – 0.00165 = 0.00935 mol; 100(0.00935)(253g/mol)/12.56= 18.8 = 19 (B) An oil with an iodine number of 12 has fewer multiple bonds than this oil.

6

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

9. EMnO4- = EoMnO4- -

0.05916 5



C. McLaughlin

   0.05916 1  ; = 1.51V log   = 1.32 V 8 5  (H + )8   (0.010 ) 

log 

1

10. (A) The electron change for the Fe2+ sample in this reaction is one electron per mole. For the MnO4- there is a five-electron change. For Cr2O72- there is a six-electron change per mole.

Therefore, each mole of dichromate will oxidize 6 moles of the iron. This means less dichromate will be required to oxidize all of the iron in the sample compared to the permanganate (where each mole oxidizes five moles of the iron in the sample). More permanganate will be required.

(B) The stoichiometry of the two reactions shows that the permanganate reaction involves 8 H+ per mole of Fe2+ present. The dichromate reaction shows a dependency of 14H+ per mole of Fe2+ present. When calculated in the Nernst equation, the exponent on the hydrogen activity for dichromate will be 14 compared to only 8 for the permanganate Nernst equation.

(C) The calculation for the EInd for dichromate will have a product component (0.05916 log [Cr3+] because the stoichiometry from Cr2O72- (reactant) to Cr3+ (product) is not 1:1.

7

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 16: Coulometry, Voltammetry, and Related Methods 1. (A) Which analytical technique requires four electrodes? (i) electrogravimetry (ii) direct coulometry (iii) direct current voltammetry (iv) amperometric titration (v) anodic stripping voltammetry (B) Which analytical technique combines coulometry and voltammetry to analyze trace metals? (i) electrogravimetry (ii) direct coulometry (iii) direct current voltammetry (iv) amperometric titration (v) anodic stripping voltammetry (C) Which analytical technique typically utilizes platinum gauze? (i) electrogravimetry (ii) direct coulometry (iii) direct current voltammetry (iv) amperometric titration (v) anodic stripping voltammetry 2. (A) Mining of metals involves engineering decisions, possible environmental impact, and possible profits. The percent of the metal in the obtainable ore will determine the possible profit to be made. Copper deposits in mining areas are typically small with as little as 0.4% copper.

Suppose a 3.5684 gram ore sample was properly treated for analysis via electrogravimetry.

The cathode in the process gained 0.01271 grams of mass. What is the percent of copper in the ore?

(B) In some cases electrogravimetry analysis oxidation of an analyte will form a product that will adhere to the anode. Such is the case when Pb2+ is oxidized to PbO2. How many grams of Pb2+ are oxidized from a solution when a current of 2.58 amps is applied for 5.00 minutes?

3. Green bell peppers have proven to contain significant amounts of vitamin C. Suppose one green pepper has a mass of 125 grams. The pepper was homogenized with water and titrated coulometrically (constant current) with iodine. After 15 minutes, at 70.00 mA the starch end point was reached. What percent of vitamin C was found in the pepper?

4. Some of the ground around older homes must be analyzed for lead contamination due to the previous use of lead-based paints on the houses.

(A) Suppose, after proper treatment, Pb2+ are detected using a DC voltammetry technique. A 9.07 x 10-3 M Pb2+ standard sample provided a limited diffusion current at 35.0 mA. What is the molarity of Pb2+ in an unknown soil extract (same background as the standard) that provided a limited diffusion current of 15.0 mA? (B) What should the value be reported as ppm?

5. In some communities the amount of arsenic (As3+) in the water supply has become an important health issue. One way to analyze this low concentration is via anodic stripping voltammetry (ASV.) A 100.0 mL water sample is prepared and the arsenic in the water is reduced to As3+. Then a potential is used at the working electrode to reduce the As3+ to Aso. 1

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Later, when the Aso was stripped from the electrode, 1.00 x 10-4 coulombs were required. What is the ppb of arsenic in the water?

6. What is the connection between the “double layer” and current in voltammetry?

2

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS TO TEST BANK CHAPTER 16 1. (A) (ii) (B) (v) (C) (i) 2.(A) 0.01271/3.5684 x 100 = 0.3562% (B) 2.58 amp x 5.00 min x 60sec/1min x 1mol e-/96500 coul x 1mol Pb2+/2 mol e- x 207.2 g/1mol = 0.8309 = 0.831 g of Pb2+ 3. 70.00 mA x 1 A/1000 x 15min x 60s/1min x 1mol e-/96500coul x 1mol I2/2mol e- x 176g/1mol = 0.05745 g vitamin C; 0.05745 g/125 g x 100 = 0.0460% 4. (A) Id = k x C; For this experimental design, from the standard solution, k = Id/C = 35.0/9.07 x 10-3 = 3859 mA/M; For the unknown; Id = 3859 x C; 15.0/3859 = 3.89 x 10-4 M (B) 3.89 x 10-3 mol/1L x 207 g/mol = 0.805 g/1000mL; 0.805g/1000 =X/106; X = 805 ppm 5. 1.00 x10-4 C x 1F/96500 C x 1molAs/3 mol F x 74.92 g/1mol As = 2.59 x10-8 g; 2.59 x 10-8g/100.0 g = X/1 x109 = 0.259 ppb 6. The double layer creates non-Faradaic current. This takes place when the applied potential is changed. A negative potential will attract positive ions from the solution at the electrode interface. This layer will then attract negative ions (hence the term double layer). Since there is no actual oxidation/reduction here, the current is not related to the Faradaic current used for the analyte measurement.

3

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chaper 17: An Introduction to Spectroscopy 1. The cover of our A-Chem textbook displays the false-color image of a forest fire in the Bitterroot National Forrest in Montana. These images can be used to track fire direction and intensity. The spectroscopic measurements were based on Infrared Radiation (IR) detected by an orbiting satellite.

(A) If some of the IR waves detected had a wavelength () of 4.2 m, what would be the energy of just one photon of the IR radiation?

(B) The enhanced image produced by using IR spectrometry can be converted to visible colors.

(As was done on the textbook cover.) Properly fill in the following blanks by circling the correct choice in each: (i) Visible light will have a (higher or lower) frequency than IR. (ii) The wavenumber of this IR radiation will be ( higher or lower) than that of visible light. (iii) In the visible spectrum, which has the longer wavelength? ( Red or Blue) 2. Measuring atmospheric carbon dioxide levels is critical to any decisions about regulating greenhouse gases. Several EPA studies have shown that greenhouse gas pollution is a serious problem now and for the future. CO2 absorbs strongly in the infrared range.

(A) If one photon of this absorbed energy is 4.62 x 10-20 Joules, what would be the frequency and wavelength?

(B) Which type of excitation is most likely caused when CO2 absorbs this infrared radiation? (circle choice) Bond breaking Electron excitation Vibration-rotation 3. Ultraviolet light (UV) has enough energy to excite electrons and, in some cases, even enough to break chemical bonds. Analytical chemists determine this absorbance using UV spectrophotometers to measure specific analytes.

(A) Suppose an absorbance measurement is made using a max of 285 nm. What is the energy per photon of the radiation?

(B) As solar-generated UV strikes Earth’s atmosphere, some is absorbed by ozone and some penetrates through to Earth’s surface. UV energy can damage skin, especially in high altitudes.

This UV is sometimes divided into classes.

Energy with a wavelength of 359 nm is classed in UV-A range.

Energy with a wave length of 291nm is classed in UV-B range.

When absorbed, which UV would provide the most energetic electron transition?

(C) While waiting in a grocery checkout line, you might notice the laser light being used to scan those UPC barcodes on products.) The laser light may have a  of 4.6 x 1014 Hz. The following list provides molecular processes related to absorbed energy of certain wavelengths. Which of the following transitions could this beam best match? (i) bond breaking (10-9 m) (ii) electron excitation (10-7 m) (iii) vibration (10-5 m) (iv) rotation (10-2 m) 4.(A) Typical spectrophotometric measurements are made with cell lengths of 1.00 cm.

However, some special cells are designed much longer. There are two common reasons for doing this. One is because it may not be possible, due to solubility limitations, to create solutions with high enough concentration to absorb measurable amounts of light. What is the other reason for using a longer cell?

(B) Beer’s law allows the use of light absorption or emission to make quantitative observations.

For example, a solution of the drug tetracycline showed a percent transmittance of 39.6% at a wavelength of 254 nm. What is the absorbance of the solution?

1

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

(C) If a 1.42 x10-3 M solution of tetracycline was known to have that absorbance calculated in part “B,” taken in a 2.00 cm cell, what would be reported as the molar absorptivity (or extinction coefficient) of tetracycline? (Include the proper label.)

5. (A) The velocity of light in a vacuum is 299 792 458 m/s. During this measurement the refractive index being used is 1.0000. If light is traveling through air where the refractive index is 1.0003, what percent error could using 1.0000 instead of 1.0003 cause?

(B) What percent of light (589 nm) will be reflected when it moves from air (n = 1.0003) to the side of a flint glass (n = 1.575) test tube?

6. (A) One of the following statements about the physical interaction of light with matter contains an error. Circle the one with the error: (i) Reflection occurs when light encounters a boundary between two regions that have different refractive indices and part of the light changes its direction of travel and returns to its original medium of travel. (ii) Refraction is the process that happens when a beam of light changes directions as it passes through a boundary between two media with different indices of refraction. (iii) Scattering takes place when a photon collides with another particle such as an atom or molecule causing a change in the direction of travel for the photon. (iv) Diffraction happens when waves spread around an object and the waves recombine in a Rayleigh scattering pattern.

(B) If the interplanar distance in a diamond was known to be 2.06 x 10-10 m, what would be the expected first order constructive interference band observed from exposing the crystal to X-rays with a 5.5 x10-11 m?

7. Spice made from saffron has been used for thousands of years. It can impart a distinct bittersweet taste and it is used in some dyes. It may be the world’s most expensive herb.

Ultraviolet and visible spectroscopy can be used to analyze extracts to determine if cheaper adulterants have been added to the expensive spice.

(A) The following absorbance data (at 495 nm) was collected on one specific adulterant: M A 0.0010 0.20 0.0020 0.45 0.0040 0.80 Unknown 0.60 Prepare a Beer’s Law graph, report the approximate b value for the compound, and report the approximate concentration in the unknown.

0.0010 M 0.0020 0.0030 0.0040 (B) Assume that the following graph represents the absorbance spectrum for the adulterant. A spectrochemical analysis is to be performed in a solution that contains both the adulterant and the key saffron pigment. Sketch a dotted line on the graph representing the absorbance of the pigment that would enable a reasonable simultaneous determination.

8. (A) Here, again, is the same graph as was used in question #4. Instead of measuring absorbance at 495 nm, someone chooses to make their analysis at 550 nm. Address these two effects would be noted: (i)  would : be greater, be smaller, remain constant (ii) Selecting the measuring wavelength in this region would cause less absorbance and perhaps less sensitivity, but what other problem about the 2

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

measurement with this light could also arise?

(B) Which instrument set up (schematics shown) depicts an emission type measurement device?

(C) An absorbance measurement was made on a saffron spice sample. Complete any missing information: A = 1.05

%T = ______

 = 808 _____

M = 0.00026

b = ______ cm

(missing label)

9. Determining the level of phosphorus in urine is an important health indicator. Phosphate can be reacted with a molybdenum compound to form a colorful blue compound (absorbs at 690 nm). A patient produced 1350 mL of urine in a day. A one mL sample of this was treated and diluted to 50.0 mL. This colorful, diluted solution gave an absorbance reading of 0.635. (The molar absorptivity value, , = 6610 M-1cm-1.)

(A) What is the molarity of phosphorus in the diluted sample?

(B) What is the molarity of phosphorus in the urine?

(C) How many actual moles are in the entire urine sample?

10. A recent court case awarded residents near a zinc mine run a nearly $200 million settlement. Zinc in the soil may have been determined by absorption analysis.

(A) A series of standard zinc solutions were prepared from properly treated zinc samples. The absorbance of each of these standards (at max) were recorded as: Absorbance Standard concentrations 0 0M 0.10 1.0 x10-4 M 0.50 5.0 x10-4 M 1.00 1.0 x10-3 M What is the molar absorptivity () of the zinc compound being used in this technique?

(B) A soil sample was extracted and prepared in a 1.00 L solution. (The same method was used to prepare zinc standards.) This soil extract solution was diluted in a 1:25 ratio. (That is, 1 mL of the soil solution was added to enough water to produce 25.0 mL of diluted solution.) An aliquot of this solution was put in a spectrophotometer. An absorbance of 0.65 was obtained.

What is the molarity of this diluted solution?

(C) What is the molarity of the zinc in the original 1.00 L of soil solution?

(D) Convert the molarity of the 1.00 L solution to ppm: (1 mol of Zn = 65.4 g/mol) (E) Two zinc samples from the same soil extract, with the identical Zn ppm values, produced different absorbance readings. Describe three possible explanations.

3

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

SOLUTIONS FOR CHAPTER 17 TEST BANK 1. (A) E = hc/; 6.626 x10-34(3.00 x108)/(4.2m x 10-6m/1m) = 4.73 x10-20 J (B) higher ; lower wavenumber; red 2. (A) 4.62 x10-20 J = h; 4.62 x10-20J/6.626 x10-34 J s =  = 6.97 x1013 Hz; c=  x ; c/ = ; 3.00 x108 m/s x 6.97 x 1013Hz = 4.3 x10-6 m (B) IR typically has an energy range to affect absorption - vibration modes.

3. (A) 285 nm x 1m/109nm = 2.85 x 10-7m; E = hc/ = 6.626 x10-34(3.00 x108m/s)/2.85 x10-7m = 6.97 x10-19 J (B) Shorter  will have higher ; higher  will provide more E; Therefore, 295 nm has higher E (C) (ii) c = ; 3.00 x108/4.6 x1014 =  = 6.5 x 10-7 m 4. (A) If the molar absorptivity () of an analyte is very low, or if the concentration is low such as with a gas, a longer light path could produce enough change in absorbance to be used in an analysis.

(B) A = - logT = -log(0.396) = 0.402 (C) A = bC; A/bC =  = 0.402/(2 x 1.42 x10-3) = 142 M-1cm-1 5. (A) [(1.0000 – 1.0003)/1.0000 ] x 100 = 0.03%

( (

) )

2 PR n2 − n1 (1.525 − 1.0003)2 = 0.04317 x 100 = 4.317% = (B) Fresnel’s equation: ; X = 2 PO (1.525 + 1.0003)2 n2 + n1

6. (A) (iv) is false, Rayleigh scattering is classified as small particle scattering, while diffraction describes light passing around objects.

(B) sin  =

nλ 2d

=

- 11 1(6.3x10 m ) 2(2.06 x10

- 10

= 0.152;  = 8.75o

m)

7. (A) b = slope ~ 0.6/0.003 ~ 200 C~ A/b =0.60/200~ 0.003M

(B)

8. (A) b would be smaller; the measurements could have less accuracy since 550 represents a wavelength on the side of the absorbance peak where / is larger. Polychromatic light is more likely part of the measurement than if the measurement is done in an area where  is 4

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

slight.

(B) Circle first choice (C) A = - logT; 10-1.05 = T; 0.0891; %T = 8.9  = 810

L mol cm

A = bC; A/C = b = 1.05/(808x0.00026) = 5.0 cm 9. (A) A = bC; 0.635 = 6610 M-1cm-1(1cm)C; 0.635/6610 = 9.61 x10-5M (B) 9.61 x10-5 M is present in a solution that was diluted 1:50, therefore the original solution would be 50 x diluted M; 50 x 9.61 x10-5M = 4.80 x10-3 M (C) 4.80 x10-3 mol/L x 1.35L = 6.484 x10-3 moles 10. (A) slope = Y/X = 1.0 x103 M-1cm-1 (B) A = ebC; 0.65 = 1.0 x103 M-1cm-1 (1cm) C; 0.65/ 1.0 x103 M-1cm-1 = 6.5 x10-4M in the diluted sample (C) 25.0 x 6.5 x10-4M = 1.6 x 10-2 M in original sample (D) 1.6 x10-2 mol/L x 65.4 g/mol = 1.05 g/L; 1.0 x 103 ppm (E) (i) Perhaps the zinc samples were reacted with different color-complex forming reactants with different molar absorptivity values. (ii) Perhaps the two samples were measured at different wavelengths where the molar absorptivity of a complex would be different. (iii) Perhaps the wavelength and complex forming reactant were the same, but different cell paths were used.

5

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 18: Molecular Spectroscopy 1. Although both involve exciting ground state conditions to excited molecular states, UV-vis and IR spectroscopy do have unique properties. Read each of the following descriptions, then indicate which apply to UV-vis only, IR only, or both: (A) Typically causes electronic transitions (B) Absorption bands tend to be broad as exciting wavelength changes (C) Requires a source of light (D) Cannot use glass sample holders (E) Uses a grating to separate light 2. (A) A visible light spectrophotometric determination of penicillin makes use of a color-forming reaction of the penicillin. If this compound absorbs at 475 nm with an  = 1560 M-1cm-1, what would be calculated as the limit of detection with an instrument, using a 1.00 cm cuvette, with its smallest absorbance measurement at 0.005?

(B) Would the lower limit of detection be unchanged, smaller, or larger if  were larger?

(C) Would the lower limit of detection be unchanged, smaller, or larger if the cell length (b) were increased?

3. (A) Suppose you develop a new method for using visible light to investigate dextromethorphan (found in some over-the-counter cough suppressants.) Using max of 440 nm with a 1.00 cm cuvette, the  value of the colored complex used in the preparation is 0.0366 mg/L . What is the dextromethorphan concentration when an absorbance value of 0.658 is obtained? (Note; the data is fictitious, only to be used for this problem.)

(B) Another analysis is attempted on a different sample. 10.0 mL of this sample produced, in the same reaction as in part “A,” an absorbance of 0.405. When a 1.00 mL spike, was added, with a concentration of 25 mg/L dextromethorphan, the absorbance became 0.453. What is the dextromethorphan concentration of the original sample?

4. The color plate in the middle of the textbook has an example of a spectrophotometric titration for the determination of the end point in a titration of Cu2+ with triethylenetetramine (trien). The reaction was followed at 580 nm as a purple color in the nine flasks (each containing 1.43 x10-3 M Cu2+) became more intense with increased additions of the trien compound to each of the flasks. Based on the following data, determine the end point of the titration: Vol of titrant Abs 0.00 0.00 1.00 0.031 2.00 0.063 3.00 0.093 5.00 0.158 8.00 0.244 10.00 0.238 15.00 0.226 20.00 0.217 (B) Which type of spectrophotometric titration is the reaction? (A = analyte; T = titrant; P = product) (i) Only A absorbs light (ii) Only T absorbs light (iii) Only P absorbs light (iv) A and T absorb light 1

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

(v) T and P absorb light 5. In this problem, assume that a simultaneous spectroscopic method will be used to determine the cobalt and nickel content of a soil extraction sample. After proper treatment with a color producing complex, the absorbance of the aqueous sample was measured from 450 to 490 nm.

(A) Assuming that the nickel complex has a max at 475nm, and the max for the cobalt complex to be at 468 nm, after obtaining an absorbance at each of the two wavelengths, you would still need to obtain four other factors before you could determine the concentrations of nickel and cobalt in the unknown mixture. One is provided here, now identify the other three: 468Co; ____________; ____________; _____________; (B) The total absorbance value of a sample, at 468nm, would be equal to: Abs468 = _____________ + _______________ (fill in the blanks, no actual values needed) (C) If the measurements for this A vs Concentration graphs were made with a 2 cm cell instead of 1 cm, which, if any, would get larger: A, , neither, both. (circle choice) 6. There are two main forms, differing only slightly, of the important photosynthesis compound chlorophyll. In a certain solvent, chlorophyll-a solutions have a strong absorbance at 663 nm.

Chlorophyll-b solutions have a strong absorbance at 649 nm. Suppose the following experimental data were collected. At 663 nm, chlorophyll-a has an  = 8.5 x105 M-1cm-1 while chlorophyll-b has an  = 1.8 x105M-1cm-1. At 649 nm, chlorophyll-a has an  = 1.5 x105 M-1cm-1 while chlorophyll-b has  =9.5 x 105 M-1cm-1. In an experiment the following (fictitious) values were obtained. Absorbance, in a one centimeter cell, at 663 nm of a mixture of chlorophyll-a and chlorophyll-b was1.02. At 647 nm, the same mixture produced an absorbance of 0.541.

What are the concentrations of chlorophyll-a and chlorophyll-b in the mixture?

7. (A) The most obvious difference between the UV-vis absorption spectra and IR spectrum for a compound is the number of peaks. Why are there so many more peaks present in the IR result?

(B) Why are the sample holders for IR absorbance experiments often made from ionic solid pressed disks instead of glass vials?

(C) Unfortunately, the labels of two flasks, both containing colorless liquids, have fallen off. One of the flasks contains cyclohexane while the other contains cyclohexene. What advantage would an IR analysis of the liquids provide in correctly replacing the two labels?

8. Fluorescence and phosphorescence are used as markers for DNA and amino analysis.

(A) Of the following comparisons, which (if any) is accurate? (Circle one best choice) (i) Fluorescence in a sample has a longer lifetime than phosphorescence. (ii) Fluorescence takes place from a singlet (S1) state. Phosphorescence from a triplet (T1) (iii) Fluorescence emission is generally in a longer  region than phosphorescence. (iv) Fluorescence emissions are generally observed at a lower  than phosphorescence. (v) None of the above statements makes an accurate comparison.

(B) Fluorimeters (fluorescence spectrophotometers, and UV-vis spectrophotometers) have several, but not all, design aspects in common. Which, if any, of the following is NOT true? (i) Both use an incoming light source to cause excitation. (ii) Both use monochromators. (iii) Both have sample holders (iv) Both have their detectors positioned at 90o away from the incoming light source. (v) All of the above are true.

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C. McLaughlin

(C) Of the following, which offers the best scientific reasoning for explaining why fluorescent measurements are typically more sensitive than absorbance measurements? (i) Fluorescent measurements use longer , allowing more of an energy change to be noticed. (ii) Energy levels are closer together for fluorescent states, allowing greater sensitivity. (iii) The luminescence of fluorescent measurements is detected against a dark background. (iv)The excited state needed for fluorescence is achieved through vibrational relaxation which is very selective. (v) Absorbance takes longer than fluorescence so it is not as sensitive.

9. The nutritional importance of various forms of selenium have been investigated for their roles as a micronutrient and possibly helping preventing cancer and other diseases. Brazil nuts are an excellent source of selenium. The concentration of selenium can be determined by combining the selenium with a fluorescent component. There is an analogous Beer’s Law equation for fluorescence intensity: I = KC, where K is a proportionality constant, and C is the concentration.

(A) What is K, if I = 512 at 370 nm for a 1.5 X 10-6 M solution of the component? (i) 3.4 x 108 M (ii) 3.4 x 108 M-1 (iii) 2.9 x 10-9 M-1 (iv) None of these (B)Select ONE of the following aspects of fluorescence to briefly explain: (i) Why is the emission spectra of fluorescence lower in energy than the excitation spectrum? (ii) Why does phosphorescence last longer than fluorescence? (iii) Why does phosphorescence of a sample increase with colder temperatures?

10. (A) Soy-based proteins are found in many flours, infant formulas and other meat products.

Suppose a fluorescence analysis of soy-based flour deterioration used a calibration curve based on fluorescence intensity (I) and concentration of an oxidative product (mg/mL). The slope of the calibration line was 3600. An unknown sample was then treated the same way to determine its concentration. If the unknown produced an intensity (I) of 32 (arbitrary units) what would be the concentration of the oxidative product in the unknown?

(B) Another researcher uses the same procedure, as stated above, on some soy samples.

However, a matrix problem occurred that could be solved using the process of standard addition. A 10.0 mL sample of an unknown gave a fluorescence reading (I) of 57.

1.0 mL spike (1.0 mg/mL of the compound) when added to the sample produced I = 125.

What is the concentration of the unknown compound in this analysis?

11. (A) Suppose an experiment is performed where plants are grown in higher atmospheric CO2 levels. The chemical profiles of the plants could be monitored via fluorescence. (A profile may measure fructose, aldehydes, and chlorophyll, etc.) The following (fictitious) data were collected during one study. Fluorescence measurements, detected at 685 nm, on various chlorophyll solutions of known concentrations produced a calibration curve with a slope (intensity vs concentration) of 25. An extract of chlorophyll from spinach was collected and treated to make a 1:10 dilution. The diluted extract provided an intensity reading of 7.8. What was the concentration of chlorophyll in the original (undiluted) extract?

Intensity

slope = 25/g/L

Concentration (g/L)

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

(B) As part of Earth’s carbon cycle, blue-green algae in the oceans take CO2 from the atmosphere and, through photosynthesis, form sugars. This is currently being studied via attaching fluorescent tags to specific proteins in the cells. Answer the following fluorescent related questions: (i) Which has a longer , the light used to excite fluorescent molecules or the light emitted by fluorescence? (ii) Which is the faster process: fluorescence or phosphorescence? (iii) Which involves the triplet state: fluorescence or phosphorescence? (iv) Which type of molecule, rigid or flexible, typically is more likely to demonstrate fluorescence? (v) Which typically emits energy with the higher frequency: fluorescence or phosphorescence?

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SOLUTIONS TO CHAPTER 18 TEST BANK 1. (A) UV-vis; (B) UV-vis; (C) Both; (D) IR; (E) Both 2. (A) A = ebC; 0.005 = 1560M-1cm-1(1.00 cm)C; 0.005/(1.00 x1560) = 3 x10-6 M (B) If  were greater, more light would be absorbed per mole of penicillin, LOD would be lower (C) If “b” were increased, more chromophores would interact lowering the value for LOD 3. (A) A = bC; A/b = C = 0.658/0.0366 = 18.0 mg/L Ao Co (Vo + Vs ) (B) Standard Addition: = A sp Cs Vo + Cs Vs ) Co (10.0 + 1.00) 0.405 = ; C = 22.1 mg/L 0.453 (25.0 x10.0) + (25.0 x1.00 ) 4.Abs .5

end point ~ 8.0 mL; (B) Only P absorbs appreciably

0 5

10

15

20

mL of Trien added 5. (A) 468Co; 475 Co; 475 Ni; 468 Ni (B) Abs468 = 468 CobCCo + 468 NibCNi (C) If the cell path was increased, more chromophores of both nickel and cobalt would absorb ligh, so A would increase 6. A647 = 1.5 x105(1)Ca + 9.5 x105(1)Cb = 0.541 A663 = 8.5 x105(1)Ca + 1.8 x105(1)Cb = 1.02

(0.541− (9.5x105 Cb )) (1.5x105 ) Then substitute that expression into equation #2 and solve for C : (0.541− (9.5x105 Cb )) + 1.8 x10 C ; 1.02 = 8.5 x10 x (1.5x105 ) Solve equation #1 in terms of Ca =

b

5

1.02 8.5 x10

5

5

b

5 x 1.5 x10 = 81150 – 1.42 x1011Cb + 2.70 x1010Cb

- 81150 = -1.15 x1011Cb; Cb = 7.06 x10-7 M; 1.5 x105(1)Ca + 9.5 x105(1)7.06 x10-7 = 0.541 Ca = [0.541(9.5 x105 x 7.06 x 10-7)] /1.5 x105 = 2.4 x10-6M

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C. McLaughlin

7. (A) IR absorption excites vibrational and rotational states in molecules. UV-vis typically excites energy level transitions for electronic states. There are more available rotational and vibrational quantum states than there are electronic states.

(B) The glass used to construct traditional cuvettes absorbs IR radiation. If those glass vials can be replaced by a non-absorbing material such as pressed ionic disks, the absorption can be related to the analyte.

(C) Cyclohexane will have no strong absorption of IR in the alkene regions (1680 -1620 cm-1) The liquid that shows strong absorption in that region is cyclohexene (unsaturated bonding).

8. (A) (ii); (B) (iv); (C) (iii) 9. (A) I = KC; 512/1.5x10-6 = K = 3.4 x108 M-1 (B) (i) When a fluorescent analyte is excited, an electron moves from a lower level to a higher energy level. The vibrational level in the excited state is also excited. That quantum change requires a specific amount of energy. When the electron begins to relax it will move from a high vibrational level in the excited state to a lower vibrational state in the excited electronic state before relaxing to the lower electronic state. That change means that the energy released from the excited electronic state to the lower electronic state will be less than the energy that was used to achieve the excited state. (see Figure 18.18) (ii) Fluorescence takes place from a singlet excited state to a singlet ground state. For phosphorescence, an electron moves from a singlet state to a triplet state. Then the electron must undergo a “spin forbidden” transition from a triplet state to a singlet state. This latter process is less likely to occur and is slower. (iii) The triplet to singlet transition in the relaxation process of phosphorescence is slower than the singlet to singlet state relaxation process of fluorescence. Since more time is involved, collisions that transfer energy without emission are most likely to take place. By lowering the temperature, fewer molecular collisions of this nature will be taking place. Therefore, other energy releasing processes (such as phosphorescence) have greater likelihood.

10. (A) I = KC; 32/3600 = 8.9 x10-3 mg/mL (B) Standard addition: Co (10.0 + 1.00) Io C (V + Vs ) 57 ; = ; 5.016 = 11.0Co; Co = 0.456 = 0.46 mg/mL = o o Isp CsVo + CsVs ) 125 (1.0 x10.0) + (1.00 x1.00 ) 11. (A) I = KC; 7.8/25 = 0.312 ug/L; x 10 dilution factor = 3.1 g/L (B) (i) emitted (ii) fluorescence (iii) phosphorescence (iv) rigid (v) fluorescence

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C. McLaughlin

Chapter 19: Atomic Spectroscopy 1. (A) Volatilization, atomization, and ionization are all important processes to consider in atomic spectroscopy. How is energy used in these three processes?

(B) Which is not a desirable process for most atomic absorption analysis? Explain your choice.

(C) What are the two considerations for using a certain temperature in the heat source in AAS? 2. Samples being prepared for atomic spectroscopy could be subject to AAS, AES, or AFS techniques. Briefly state what atomic property is related to concentration in each technique.

3. Two methods commonly used in atomic absorption spectroscopy to generate energy used to produced analytes in an excited state are the hallow cathode lamp and graphite furnace. Each can be effective, but each also has limitations. Cite one advantage and one problem presented by each of these important energy sources.

4. The use of NaCl in processed foods has recently come under new study. Suppose atomic emission spectroscopy was going to be employed for an analysis of sodium concentration in a food sample. After extracting the sodium from the sample, a flame ionization technique used a fuel mixture that produced a temperature of 2267 K (propane). Another lab analyzed the same sample but had a 3342 K flame (acetylene). By what factor did the sodium atoms in the excited state increase? (Use Pi = 3; Po = 1 and 3.37 x 10-19 Joule for the energy transition.)

5. Two complications found in flame-based atomic measurements arise when refractory oxides and other refractory compounds form. Explain how each problem can be minimized. + 6. Using the reaction of Na →  Na + e to represent undesirable ionization during an atomic

absorption measurement, how could undesirable ionization in a sample be reduced?

7. Copper has many world-wide applications from jewelry to the electronics industry. An atomic emission spectroscopy experiment is set up to determine copper measurements. The copper level is detected at 216.6 nm. In this experiment, a blank sample with the same matrix as the samples to be measured provided an emission intensity of 0.5. A 5.00 ppm standard provided an emission intensity of 25.0. An unknown sample, treated the same way in the instrument, provided an emission intensity of 20.0. What is the concentration of copper in the unknown?

8. An atomic absorption analysis is to be done on the calcium in a sample of powdered milk. A 12.55 gram sample of the powder is wet ashed in nitric acid and diluted to 50.00 mL with distilled water. A 1.00 mL aliquot of this is transferred to a 50.00 mL solution using EDTA. Using a wavelength of 422.6 nm in an atomic absorption experiment, this diluted sample showed an absorbance of 0.375. A standard calcium sample (treated the same way) with a known concentration of 20.0 ppm provided an absorbance of 0.859. What is the ppm of calcium in the powdered milk?

9. (A) A hollow cathode tube spectrophotometer is fitted with a chopper. Describe the purpose of this part of the spectrophotometer.

(B) Suppose a chopper in such an instrument allows a detected signal of 56 in the open position, but only 42 when the chopper is in the closed position. A blank with the same matrix as the sample provides a signal of 79 in the open position and 1.0 in the closed position. What is the absorbance being measured for this sample?

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

10. A brass sample is to be analyzed for zinc. The sample is dissolved and the zinc separated.

The resulting 50.0 mL zinc solution has an absorbance of 0.865 in an atomic absorption analysis. An identical 50.0 mL sample of the zinc solution receives a 5.00 mL spike with a 5.00 ppm zinc concentration. This second solution provided an absorbance of 1.075. What is the concentration of zinc in the original sample?

SOLUTIONS FOR CHAPTER 19 TEST BANK 1. (A) Volatilization uses added energy to change non-gas samples into gas to enter flames.

Atomization requires heat to turn compounds into atoms. Ionization uses energy to strip electrons from atoms.

(B) Ionization produces ions that may have different absorption and emission qualities than neutral atoms. This can cause errors when trying to relate atom concentration (or identification) to absorption of specific wavelengths.

(C) The temperature should be high enough for volatilization and atomization, but not so high that ionization occurs.

2. In AAS, samples are subjected to exciting light. The measurement that relates to concentration is concerned with the light that does not strike a detector. The analyte absorbs some light in an excitation process. With AES, light from an excited analyte is measured by a detector. The intensity of this emitted light is related to the concentration of analyte in a sample.

In AFS, light is used to excite an analyte, but the light emitted is of a different frequency. The intensity of emitted light is detected and directly related to concentration.

3. The hollow cathode lamp source provides light of the same frequency as that needed to promote the analyte from ground to excited state. However, when excited atoms collide with other atoms, they lose some energy, thus shortening their time in the excited state. If this time decreases the change in energy will increase (Heisenberg’s uncertainty principle). A change in energy widens the bands used in measurements. The graphite furnace advantage provides a longer time for analyte atoms to be exposed to exciting radiation so more atoms can participate in the analysis, thus improving the limit of detection and having the small sample option available. The disadvantage here lies in a drop in precision and contamination problems.

4. Use the Boltzmann distribution equation for both temperatures to determine ratio of excited to ground at both temperatures: Ni

=

Pi

No

Po

Ni

= 3e

No

e

( − ΔE/(kT))

;

Ni

= 3e

( −3.37x10- 19 /(1.38x10- 23 (2267))

= 6.29 x10-5 for propane flame

No

( −3.37x10- 19 /(1.38x10- 23 (3342)))

= 2.01 x19-3 for acetylene flame

The ratio of these two values is nearly 32 to 1 indicating that the increase in the ratio of excited to ground state atoms is significant (~ 32 times).

5. Refractory oxides such as Al2O3 can be prevented from forming in aluminum measurements by not allowing oxygen in the fuel-oxidant mixture. Replacing oxygen with N2O removes oxygen in the product of combustion so it is not available to form refractory oxides.

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Other refractory compound formation can be prevented by adding, to both samples and standards, releasing or protective agents. If EDTA is added to calcium-containing samples, a stable Ca-EDTA complex can form. This protects the calcium from reacting with phosphates or sulfates that would not allow atomic calcium to be in the flame. The EDTA is later burned off of the complex, thus freeing the calcium for analysis.

6. If lithium is added to the sample (and standards) in the sodium analysis it will provide a constant source of electrons at a high enough level to keep the sodium ionization reaction shifted, via LeChatelier principle, further to the reactant side of the reaction.

7. After correcting each measurement for the blank reading, the intensity ratio provides the concentration ratio.

X

=

5.00ppm

(20.0 − 0.5) ; X = 3.98 ppm (25.0 − 0.5)

8. Use the standard to solve for k in A = kC; 0.859 = k(20.00 ppm); k = 0.0430 Next, use that value of k to solve for ppm in diluted sample: 0.375= 0.0.0430C ; C = 8.73 ppm Finally, correct for the 1:50 dilution: 8.73 ppm x 50 = 437 ppm in the original solution.

9. (A) This device alternately blocks and opens the light coming from the light source before it approaches the flame and sample. In this way the detector receives light from the light source and the light from atoms that are relaxing from their excited state when the chopper is in the “open” position, but the detector receives only emission light from the analyte when the light source is temporarily blocked. The difference in the two signals is the signal for only the analyte.

(B) First determine the values of I and Io; 56-42 = I = 14; Io = 79-1 = 78 Next, determine the value of transmittance (T); T = 14/78 = 0.179; Finally determine A; A = -log(0.179) = 0.746 = 0.75 10. Using the standard addition equation and making appropriate substitutions: 0.865 1.075

=

Co (50.0 + 5.00) Co (50) + (5.00x5.00)

; Co = 1.36 ppm

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 20: An Introduction to Chemical Separations 1. In the approximately 40 years since the first Earth Day, we have seen that chemistry has done much “for us and to us.” Perfluorooctanoic acid (HPFOA), used to make Teflon-coated cookware, has caused some controversy. When 80.0 mL of an aqueous solution of HPFOA was extracted with 50.0 mL of an organic solvent, a fraction of 0.75 of the original acid was extracted.

(A) The extraction can be represented as: HPFOA(aq) →  HPFOA(organic) What is the expression, using this system, for the partition constant and partition ratio?

(B) Calculate the KD for this system. (For the calculation assume that the weak acid nature of the compound is ignored so that essentially only the HPFOA is considered (rather than including PFOA- form)).

(C) This weak acid has an approximate Ka = 1.6 x 10-4. Now calculate Dc for the same experiment with an 80.0 mL solution of the acid (pH = 2.40) when it was extracted with a new 50.0 mL sample of the same organic solvent from part “A” (D) Using the value of Dc instead of KD for the extraction predicts what fraction will be extracted into the organic phase?

2. (A) Lycopene and beta carotene are important antioxidants found in some fruits and vegetables. Suppose an organic solvent was going to be used to extract and measure lycopene from a new hybrid tomato. A sample of aqueous tomato paste was introduced into a separatory funnel. Then an organic solvent was added. The funnel contents were swirled for several minutes. After extraction, the concentration of lycopene in the organic solvent was found to be 6.5 x 10-4 M while the aqueous phase maintained a concentration of 1.5 x 10-5. What is the value of the partition ratio (KD) with this extraction?

(B) Using the value of KD from “A,” what fraction of lycopene would be extracted from the aqueous tomato paste into the organic phase if you began with 100.0 mL of tomato paste and added 50.0 mL of the organic solvent?

(C) What fraction would be extracted from 100.0 mL of aqueous tomato paste if the extraction was set up to use two 25.0 mL portions of the organic solvent?

3. (A) Tomatoes also contain vitamin C (ascorbic acid; Ka = 7.9 x 10-5). At a pH of 3.50, what is the value of the distribution coefficient (DC) for ascorbic acid using an organic solvent where the KD value is 7.2?

(B) Now, suppose that solvent is going to be used to extract ascorbic acid from a pH = 3.50 aqueous tomato paste solution. What fraction of the ascorbic acid would remain in the aqueous phase if 100.0 mL of paste is extracted once with 50.0 mL of the solvent?

(C) If the pH of the tomato paste was 2.5, would the expectation be that more or less of the ascorbic acid would be extracted into the organic phase? What is the chemical basis of this prediction?

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C. McLaughlin

4. An episode of a popular crime scene investigation television episode used a separation technique as part of the plot. Searching for clues of a crime, an extract from the vitreous humor of the eye (liquid in the eyeball) is examined for the presence of a particular drug. The procedure uses non-polar toluene to extract the drug. (A) When toluene, in contact with the vitreous humor fluid, reached equilibrium it was determined that 0.0100 moles of the drug was found in 0.020 L of toluene and 0.0010 moles of the drug was in 0.010 L of the aqueous vitreous humor. What is the value for the partition ratio?

(B) Using that value of KD from part “A” in another experiment, what fraction of the drug would remain in 0.0050 L of the vitreous humor being extracted with 0.010L of toluene?

(C) What fraction is extracted into the 10.0 mL of toluene?

5. An episode of a popular crime scene investigation television episode used a separation technique as part of the plot. Searching for clues of a crime, an extract from the vitreous humor of the eye (liquid in the eyeball) is examined for the presence of a particular drug. The procedure uses 0.010 L of non-polar toluene to extract the drug from 0.0050 L of vitreous humor?. However, they also find that the drug has the properties of a weak base. + -5 B + H2O →  BH + OH ; Kb = 1.0 x 10 .

(A) At pH = 7.80 (and using a KD value of 5.8) what fraction of the basic drug would remain in 0.0050 L of the vitreous humor? What fraction would appear in the toluene phase?

(B) If the pH was raised to 8.98 would the extraction of the base into non-polar toluene be more or less effective? Using equilibrium expressions and LeChatelier’s Principle, provide a brief explanation for your answer:

6. (A) A solute with a partition ratio (KD) of 2.0 is extracted from 10.0 mL of aqueous solution (phase 1) into an organic solvent (phase 2.) What volume of phase 2 is needed to extract 99.0% of the solute into phase 2?

(B) The same extraction set up of the solute from 10.0 mL of phase 1 is repeated using three equal volume extractions. What is the volume that should be used (of phase 2) in each of the three extractions to achieve 99.0% extraction?

7. Decabromodiphenyl ether (DECA) has been used as a flame retardant in carpets and furniture. However, due to some concerns over health problems, the industry has begun phasing out its use. Suppose that the following chromatogram was produced when analyzing compounds found in carpets.

(This diagram is used to model this problem –not actual data.)

(A) If peak #6 was DECA, what would be calculated as its adjusted retention time? (The small mark at 1.0 min represents void time.)

(B) What would be calculated as the retention factor (k)? Explain why increasing the flow rate of the mobile phase would affect the tR but not affect k.

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(C) The base width of peak #6 is 0.60 min. What is the number of theoretical plates involved in the separation of DECA? 8. (A) The compounds noted at peaks #1 and #2 are not well resolved. List two chemical reasons related to the compounds that could cause this unfortunate lack of good separation.

(This is about the molecules, not the column.)

(B) The resolution between peaks #1 and #2 (from question six) has been calculated to be about 0.98. (Obviously, less than ideal.) If N for peak #2 was 580 and the column was 5.0 meters in length, what would be the height equivalent to a theoretical plate (HETP)? If all parameters were kept the same, lengthening the column would have what effect on HETP? “no effect” “better separation” “worse separation” (circle one) (C) When peaks are detected from emerging compounds in GC work, narrow peaks are a mark of high efficiency. Many peak-broadening processes involve diffusion of the solute. Briefly describe one diffusion-related cause of peak broadening.

9. (A) Resolution can be calculated two ways. One direct method uses tR values and peak base widths (wb.) In the following sentence, match the two choices by circling one correct response in each set of parentheses that would allow the sentence to be logical for better peak resolution: “A (large or small) difference in tR1 and tR2 combined with (large or small) base widths will result in better resolution.”

(B) Some studies have shown that the fiber and other antioxidants in service berries (Saskatoon berries) may offer some health benefits for those suffering from diabetes. Suppose as part of one study the level of vitamin C was to be determined using GC analysis.

An injection was made containing 0.0010 mol/L vitamin C and 0.00050 mol/L pantothenic acid as an internal standard. The vitamin C peak had an area of 842 cm2. The internal standard had a peak area of 555 cm2. (i) What is the ratio of the detector response to these two compounds? (ii) Another experiment under identical GC operating parameters, with an unknown amount of vitamin C from a service berry extract (and the same internal standard) produced an identified vitamin C peak of 366 cm2 and a pantothenic acid peak of 495 cm2. What was the vitamin C concentration in the extract?

10. Suppose this chromatogram was run on an extract of service berry. If the retention factor of peak #8 is 28 min; and the retention factor for peak #9 is 28.5 min and a resolution of 1.5 was being sought, how many theoretical plates would be needed to achieve that resolution?

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C. McLaughlin

SOLUTIONS FOR CHAPTER 20 TEST BANK 1. (A) Partition constant: KoD =

HPFOA (o) HPFOA (aq)

; Partition ratio: KD =

[HPFOA](o) [HPFOA](aq)

(B) fphase2 = 1 - fphase1; 0.75 = 1 – fphase1; fphase1 = 0.25 KD = 0.25 = (C) Dc ~

1 1 + K D (50.0/80.0)

KD

(D) fphase 1 =

2.(A) KD =

1 + 1.6 x10 1

[Lycopene] (o)

= 4.6 / 0.0040

[6.5 x10

[Lycopene] (aq)

= [1.5 x10

1 1

1 + 43(25.0/100.0)

KD

-5

](o)

= 43

](aq)

]2 = 0.0072; fphase2 (organic) = 1 – 0.0072 = 0.993 ~ 0.99

7.2

=

+ 1 + K a /H

-4

= 0.044; fphase2 (organic) = 1 – 0.044 = 0.96

1 + 43(50.0/100.0)

(C) fphase1(H2O) = [

(B) fphase1(H2O) =

−4

=0.258; fphase2 (organic) = 1- 0.258 = 0.74

1 + 4.6(50.0/80.0)

(B) fphase1(H2O) =

3. (A) DC=

4.8

=

+ 1 + K a /H

; 0.25 + 0.156KD = 1.00; KD = 4.8

1 + (7.9 x10

−5

1 1 + 5.8(50.0/100.0)

/ 3.16 x10

-4

= 5.8 )

= 0.26

(C) As the pH is lowered, more H+ is found in the solution. This forces the acid dissociation equilibrium to shift toward the left (undissociated form of the acid). At this lower pH, more of the acid will be extracted into the nonpolar organic solvent.

4. (A) KD =

[Drug](o) [Drug](aq)

(B) f phase1 (H2O) =

=

[(0.010/0.020] (o) [(0.0010/0. 010] (aq)

1 1 + 5.0(0.0050L/0. 010L)

= 5.0

= 0.29 remaining in aqueous phase

(C) fphase2 (toluene) 1 – 0.29 = 0.79 extracted into toluene 5. (A) fphase1(H2O) =

K D base +

1 + K b [H ] / K

5.8

= w

1 + 1.0 x10

−5

(1.58 x10

fphase2 (organic) = 1 – 0.35 = 0.65 in toluene phase

4

-8

)1.0 x10

- 14

= 0.35 remains

Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

(B) Raising the pH would decrease the H+ in the solution (increase the OH-), which would cause + more of the base in the neutral B form in B + H2O →  BH + OH to be available by shifting the equilibrium to the left. Therefore, the extraction would be more effective for this base at higher pH values.

6. (A) fphase2 (organic) = 0.99; fphase1 (H2O) = 1- 0.99 = 0.01 0.01 =

1 1 + 2.0(X/10.0)

(B) 0.01 =[

; 0.01 + 0.02X/10.0 = 1; X = 500 mL

1 1 + 2.0(X/10.0)

]3; (.010)1/3 =

1 1 + 2.0(X/10.0)

; 0.215 =

1 1 + 2.0(X/10.0)

; X = 18 mL

7. (A) t’R = 8.0 – 1.0 = 7.0 min (B) k = t’R/tM = 7.0/1.0 = 7.0; Since k is a ratio of two terms that would both be affected in the same way by a faster flow rate, the k factor will not change.

(C) N = 16(tR/wb)2 = 16(8/0.60)2 = 2843 = 2800 8. (A)The molecules could have very similar polarities. The molecules could have very similar molar masses.

(B) HETP = L/N = 5.00/580 = 0.0086 m (C) If all other factors remain constant, a longer column (L) will provide for more “local equilibrium” opportunities, so N would increase also. Therefore, the HETP would remain unchanged in this situation.

(D) Longitudinal diffusion is based on the solute’s diffusion along the length of the column. This process is emphasized with slow flow rates.

Eddy diffusion is a problem when the solute has many path options as it moves through the support on the inside of a column. This depends on the size and shape of the support particles.

9. (A) “A (large) difference in tR1 and tR2 combined with (small) base widths will result in better resolution.”

0.001

(B)(i) response ratio (vit C/pant acid) =

X

842 = 1.31; (ii) 1.31 = 366 ; 0.0005 0.0005 555

495

X = 4.8 x10-4 M vitamin C 10. (A) 1.5 =

N

•

 −1



4 1.5 =

N 4

•

1.02 − 1 1.02

•

•

k2 k2 + 1

28.5 28.5 + 1

;  = k2/k1 = 28.5/28 = 1.02; 1.5 =

N 4

•

1.02 − 1 1.02

•

28.5 28.5 + 1

; 1.5 = (N).5 x 0.00474; N = 120 000 theoretical plates

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Test Bank Hage/Carr Analytical Chemistry & Quantitative Analysis

C. McLaughlin

Chapter 21: Gas Chromatography 1. Gas chromatography can be used for detection, measurement, and separation of volatile chemicals. In each of the following, decide which (there could be more than one), if any, of those terms apply to the description.

(A) Retention time could be used to investigate: (B) Volatility of compounds affects: (C) Peak height could be used to determine: (D) A chromatogram can reveal information about: (E) Can be determined when the boiling point is above 500oC 2. In the following sentences, select the proper choice, inside of each set of parentheses, so that the sentences will be accurate and logical: “In a chromatogram, the peaks that emerge first will represent compounds that are the (most or least) volatile. Those compounds will also likely have (low or high) boiling points and show the (most or least) interactions with the stationary phase of the GC column. If the emerging peaks are from a nonpolar homolog set, the first to emerge will have (lower or higher) molar masses than those emerging later. Heating the GC oven containing the column will likely (increase or decrease) the time before seeing the first peak.

3. (A) An unknown volatile organic compound has been isolated from the area near the brake pad of an automobile. When injected into a GC, the unknown had a retention time of 10.0 minutes. On the same column octane had a retention time of 9.0 minutes, while nonane had a peak at 11.5 minutes. The column void time was 1.5 minutes. What is the Kovats retention index for this compound?

(B) Assume these measurements were taken from a column with non-polar squalane as the support media. Would the calculated value of “I” likely increase, decrease, or stay the same if the same comparisons were made on a carbowax column? Briefly explain.

4. (A) As an alternative to gasoline, many car makers are researching diesel-fueled engines.

Hexadecane is one of several hydrocarbons found in diesel fuel. It is a low volatility compound with the formula C16H34. Chromatography of diesel fuel components can pose the “general elution problem.” What is meant by this term?

(B) Describe the four steps in a typical temperature programming technique that could be used in chromatography used with diesel samples.

5. (A) Even though open-tubular chromatography columns can accept only small sample sizes, they are often the choice over packed columns for analytical work. Cite three reasons behind this selection.

(B) After reading each of the following statements decide if each applies to SCOT, WCOT, or PLOT type open-tubular column: (if a statement applies to more than one, list all that apply) (i) A solid material is used as the stationary phase without further coating. (ii) Has a thin film liquid coating on a supporting medium. (iii) Has a typical column length between 10 and 100 meters (C) The ratio of the inner diameter of a packed column to a typical open-tubular column is approximately _____.

6. The use of less toxic propylene to replace ethylene glycol as an antifreeze is being proposed in many applications. The molar mass of ethylene glycol is 62 g/mol. The boiling point of ethylene glycol is in the 240oC range. Its molecular structure shows two –OH groups.

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Propylene glycol has a molar mass of 76 g/mol with a boiling point in the 189oC range.

Propylene glycol also has two –OH groups.

(A) Which compound is more volatile?

(B) In a chromatographic separation using a polar stationary phase such as 50%cyanopropylmethyl-50% phenylmethylpolysiloxane which compound would elute first?

(C) If the column was switched to a non-polar stationary phase such as dimethylpolysiloxane the two peaks would still likely be well resolved but may not be as well resolved. Explain how this could happen.

7. (A) Peaks #8. #9, and #10 in this chromatogram represent polar molecules. Peak #11 is known to be caused by a nonpolar compound but does have a large molecular weight. Suppose the analysis is shown on a nonpolar stationary phase. It was later repeated on a much more polar column. What effect could this most likely have on peak #11? ( appear in front of the other three; appear further to the right of the other three; no effect) (B) Suppose that the separation brought about by the analysis was going to be used to collect the different compounds for later work. Which type of detector (Thermal Conductivity or Flame Ionization) would be best to use to accomplish that goal? Explain.

8. Many plastic materials, from drinking bottles to toys, contain phthalates as additives. Some health questions are being raised about phthalate exposure (possible damage to reproductive organs, neurological problems) due to phthalates leaching out from those plastics. Gas chromatography can be used to separate and identify phthalates.

(A) Suppose this chromatogram represents the separation of a mixture of phthalates from a nonpolar polar column. Suppose you wish to analyze a mixture of three different phthalate compounds. Which peak shows the molecule with the highest volatility? C D E (B) Another analysis is performed on the same three phthaltes. This time the column is switched to a polar stationary phase column . The results are shown here. Although the three phthalates are only slightly polar, which is likely the most polar? C D E

D C E

9. (A) Helium is often used as a carrier gas for chromatographic separations using thermal conductivity detectors. Helium is safe and nonreactive. What is one other chief reason for using this gas?

(B) Select a detector that matches the description provided for each of these situations: (i) The analytes in the mixture contain electronegative components. (ii) Organic compounds are converted to cations, uses hydrogen, collects cations at an electode to make a current. (iv) Can serve as both a general and selective detector 10. (A) Samples that are injected into a gas-liquid chromatograph must be volatile to be swept into the carrier gas current. Provide an example of how a component with low volatility can be made more volatile without changing the temperature or pressure of the injection port.

(B) Which type of injection system for liquids matches the following descriptions or comments? (i) Most of the injected sample and carrier gas enter the column, is very good choice for trace analysis.

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(ii) Can handle relatively large samples for packed columns, not used with many open-tublular columns. (iii) Sample is injected from a cold region into the column. (iv) Less than a representative 10% of the sample gets to the column

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SOLUTIONS FOR CHAPTER 21 TEST BANK 1. (A) identification, separation (B) identification, measurement, separation (C) measurement (D) identification, measurement, separation (E) none 2. (i) most; (ii) low; (iii) least; (iv) lower; (v) decrease 3. I = 100z + 100

logt'Rx −logt'Rz ; t’Rx = 10.0 -1.5 = 8.5; t’Rz 9.0-1.5= 7.5; t’R(Z+1) = 11.5 -1.5 = logt'R(z+1) −logt'Rz

10.0; I = 100(8) + 100

log(8.5) − log(7.5) log(10.0) − log(7.5)

= 844;

(B) I would likely increase with the more polar carbowax support. More polar carbowax will have greater intermolecular interactions with all the compounds even if the interactions are small (dispersion forces) the retention times will be larger.

4. (A) The general elution problem describes the situation of balancing adequate resolution with reasonable time for elution. Heavier, nonvolatile hydrocarbons in diesel fuel require more time for elution. This time could be reduced perhaps with hotter column temperatures, but then resolution of some of the more volatile compounds may suffer.

(B) One solution to this problem is to use temperature programming that changes the column temperature, in a controlled manner, during the elution. Lower temperatures could be used in the first part of the elution to allow the more volatile compounds to separate and elute. The temperature could be ramped up to a higher temperature, while less volatile compounds could be encouraged to elute faster while still being resolved. The temperature could then be maintained to ensure all samples elute. Finally, the temperature could then be cooled back to the original value.

5. (A) Open-tubular columns tend to have (1) better efficiency (2) lower detection limits, and (3) faster separations over the packed column style (B) (i) PLOT; (ii) SCOT; (iii) SCOT, WCOT, PLOT (C) Approximately 10:1 6. (A) With its lower boiling point, polypropylene is more volatile.

(B) Both molecules will have polar interactions (likely hydrogen-bonding) with the stationary phase but the more volatile (polypropylene) will elute first.

(C) Using the non-polar stationary phase eliminates many of the stronger interactions with the two polar solutes. Since one of the differing variables is reduced, the resolution will be more dependent on molar mass where polypropylene is heavier even though it has a lower boiling point. So, the order will not likely be reversed, but the peaks may elute closer together.

7. (A) The polar compounds shown as peaks 8, 9, 10 would be retained more on a more polar column, thus peak #11 would appear sooner (to the left) of the three peaks representing polar compound.

(B) Thermal conductivity detectors do not destroy the compounds they analyze, therefore TC would make a better choice to preserve the separated components.

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8. (A) E

C. McLaughlin

(B) D

9. (A) The thermal conductivity, at 240 mW/(mK) is very high and as such will likely differ from most analytes.

(B) (i) Electron capture; (ii) flame ionization; (iii) thermal conductivity; (iv) mass spectrometry 10. (A) One strategy to make compounds more volatile is to form a derivative of the compound that is more volatile. Even large molecules can often be made more volatile by replacing polar groups with nonpolar attachments such as trimethylchlorosilane.

(B)(i) Splitless; (ii) direct injection; (iii) on-column injection; (iv) split injection

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C. McLaughlin

Chapter 22: Liquid Chromatography 1. Gas chromatography and liquid chromatography techniques share several common traits.

However, there are several important differences. Address each of these chromatography issues comparing GC to LC. (A) Compare the plate numbers in a typical GC column to a typical LC column.

(B) Role of the mobile phase in GC compared to the role of the mobile phase in LC. (C) Name a type of detector that is used both in GC and LC (D) Compare the requirements of analyte properties in GC and LC. (E) Compare the method of moving the mobile phase in both GC and LC. 2. Comparing mechanical methods of separation in LC allows five major divisions. Select the appropriate separation method (adsorption chromatography, partition chromatography, ionexchange chromatography, size-exclusion chromatography, affinity chromatography) that matches each of these brief descriptions: (A) Home water softeners use this method.

(B) This method depends on the binding strength of the analyte to the support and the surface area of the support.

(C) A biological-related ligand binds the analyte (D) Separation is based on the ability of the analytes to access the mobile phase within the pores of the support.

(E) In this type of LC the support is coated with either a polar or nonpolar liquid.

3. (A) Elutropic strength of the mobile phase in adsorption chromatography is important when planning separations of many closely related organic compounds such as tocopherols shown in the text. What does a high value for o indicate? How does a high value affect the analyte passing through a column?

(B) If an adsorption experiment to separate polar ethanol from other polar compounds that may be found in a fermentation reaction was being planned, would it best to have a mobile phase bound to silica support that has a high or a low value of o? Explain the logic of the decision.

4. (A) Normal phase liquid chromatography (NPLC) and reverse phase liquid chromatography (RPLC) are both examples of partition chromatography. Which one uses a polar stationary phase?

(B) A biochemist studying the amino acids present in a protein supplement uses partition chromatography for the separation. The mobile phase consists of a 40%/60% mixture of isopropanol (P = 3.92) and methanol (P = 5.10) What is the solvent polarity ( Ptot) of this mixture?

5. (A) Suppose two cations in a mixture are about to be analyzed with ion chromatography.

Both are competing for negatively charged binding sites on a support. Would the cation with the higher selectivity coefficient have a longer or shorter retention time than the cation with a smaller selectivity coefficient? How does the selectivity coefficient explain this answer?

(B) An environmental study involved resolving a mixture of SeO32- and SeO42- using IC. Would it be best to use a cation or anion exchange column?

(C) One problem that can arise in this analysis if conductivity is used as a detector is the presence of a high concentration of ions in the mobile phase. How is this problem resolved?

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6. (A) What is the meaning of the ratio known in size exclusion chromatography as Ko? (B) Would the Ko value of a large molar mass solute approach one or zero?

(C) The size exclusion analysis of insulin – type proteins isolated from fish samples showed molecular masses in the range of 6000-10000 Da. What would be a reasonable pore size to use for separating this size of protein? What would be the retention volume (VR) of such a protein if on a column with a VM = 16.0 mL and VE = 11.0 with a Ko = 3.78?

7. (A) Affinity columns have an “on/off” mechanism. How does this term explain the operational mechanism for this type of LC? (B) The retention of an analyte on an affinity column can be described by the association equilibrium constant that is calculated from the analyte-ligand complexation reaction. Suppose an affinity column was designed with a specific nucleic acid as the affinity ligand. An analysis of the complementary nucleic acid is to be performed. If the column contains 25 nmoles of the affinity ligand DNA and has a void volume of 1.5 mL with a KA (association equilibrium constant) for the complexation of 1.0 x 107M-1, what is the retention factor for this nucleic acid on this column?

8. (A) The terms “weak” and “strong” mobile phase are used in adsorption chromatography and affinity chromatography. How does a strong mobile phase in adsorption chromatography affect the separation of analytes? How does a strong mobile phase affect the separation of analytes in affinity chromatography?

(B) How does a strong mobile phase in RPLC compare to a strong mobile phase in affinity chromatography?

9. (A) Detectors in LC may be general or specific, or in some cases may be utilized as both.

Explain how mass spectrometry may be used as either general or specific.

(B) Explain what aspects of fluorescent detectors qualify them to be classified as a type of selective detector for LC. (C) What key aspect of Refractive Index detectors allow them to have a general application in LC? 10. (A) The liquid mobile phase for LC must be pumped into the column. Cite one key difference between the syringe and reciprocating pumps in terms of the approximate ratio of volume pumped per minute.

(B) Compare the reason for making derivatives for LC and GC analysis schemes.

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SOLUTIONS TO TEST BANK FOR CHAPTER 22 1. (A) Typical GC columns have more theoretical plates than LC (B) The mobile phase in GC is a gas that pushes the analyte mixture through the column so that it may interact with the stationary phase. In LC the mobile phase is a liquid that interacts (through polar or nonpolar interactions) with the solute mixture.

(C) Mass spectrometry can be used as a detector in both GC (GCMS) and LC (LCMS) (D) A key requirement for separating analytes in a mixture using GC is that the sample must be made volatile. A key requirement for separating analytes in a mixture using LC is that the analytes must be soluble in an injectable liquid.

(E) The mobile phase in GC is a gas that moves through the column from a pressurized tank with a control valve. In LC the mobile phase is pumped through the column in liquid form.

2. (A) ion chromatography (B) Adsorption Chromatography (C) Affinity Chromatography (D) Size Exclusion Chromatography (E) Partition Chromatography 3. (A) Solvents with a large o value will occupy many binding areas of polar silica. This will cover the silica binding areas and prevent the analytes from attaching as much to the support.

This means the analytes will elute faster than with a mobile phase with a low o value.

(B) A solvent with a higher o value would be logical to ensure that the polar alcohols did not stay on the column, attached to the polar silica support, too long. Longer retention times would lead to peak broadening.

4. (A) NPLC uses a polar stationary phase.

(B) Ptot = APA + BPB = (0.40 x 3.92) + (0.60 x 5.10) = 4.6 5. (A) A higher selectivity coefficient indicates that the cation would have a longer retention time.

The two cations must bind to the proper sites. The selectivity coefficient (kA,C) provides a ratio of bound cation to unbonded cation when a competing ion is present.

(B) Anion exchange chromatography would be used for this separation.

(C) When high ion concentrations are present in the mobile phase the conductivity detector will not be as effective. However, if the number of binding sites in the stationary phase are reduced, then fewer ions are required in the mobile phase.

6. (A) Ko is a ratio that represents the fraction of the volume between VM and VE in which the solute elutes.

(B) For large molar mass solutes the value of Ko will approach zero.

(C) A pore size of approximately 5.0 nm would function for proteins in that range Ko =

(VR − VE ) (VR − 11.0) = 3.83 = ; V = 29.9 mL (16.0 − 11.0) (VM − VE ) R

7. (A) The sample is applied to the column (with its specific affinity ligand) with an application buffer. The strong attractions hold the analyte to the support. After the nonretained solutes and solvent are washed away, an “elution buffer” is then applied that releases the analyte from the 3

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affinity ligand to be detected. The column is then returned to its original state with original mobile phase.

(B) k = KA(mL/VM) = 1.0 x107M-1(2.5 x10-8moles/0.0015L) = 167 = 170 8. (A) A strong mobile phase in adsorption chromatography adheres to the silica support and will cause polar analytes to spend more time in the mobile phase and elute quickly. A strong mobile phase in affinity chromatography is a solvent that removes the analyte from its attachment to an affinity ligand and causes lower retention.

(B) A strong mobile phase in RPLC is nonpolar. Solutes will partition between the mobile and stationary phases. Nonpolar analytes will be carried by a strong nonpolar mobile phase through the column.

9. (A) Using mass spectrometry in a full scan mode allows ion fragments from the eluants to be classified based on library matched patterns. Each analyte has a specific pattern. If, instead, the mass spectrometer is used selectively at just a few characteristic ion fragments from a known analyte, the intensity can be related to a quantitative measure of just that specific analyte.

(B) Fluorescent detectors excite electrons in analytes from one specific quantum state to another unique quantum state through absorption of a specific wavelength. The emission form the excited state back to the ground state is unique for the analyte. Only fluorescent compounds can be measured with this technique.

(C) Since the bending of light by a solution (compared to a reference) is a function of every liquid, any analyte variability in the mobile phase can be detected by the change in the refractive index of the mobile phase containing an analyte. When other characteristics are not specifically known, the RI detector can still supply useful information.

10. (A) Reciprocating pumps produce flows of in the range of mL/min, while syringe pumps have a flow in the L/min range. Therefore, the approximate ratio is 10-3L/10-6L = 1000/1 (B) A key reason for making derivatives in GC was to increase volatility. In LC derivatization is often done to cause an analyte to have specific properties related to detection. (i.e., an analyte could be attached to a fluorescent moiety to enable specific fluorescence detection)

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C. McLaughlin

Chapter 23: Electrophoresis 1. (A) Migration of analytes during an applied field is the common property used in both gel electrophoresis and capillary electrophoresis to obtain separation of analytes. However, there is a key difference in measurements of migration. What key migration property is measured in gel electrophoresis? How does this differ from the migration property measured in capillary electrophoresis?

(B) In equations for electrophoretic mobility what key factor predicts the ability of two analytes to be separated with electrophoresis?

(C) Describe two factors that would decrease the velocity of an analyte moving at a steady state in an electrophoretic system.

2. The protein profile in wheat contributes both to the bread making properties and to nutritional value. Identification of the proteins in wheat aids in comparing wheat grown in various environments. Capillary electrophoresis may be used as a tool in those separations. For example, suppose one sample provided two protein bands with the following parameters: protein A had a migration time of 11.5 minutes while protein B had a migration time of 13.7 minutes. The system was composed of a capillary tube of 42.0 cm, with a distance to the detector of 40.5 cm. The applied voltage used was 10.0 kV. What were the electrophoretic mobilities of the two proteins?

3. (A) One way to use electrophoretic mobility is to assist in separating acidic or basic analytes is to change the net charge of the analytes is by changing the pH. For example if the pKa of an acidic analyte was 6.55, what fraction of the acid would be in the neutral (HA) form in a solution with a pH of 7.04?

(B) Perhaps a series of amine compounds are going to be separated via electrophoresis. If the pH of the solution was 8.74, what fraction of an amine with a pKb of 5.45 would be in the HB+ form?

4. (A) If an analyte was presenting difficulties in separation due to longitudinal diffusion which of the following factors would help decrease the problem? (select one choice) (i) decrease the viscosity of the running buffer (ii) raise the temperature of the system (iii) use a porous support with small enough pores (iv) increase the voltage (v) All of the above (B) Reducing Joule heating will decrease band broadening during electrophoresis separations.

Which of the following would be helpful in reducing Joule heating? (select one choice) (i) reduce the current being used in the separation (ii) increase the ionic strength of the buffer to increase the current (iii) use a higher voltage to increase migration velocity so less time is spent for separation (iv) increase the time for the analysis to allow for gradual cooling of the system (v) all of the above (C) Electroosmotic mobility …. (select one choice) (i) refers to the effects of eddy diffusion in band broadening. (ii) refers to the acceleration of analyte in the direction of the running buffer regardless of the charge on the support. (iii) is directly related to the viscosity of the running buffer. (iv) is usually a small factor for gel electrophoresis (v) all of the above are true.

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5. (A) Agarose and polyacrylamide are both used in gel electrophoresis. Cite one similarity and one key difference between these two useful supports.

(B) A stacking gel and a running gel are both found in gel electrophoresis and used to decrease band width. Cite two key differences between these two gels.

(C) Identify each of the following detection methods with a specific name (either reagent stain, silver stain, Southern Blot, Northern Blot, Western Blot, Mass spectrometry) (i) detects specific sequences of DNA by using known 32P labeled DNA (ii) coomassie brilliant blue is used with proteins (iii) a LASER is part of the analysis 6. (A) Bovine serum albumin samples can be analyzed by SDS-PAGE. (i) What is the function of SDS in this method? (ii) How is a calibration curve for SDS-PAGE prepared?

(B) Isoelectric focusing electrophoresis may be used to characterize enzyme and enzyme levels in diseased and healthy patients. IEF may be the first step in two-dimensional electrophoresis. (i) When does electrophoretic mobility become zero in IEF? (ii) Explain why IEF produces very tight protein bands.

7. Capillary electrophoresis can be used to determine very low levels of dissolved cations. If sodium ion was being measured using CE, an internal standard may be added to ensure more accurate detection. The following data was collected on the sodium cation amounts in constant volume testing.

Na+ amount peak height internal standard amount peak height 1 pmol 3.0 cm 0.50 pmol 12 cm 5 pmol 18 cm 0.50 pmol 14 cm 10 pmol 38 cm 0.50 pmol 11 cm Unknown 12 cm 0.50 pmol 12 cm What is the amount of Na+ in the unknown?

8. (A) Detection methods for CE and liquid chromatography are often similar (i.e., fluorescence, conductivity UV) Cite two considerations that cause the use of those detection techniques in LC and CE to differ.

(B) Two amino acids in a sample are being compared using CE with an absorbance detector.

The following data was obtained. Amino acid #1 showed an absorbance peak area of 256 cm2 and a retention time of 12 minutes. Amino acid #2 showed absorbance peak area of 356 cm2 and a retention time of 15 minutes. What are the corrected peak areas of the two amino acids?

Which is present in the greater amount?

9. Capillary electrophoresis is well-known for its ability to have a large number of theoretical plates (N) created in a tube. Several anions and cations can be analyzed using CE. Suppose the diffusion constant for a divalent cation was 4.5 x 10-5 cm2/s while the electrophoretic mobility was 3.44 cm2/kVmin. The detector was located 25.0 cm from the injection point in a 30.0 cm tube. The analysis was conducted at 10.0 kV. What is the maximum number of theoretical plates that could be obtained in this analysis? (Assume that longitudinal diffusion is the only band-broadening factor.)

10. A specific enzyme is being analyzed by CE. As it migrates through a 45 cm column with an applied voltage of 12 kV, it produces a peak area of 863 units. The detector is located 42 cm from the injection point. What is the expected area if the same conditions are maintained except that the applied voltage is dropped to 20 kV? 2

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SOLUTIONS FOR CHAPTER 23 TEST BANK: 1. (A) In gel electrophoresis, the migration distance is measured as the analytes move through the system in a set amount of time that does not allow the analytes to reach the end of the gel.

If using capillary electrophoresis, the time required to reach the end of the system for each analyte is measured.

(B) The ratio of charge to mass (z/r) is the key factor to compare between two analytes. A larger difference indicates better opportunities for separation.

(C) Assuming other factors are constant, when the viscosity of the running buffer increases the analyte velocity decreases. When the solvated radius increases, the velocity will also decrease.

 Ld   40.5      v  tm  v 11.5  2.  = = = = = 14.8 cm2/kVmin for protein A; E  V  E  10.0      L  42.0   40.5    v  13.7  = = = 12.4 cm2/kVmin for protein B E  10.0     42.0  3. (A) pH = pKa + log (A-/HA); 6.55 = 7.04 + log (A-/HA); 10-0.49 = 0.32/1 or 76% HA (B) 14.00- pKb = pKa; 14.00 – 5.45 = 9.55; 8.74 = 8.55 + log (B/HB+); 100.19 = (B/HB+) = 1.55/1; 39% HB+ 4. (A) (iii); (B) (i); (C) (iv) 5. (A) Both agarose and polyacrylamide have low, nonspecific binding to biological compounds.

Agarose typically has relatively wide pores, while pores found in acrylamide tend to be smaller (and can be adjusted).

(B)The running gel is the support used in the separation. The stacking gel is on top and is where the wells are found. Analytes travel quickly through the stacking gel but slow down when they meet the boundary with the running gel. The stacking gel also has larger pores.

(C) (i) Southern Blot; (ii) reagent stain; (iii) Mass Spectrometry (MALDI-TOF MS) 6. (A) (i) SDS has a nonpolar “tail” that coats each segment of a denatured protein, forming rods with a negative coating from the sulfate group at the “head” of the SDS molecule. (ii) Plotting the log of MW versus either migration distance or retardation factor produces a curve with a linear portion that can be used to create a calibration curve that can be used with unknowns.

(B) (i) Electrophoretic mobility becomes zero when the zwitterion migrates to a pH in the support that equals the pI of the zwitterion. (ii) If a zwitterion starts to migrate away from the area where pH = pI, the system actually brings the zwitterions back to the appropriate region. This takes place because the gradient has high pH near the negative electrode and low pH near the positive electrode. If a solute diffuses slightly out of its band toward the negative region (high pH), it will take on a more negative

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charge and be attracted back to the relatively more positive region back near its band. (These would be reversed for a solute moving toward the positive electrode.)

7. When the peak height of the Na+ divided by the peak height of the internal standard is plotted versus the Na+ amount in the constant volume samples, the slope = 0.36 while the Y-intercept = - 0.25. For the unknown, Na+ peak height divided by internal standard peak height = 1. Using Y = aX + b, the amount of Na+ = 3.5 pmol.

8. (A) Detection in CE must be well-suited for very small samples compared to the larger samples in most LC techniques. Also, in CE, analytes with different retention times will spend different amounts of time in the detector. In LC, all samples spend the same time in the detector.

(B) Amino acid #1 corrected area = 256/10 = 26; Amino acid #2 corrected area = 356/15 = 24 Amino acid #1 is present in a slightly greater amount 9. N =

μVL d 2DL

=

(3.44)(10.0)(25.0) = 320 000 plates 2 4.5x10- 5 (30.0)

(

)

10. 20/12 = 1.67; The migration time will decrease by 20/12 = 1.7. The peak area should decrease to 863/1.7 = 510 units.

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