Chemical principles 8th edition zumdahl test bank download by rex.young703

Solution Manual for Chemical Principles 8th Edition by Zumdahl

DeCoste ISBN 1305581989 9781305581982

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1. A solution contains 0.50 (Ka = 2.0 × 10-8) and 0.22 M NaA. Calculate the pH after 0.05mol of NaOH is added to 1.00 L of this solution.

a. 7.48

b. 7.70

c. 7.34

d. 7.92

e. 7.19

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.1

KEYWORDS: acids and bases | buffer | general chemistry | pH of a buffer | solutions of a weak acid or base with another solute

2. Calculate the pH of a solution that contains

NaCN.

a. 8.28

b. 7.46

c. 9.25

d. 8.86

e. none of these

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.1

KEYWORDS: acids and bases | Brønsted-Lowry concept of acids and bases | general chemistry | pH of a buffer | solutions of a weak acid or base with another solute

3. How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 × 10–5.) Assume negligible volume of the HCl.

a. 6.8 × 10–3 mol

b. 7.5 × 10–2 mol

Chapter 08 - Applications of Aqueous Equilibria

3.25 M HCN (Ka = 6.2 × 10–10), 1.00 M NaOH and 1.50 M

Chapter 08 - Applications of Aqueous Equilibria

c. 1.0 × 10–3 mol

d. 5.0 × 10–1 mol

e. none of these

ANSWER: e

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.1

KEYWORDS: acids and bases | adding an acid or base to a buffer | buffer | general chemistry | solutions of a weak acid or base with another solute

4. Calculate the pH of a solution made by mixing 46.0 mL of 0.350 M NaA (Ka for HA = 1.0 × 10–9) with 26.0 mL of 0.190 M HCl.

a. 9.00

b. 9.51

c. 8.65

d. 9.07

e. 9.35

ANSWER: e

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.1

KEYWORDS: acids and bases | Brønsted-Lowry concept of acids and bases | general chemistry | solutions of a weak acid or base with another solute

5. Consider a solution of 2.0 M HCN and 1.0 M NaCN (Ka for HCN = 6.2 × 10–10). Which of the following statements is true?

a. The solution is not a buffer because [HCN] is not equal to [CN–].

b. [OH–] > [H+]

c. The pH will be below 7.00 because the concentration of the acid is greater than that of the base.

d. The buffer will be more resistant to pH changes from addition of strong acid than to pH changes from addition of strong base.

e. All of these statements are false.

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.1 8.2

KEYWORDS: acids and bases | buffer | general chemistry | solutions of a weak acid or base with another solute

6. Buffers in the human body

Chapter 08 - Applications of Aqueous Equilibria

a. precipitate proteins so enzymes are inactive.

b. help to keep the body temperature constant.

c. help change the blood plasma pH when foods are eaten.

d. help to maintain a constant blood pH.

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.2

KEYWORDS: acids and bases | buffer | general chemistry | solutions of a weak acid or base with another solute

7. You are given a solution of the weak base Novocain, Nvc. Its pH is 11.00. You add to the solution a small amount of a salt containing the conjugate acid of Novocain, NvcH+ . Which statement is true?

a. The pH and the pOH remain unchanged.

b. The pH and the pOH both decrease.

c. The pH and the pOH both increase.

d. The pH increases and pOH decreases.

e. The pH decreases and the pOH increases.

ANSWER: e

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.2

KEYWORDS: acids and bases | adding an acid or base to a buffer | buffer | general chemistry | solutions of a weak acid or base with another solute

8. Calculate [H+] in a solution that is 0.30M in NaF and 0.46 M in HF. (Ka = 7.2 × 10–4)

a. 1.2 × 10–4 M

b. 5.7 × 10–3 M

c. 231.5 × 10–2 M

d. 4.7 × 10–4 M

e. 0.46 M

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.2

KEYWORDS: acids and bases | buffer | general chemistry | solutions of a weak acid or base with another solute

9. Which of the following will not produce a buffered solution?

Chapter 08 - Applications of Aqueous Equilibria

a. 100 mL of 0.1 M Na2CO3 and 50 mL of 0.1 M HCl

b. 100 mL of 0.1 M NaHCO3 and 25 mL of 0.2 M HCl

c. 100 mL of 0.1 M Na2CO3 and 75 mL of 0.2 M HCl

d. 50 mL of 0.2 M Na2CO3 and 5 mL of 1.0 M HCl

e. 100 mL of 0.1 M Na2CO3 and 50 mL of 0.1 M NaOH

ANSWER: e

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.2

KEYWORDS: acids and bases | buffer | general chemistry | solutions of a weak acid or base with another solute

10. For ammonia, Kb is 1.8 × 10–5 . To make a buffered solution with pH 10.0, the ratio of NH4Cl to NH3 must be

a. 1.8 : 1.

b. 0.18 : 1.

c. 1 : 1.8.

d. 1 : 0.18.

e. none of these

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.2

KEYWORDS: acids and bases | buffer | general chemistry | Henderson-Hasselbalch equation | solutions of a weak acid or base with another solute

11. Calculate the pH of a solution prepared by mixing 50 mL of a 0.10 M solution of HF with 25 mL of a 0.20 M solution of NaF. pKa of HF is 3.14.

a. 5.83

b. 12.00

c. 3.14

d. 7.35

e. 10.80

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.2

KEYWORDS: acids and bases | buffer | general chemistry | pH of a buffer | solutions of a weak acid or base with another solute

Chapter 08 - Applications of Aqueous Equilibria

12. Consider a solution consisting of the following two buffer systems:

H2CO3 HCO3 –+ H+ pKa = 6.4

H2PO4 – HPO4 2–+ H+ pKa = 7.2

At pH 6.4, which one of the following is true of the relative amounts of acid and conjugate base present?

ANSWER: e

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.2

KEYWORDS: acids and bases | buffer | general chemistry | pH of a buffer | solutions of a weak acid or base with another solute

13. How much solid NaCN must be added to 1.0 L of a 0.5 M HCN solution to produce a solution with pH

7.0? Ka = 6.2 × 10–10 for HCN.

a. 0.0034 g

b. 0.15 g

c. 160 g

d. 24 g

e. 11 g

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.2

KEYWORDS: acids and bases | buffer | general chemistry | solutions of a weak acid or base with another solute

14. For 120.0 mL of a buffer that is 0.40M in HOCl and 0.42 M in NaOCl, what is the pH after 17.5 mL of 1.1 M NaOH is added? Ka for HOCl = 3.5 × 10–8 . (Assume the volumes are additive.)

a. 7.12

b. 7.48

c. 7.43

d. 7.07

e. 7.84

ANSWER: e

POINTS: 1

DIFFICULTY: moderate

2CO3] > [HCO3 –]

[H2PO4 –] > [HPO4 2–]

a. [H

and

[HCO3 –] > [H2CO3] and [HPO4 2–] > [H2PO4 –]

[H2CO3] > [HCO3 –]

[HPO4 2–] > [H2PO4 –]

[H2CO3] = [HCO3 –] and [HPO4 2–] > [H2PO4 –]

[H2CO3] = [HCO3 –] and [H2PO4 –] > [HPO4 2–]

and

Chapter 08 - Applications of Aqueous Equilibria

TOPICS: 8.2

KEYWORDS: acids and bases | adding an acid or base to a buffer | buffer | general chemistry | solutions of a weak acid or base with another solute

15. What is the pH of a solution that results when 0.012 mol HNO3 is added to 610.0 mL of a solution that is 0.26 M in aqueous ammonia and 0.46M in ammonium nitrate. Assume no volume change. (Kb for NH3 = 1.8 × 10–5.)

a. 8.96

b. 9.56

c. 5.04

d. 9.01

e. 4.44

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.2

KEYWORDS: acids and bases | buffer | general chemistry | pH of a buffer | solutions of a weak acid or base with another solute

16. What combination of substances will give a buffered solution that has a pH of 5.05? Assume each pair of substances is dissolved in 5.0 L of water. (Kb for NH3 = 1.8 × 10–5 ; Kb for C5H5N = 1.7 × 10–9)

a. 1.0 mol C5H5N and 1.5 mol C5H5NHCl

b. 1.0 mol NH3 and 1.5 mol NH4Cl

c. 1.5 mol NH3 and 1.0 mol NH4Cl

d. 1.5 mol C5H5N and 1.0 mol C5H5NHCl

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.2

KEYWORDS: acids and bases | buffer | general chemistry | pH of a buffer | solutions of a weak acid or base with another solute

17. You have solutions of 0.200 M HNO2 and 0.200 M KNO2 (Ka for HNO2 = 4.00 × 10–4). A buffer of pH 3.000 is needed. What volumes of HNO2 and KNO2 are required to make 1 L of buffered solution?

a. 286 mL HNO2; 714 mL KNO2

b. 714 mL HNO2; 286 mL KNO2

c. 413 mL HNO2; 587 mL KNO2

d. 500 mL of each

Chapter 08 - Applications of Aqueous Equilibria

e. 587 mL HNO2; 413 mL KNO2

ANSWER: b

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.2

KEYWORDS: acids and bases | buffer | general chemistry | pH of a buffer | solutions of a weak acid or base with another solute

18. 12.4 mL of 0.25 M HCl is added to a 120.0-mL sample of 0.220 M HNO2 (Ka for HNO2 = 4.0 × 10-4). What is the equilibrium concentration of NO2- ions?

a. 3.4 × 10–3 M

b. 2.3 × 10–2 M

c. 3.9 × 10–3 M

d. 4.7 × 10–4 M

e. 9.4 × 10–3 M

ANSWER: a

POINTS: 1

DIFFICULTY: Difficult

TOPICS: 8.2

KEYWORDS: general chemistry | acids and bases | solutions of a weak acid or base | acid-ionization equilibria | calculations with Ka

19. A student uses 16.60 mL of 0.100 M NaOH to titrate a 0.2000-g sample of an unknown acid. Which of the following acids is the unknown most likely to be? Assume that only the hydrogens bonded to oxygen are titrated by the sodium hydroxide and that, because of experimental error, the student's value may not be identical to the theoretical value.

a. oxalic acid, HOOCCOOH, molar mass 90

b. succinic acid, HOOCCH2CH2COOH, molar mass 118

c. benzoic acid, C6H5COOH, molar mass 122

d. phthalic acid, C6H4 (COOH)2, molar mass 166

e. not enough information

ANSWER: c

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.2

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

20. How many mmoles of HCl must be added to 155.0 mL of a 0.28 M solution of methylamine (pKb = 3.36) to give a buffer having a pH of 11.42?

Chapter 08 - Applications of Aqueous Equilibria

a. 19 mmol

b. 0.17 mmol

c. 6.2 mmol

d. 3.7 mmol

e. 0.61 mmol

ANSWER: c

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.2

KEYWORDS: acids and bases | adding an acid or base to a buffer | buffer | general chemistry | solutions of a weak acid or base with another solute

21. Which of the following solutions will be the best buffer at a pH of 9.26? (Ka for HC2H3O2 is 1.8 × 10–5 ; Kb for NH3 is 1.8 × 10–5.)

a. 0.20M NH3 and 0.20M NH4Cl

b. 5.0M HC2H3O2 and 5.0M NH4Cl

c. 0.20M HC2H3O2 and 0.20M NaC

d. 5.0M NH3 and 5.0M NH4Cl

e. 5.0M HC2H3O2 and 5.0M NH3

ANSWER: d

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.2 8.3

KEYWORDS: acids and bases | buffer | general chemistry | solutions of a weak acid or base with another solute

22. One milliliter (1.00 mL) of acid taken from a lead storage battery is pipetted into a flask. Water and phenolphthalein indicator are added, and the solution is titrated with 0.38 M NaOH until a pink color appears; 15.8 mL is required. Find, to within 5%, the number of grams of H2SO4 (formula weight = 98) present in 1 L of the battery acid.

a. 480 g

b. 240 g

c. 580 g

d. 290 g

e. 750 g

ANSWER: d

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

2H3O2

Chapter 08 - Applications of Aqueous Equilibria

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a strong acid by a strong base

23. A solution of hydrochloric acid of unknown concentration was titrated with 0.12 M NaOH. If a 250-mL sample of the HCl solution required exactly 24 mL of the NaOH solution to reach the equivalence point, what was the pH of the HCl solution?

a. 12.06

b. 1.98

c. 0.12

d. 1.94

e. –0.10

ANSWER: d

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a strong acid by a strong base

24. You are given 5.00 mL of an H2SO4 solution of unknown concentration. You divide the 5.00-mL sample into five 1.00-mL samples and titrate each separately with 0.1000 M NaOH. In each titration, the H2SO4 is completely neutralized. The average volume of NaOH solution used to reach the endpoint is 15.3 mL. What was the concentration of H2SO4 in the 5.00-mL sample?

a. 0.306 M

b. 0.765 M

c. 3.06 M

d. 1.53 × 10–3 M

e. 1.53 M

ANSWER: b

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acid-base properties of salt solutions | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a strong acid by a strong base

25. What is the molarity of a sodium hydroxide solution if 25.0 mL of this solution reacts exactly with 22.30 mL of 0.253 M sulfuric acid?

a. 0.284 M

b. 0.451 M

c. 0.567 M

d. 0.226 M

e. 0.113 M

ANSWER: b

Chapter 08 - Applications of Aqueous Equilibria

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a strong acid by a strong base

26. How many moles of HCl(g) must be added to 1.0 L of 2.0 M NaOH to achieve a pH of 0.00? (Neglect any volume change.)

a. 3.0 mol

b. 10. mol

c. 1.0 mol

d. 2.0 mol

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acids and bases | general chemistry | pH of a solution | solutions of a strong acid or base

27. Equal volumes of 0.1 M HCl and 0.1 M HC2H3O2 are titrated with 0.1 M NaOH. Which of the following would be equal for both titrations?

a. the pH at the equivalence point

b. the volume of NaOH added to reach the equivalence point

c. the initial pH

d. the pH at the halfway point

e. two of the above

ANSWER: b

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute

28. A 56.20-mL sample of 0.0782 M HCN (Ka = 6.2 × 10–10) is titrated with 0.745 M NaOH. What volume of 0.745 M NaOH is required to reach the stoichiometric point?

a. 5.90 mL

b. 56.2 mL

c. 535 mL

d. 0.00187 mL

e. 4.39 mL

ANSWER: a

Chapter 08 - Applications of Aqueous Equilibria

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

29. Which of the following is the net ionic equation for the reaction that occurs during the titration of nitrous acid with potassium hydroxide?

a. HNO2 + KOH → K+ + NO2 –+ H2O

b. H+ + OH–→ H2O

c. HNO2 + OH–→ NO2 –+ H2O

d. HNO2 + H2O → NO2 –+ H3O+

e. HNO2 + K+ + OH–→ KNO2 + H2O

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

30. The pH at the equivalence point of a titration of a weak acid with a strong base is

a. More data are needed to answer this question.

b. equal to 7.00.

c. greater than 7.00.

d. less than 7.00.

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

31. A 10-mL sample of tartaric acid is titrated to a phenolphthalein endpoint with 20. mL of 1.0 M NaOH. Assuming tartaric acid is diprotic, what is the molarity of the acid?

a. 10.

b. 1.0

c. 2.0

d. 4.0

e. impossible to determine

ANSWER: b

POINTS: 1

Chapter 08 - Applications of Aqueous Equilibria

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

A titration of 100.0 mL of 1.00 M malonic acid (H2A) was done with 1.00 M NaOH. For malonic acid, Ka1 =

1.49 × 10–2 , Ka2 = 2.03 × 10–6

32. Calculate the [H+] after 50.00 mL of 1.00 M NaOH has been added.

a. 2.00 × 10–6 M

b. 1.41 × 10–10 M

c. 1.49 × 10–2 M

d. 3.87 × 10–2 M

e. none of these

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

33. Calculate [H+] after 150.0 mL of 1.00 M NaOH has been added.

a. 2.03 × 10–6 M

b. 1.41 × 10–10 M

c. 1.49 × 10–2 M

d. 1.00 × 10–7 M

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

34. Calculate [H+] after 300.0 mL of 1.00 M NaOH has been added.

a. 1.00 × 10–7 M

b. 4.00 × 10–14 M

c. 1.41 × 10–10 M

d. 1.00 M

Chapter 08 - Applications of Aqueous Equilibria

e. none of these

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

35. If 16 mL of 0.78 M HCl is added to 109 mL of 0.20 M NaOH, what is the final pH?

a. 0.76

b. 13.24

c. 12.87

d. 7.00

e. 1.13

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a strong acid by a strong base

36. A 50.00-mL sample of 0.100 M KOH is titrated with 0.100 M HNO3. Calculate the pH of the solution after the 52.00 mL of HNO3 is added.

a. 6.50

b. 3.01

c. 2.41

d. 2.71

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a strong acid by a strong base

37. What quantity of NaOH(s) must be added to 1.00 L of 0.200 M HCl to achieve a pH of 12.00? (Assume no volume change.)

a. 0.200 mol

b. 0.010 mol

c. 0.210 mol

d. 0.420 mol

Chapter 08 - Applications of Aqueous Equilibria

e. none of these

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a strong acid by a strong base

38. A 73.5-mL sample of 0.18 M HNO2 (Ka = 4.0 × 10–4) is titrated with 0.12 M NaOH. What is the pH after 25.2 mL of NaOH has been added?

a. 10.07

b. 7.00

c. 3.93

d. 2.76

e. 2.87

ANSWER: e

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

39. What volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to achieve a pH of 8.00? Ka for HOCl is 3.5 × 10–8

a. 1.2 L

b. 1.0 L

c. 3.9 L

d. 5.0 L

e. none of these

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

40. A 75.0-mL sample of 0.0500 M HCN (Ka = 6.2 × 10–10) is titrated with 0.500 M NaOH. What is [H+] in the solution after 3.0 mL of 0.500 M NaOH has been added?

a. 9.3 × 10–10 M

b. 5.2 × 10–13 M

c. 1.0 × 10–7 M

Chapter 08 - Applications of Aqueous Equilibria

d. 4.1 × 10

10 M

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

41. Consider the titration of 100.0 mL of 0.250 M aniline (Kb = 3.8 × 10–10) with 0.500 M HCl. For calculating the volume of HCl required to reach a pH of 8.0, which of the following expressions is correct? (x = volume, in milliliters, of HCl required to reach a pH of 8.0)

d. [H+] = x

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

42. A 100.-mL sample of 0.10 M HCl is mixed with 50. mL of 0.10 M NH3. What is the resulting pH? (Kb for NH3 = 1.8 × 10–5)

a. 7.85

b. 1.30

c. 3.87

d. 1.48

e. 12.52

ANSWER: d

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

–

Chapter 08 - Applications of Aqueous Equilibria

43. After adding 25.0 mL of 0.100 M NaOH to 100.0 mL of 0.100 M weak acid (HA), the pH is found to be 5.90. Determine the value of Ka for the acid HA.

a. 4.2 × 10–7

b. 3.5 × 10

c. 1.6 × 10–11

d. 2.1 × 10

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak acid by a strong base

44. A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein endpoint with 25.0 mL of 0.100 M NaOH. The student then adds 13.0 mL of 0.100 M HCl. The pH of the resulting solution is 4.7. Which of the following statements is true?

a. At pH 4.7, half of the conjugate base, A–, has been converted to HA.

b. The pKa of the acid is 4.7.

c. The pKa of the acid is less than 4.7.

d. The pKa of the acid is greater than 4.7.

e. More than one of the above statements are correct.

ANSWER: d

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

KEYWORDS: acids and bases | buffer | general chemistry | Henderson-Hasselbalch equation | solutions of a weak acid or base with another solute

45. 48.8 mL of a 1.42 M NaOH solution is titrated with a 1.86 M HCl solution. What is the final volume of the solution when the NaOH has been completely neutralized by the HCl? (Assume the volumes are additive.)

a. 37.3 mL

b. 86.1 mL

c. 63.9 mL

d. 49.6 mL

e. 113 mL

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.4

–

Chapter 08 - Applications of Aqueous Equilibria

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a strong acid by a strong base

46. Consider the titration of 100.0 mL of 0.250 M aniline (Kb = 3.8 × 10–10) with 0.500 M HCl. Calculate the pH of the solution at the stoichiometric point.

a. 8.70

b. 2.68

c. 11.62

d. –0.85

e. none of these

ANSWER: b

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

47. Calculate the pH at the equivalence point for the titration of 1.0 M ethylamine, C2H5NH2, by 1.0 M perchloric acid, HClO4. (pKb for C2H5NH2 = 3.25)

a. 6.05

b. 2.24

c. 2.09

d. 5.38

e. 5.53

ANSWER: e

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

48. A 100.0-mL sample of 0.2 M (CH3)3N (Kb = 5.3 × 10–5) is titrated with 0.2 M HCl. What is the pH at the equivalence point?

a. 5.4

b. 7.0

c. 10.3

d. 3.1

e. 9.9

ANSWER: a

POINTS: 1

DIFFICULTY: difficult

Chapter 08 - Applications of Aqueous Equilibria

TOPICS: 8.4

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

49. In the titration of a weak acid HA with 0.100 M NaOH, the stoichiometric point is known to occur at a pH value of approximately 10. Which of the following indicator acids would be best to use to mark the endpoint of this titration?

a. indicator A, Ka = 10–14

b. indicator C, Ka = 10–8

c. indicator D, Ka = 10–6

d. indicator B, Ka = 10–11

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.5

KEYWORDS: acids and bases | general chemistry | solutions of a weak acid or base

50. In the titration of a weak acid HA with 0.100 M NaOH, the stoichiometric point is known to occur at a pH value of approximately 11. Which of the following indicators would be best to use to mark the endpoint of this titration?

a. an indicator with Ka = 10–11

b. an indicator with Ka = 10–14

c. an indicator with Ka = 10–10

d. an indicator with Ka = 10–8

e. an indicator with Ka = 10–12

ANSWER: e

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.5

KEYWORDS: acids and bases | general chemistry | solutions of a weak acid or base

51. An indicator HIn has Ka = 1 × 10–8 At pH = 6.0, what is the ratio HIn/In–?

a. 1/100

b. 10/1

c. 1/1

d. 100/1

e. none of these

ANSWER: d

POINTS: 1

Chapter 08 - Applications of Aqueous Equilibria

DIFFICULTY: easy

TOPICS: 8.5

KEYWORDS: acids and bases | buffer | general chemistry | Henderson-Hasselbalch equation | solutions of a weak acid or base with another solute

52. A certain indicator HIn has a pKa of 9.00, and a color change becomes visible when 7.00% of it is In–. At what pH is this color change visible?

a. 6.15

b. 7.88

c. 10.2

d. 3.85

e. none of these

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.5

KEYWORDS: acids and bases | general chemistry | solutions of a weak acid or base

53. Methyl orange is an indicator with a Ka of 1 × 10–4 Its acid form, HIn, is red, while its base form, In–, is yellow. At pH 6.0, the indicator will be

a. yellow.

b. orange.

c. red.

d. blue.

e. not enough information

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.5

KEYWORDS: acids and bases | buffer | general chemistry | Henderson-Hasselbalch equation | solutions of a weak acid or base with another solute

54. Calculate the pH when 200.0 mL of a 1.00 M solution of H2A (Ka1 = 1.0 × 10–6 , Ka2 = 1.0 × 10–10) is titrated with the following volumes of 1.00 M NaOH.

100.0 mL of 1.00 M NaOH

a. 3.35

b. 8.00

c. 3.00

d. 6.00

e. none of these

ANSWER: d

Chapter 08 - Applications of Aqueous Equilibria

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

55. A solution containing 10. mmol of CO3 2– and 5.0 mmol of HCO3 – is titrated with 1.0 M HCl.

What volume of HCl must be added to reach the first equivalence point?

a. 5.0 mL

b. 15 mL

c. 10. mL

d. 25 mL

e. 20. mL

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

Consider the following information about the diprotic acid ascorbic acid (H2As for short, molar mass = 176.1).

The titration curve for disodium ascorbate, Na2As, with standard HCl is shown below:

56. What major species is(are) present at point III?

a. HAs– only

b. H2As and H+

c. H2As only

d. As2– and HAs–

e. HAs– and H2As

ANSWER: e

POINTS: 1

Ka1 pKa H2As H+ + HAs– 7.9 × 10–5 4.10 HAs H+ + As2– 1.6 × 10–12 11.79

Chapter 08 - Applications of Aqueous Equilibria

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

57. Which of the following is a major species present at point IV?

a. As2–

b. H+

c. H2As

d. HAs–

e. none of these

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

A 200.0-mL sample of the weak acid H3A (0.100 M) is titrated with 0.200 M NaOH. What are the major species at each of the following points in the titration? (Water is always assumed to be a major species.)

58. After 0 mL of 0.200 M NaOH is added

a. H3A, H2A–, HA2–, A3–

b. H3A, H2A–

c. H3A

d. H2A–, HA2–

e. H2A–

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

59. After 350.0 mL of 0.200 M NaOH is added

a. A3–

b. HA2–, A3–

c. HA2–

d. OH–, A3–

Chapter 08 - Applications of Aqueous Equilibria

e. H2A–, HA2

ANSWER: d

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

60. After 200.0 mL of 0.200 M NaOH is added

a. HA2–, A3

b. HA2–

c. H2A

d. H2A–, HA2–

e. A3–

ANSWER: b

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

61. After 75.0 mL of 0.200 M NaOH is added

a. H2A

b. HA2–

c. H3A, H2A

d. H2A–, HA2

e. H3A

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

62. After 220.0 mL of 0.200 M NaOH is added

a. A3

b. OH

c. H2A

, A3

, HA2

–

Chapter 08 - Applications of Aqueous Equilibria

d. HA2–

e. HA2–, A3–

ANSWER: e

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

63. After 100.0 mL of 0.200 M NaOH is added

a. H3A, HA2–, A3–

b. H2A–, HA2–

c. H3A, H2A–

d. H2A–

e. HA2–

ANSWER: d

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

64. After 300.0 mL of 0.200 M NaOH is added

a. A3–

b. OH–, A3–

c. H3A, H2A–, HA2–, A3–

d. H2A–, OH

e. OH–, HA2–, A3

ANSWER: a

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

65. In the titration of 100.0 mL of a 0.200 M solution of H2A

of 0.400 M NaOH must be added to reach a pH of 5.00?

a. 25.0 mL

b. 100.0 mL

–

(Ka1 = 1.0 × 10–5 , Ka2 = 1.0 × 10–8), what volume

Chapter 08 - Applications of Aqueous Equilibria

c. 0 mL

d. 150.0 mL

e. 50.0 mL

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

Consider the following information about the diprotic acid ascorbic acid (H2As for short, molar mass = 176.1).

The titration curve for disodium ascorbate, Na2As, with standard HCl is shown below:

66. What is the pH at point I (V1/2 HCl added)?

a. 7.95

b. 12.39

c. 10

d. 11.79

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

67. What is the pH at point III?

a. 7.95

b. 11.79

c. 12.39

d. 4.10

e. none of these

ANSWER: d

POINTS: 1

Ka1 pKa H2As H+ + HAs– 7.9 × 10–5 4.10

H+ + As2– 1.6 × 10–12 11.79

HAs

Chapter 08 - Applications of Aqueous Equilibria

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

68. A solution contains 10. mmol of H3PO4 and 5.0 mmol of NaH2PO4. How many milliliters of 0.10 M NaOH must be added to reach the second equivalence point of the titration of the H3PO4 with NaOH?

a. 2.0 × 102 mL

b. 150 mL

c. 250 mL

d. 1.0 × 102 mL

e. 50 mL

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

69. A solution contains 25 mmol of H3PO4 and 10. mmol of NaH2PO4. What volume of 2.0 M NaOH must be added to reach the second equivalence point of the titration of the H3PO4 with NaOH?

a. 60. mL

b. 30. mL

c. 5.0 mL

d. 12 mL

e. 25 mL

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

70. A 133.4-mL sample of a 0.21 M solution of H3PO4 is titrated with 0.14 M NaOH. What volume of base must be added to reach the third equivalence point?

a. 4.0 × 102 mL

b. 2.0 × 102 mL

c. 6.0 × 102 mL

d. 2.7 × 102 mL

e. 67 mL

Chapter 08 - Applications of Aqueous Equilibria

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

71. For carbonic acid (H2CO3), Ka1 = 4.30 × 10–7 and Ka2 = 5.62 × 10–11 . Calculate the pH of a 0.50 M solution of Na2CO3.

a. 11.97

b. 8.31

c. 10.67

d. 2.03

e. 3.33

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base properties of salt solutions | acids and bases | general chemistry | pH of a salt solution | solutions of a weak acid or base

72. A solution containing 10. mmol of CO

and 5.0 mmol of HCO3 – is titrated with 1.0 M HCl.

What total volume of HCl must be added to reach the second equivalence point?

a. 20. mL

b. 5.0 mL

c. 30. mL

d. 10. mL

e. 25 mL

ANSWER: e

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of a weak base by a strong acid

73. You dissolve 1.22 g of an unknown diprotic acid in 155.0 mL of H2O. This solution is just neutralized by 6.22 mL of a 1.23 M NaOH solution. What is the molar mass of the unknown acid?

a. 1.33 × 102 g/mol

b. 3.19 × 102 g/mol

–

Chapter 08 - Applications of Aqueous Equilibria

c. 3.09 × 102 g/mol

d. 1.59 × 102 g/mol

e. 1.96 × 102 g/mol

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

74. A 5.95-g sample of an acid, H2X, requires 45.0 mL of a 0.500 M NaOH solution for complete reaction (removing both protons). What is the molar mass of the acid expressed in grams per mol?

a. 529

b. 264

c. 178

d. 132

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

75. Consider the titration of 200.0 mL of a 0.100 M solution of the weak acid H2A with 0.200 M NaOH. The first equivalence point is reached after 100.0 mL of 0.200 M NaOH has been added, and the pH is 6.27. The pH after 65.0 mL of 0.200 M NaOH has been added, the pH is 4.95.

Calculate the pH of the original acid (no NaOH added).

a. 3.19

b. 2.84

c. 2.55

d. 4.68

e. none of these

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-ionization equilibria | acids and bases | experimental determination of Ka | general chemistry | solutions of a weak acid or base

Chapter 08 - Applications of Aqueous Equilibria

76. A 0.210-g sample of an acid (molar mass = 192 g/mol) is titrated with 30.5 mL of 0.108 M NaOH to a phenolphthalein endpoint. The formula of the acid is

a. H3A.

b. H2A.

c. H4A.

d. HA.

e. not enough information given

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

Calculate the pH when 200.0 mL of a 1.00 M solution of H2A (Ka1 = 1.0 × 10–6 , Ka2 = 1.0 × 10–10) is titrated with the following volumes of 1.00 M NaOH.

77. 0 mL of 1.00 M NaOH

a. 3.00

b. 6.00

c. 0

d. 3.35

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

78. 150.0 mL of 1.00 M NaOH

a. 6.00

b. 6.48

c. 8.00

d. 8.50

e. none of these

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

Chapter 08 - Applications of Aqueous Equilibria

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

79. 200.0 mL of 1.00 M NaOH

a. 9.50

b. 6.48

c. 6.00

d. 8.00

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

80. 250.0 mL of 1.00 M NaOH

a. 10.00

b. 9.52

c. 8.00

d. 9.00

e. none of these

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

81. 300.0 mL of 1.00 M NaOH

a. 10.00

b. 11.21

c. 9.50

d. 9.30

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

Chapter 08 - Applications of Aqueous Equilibria

82. 600.0 mL of 1.00 M NaOH

a. 11.70

b. 10.00

c. 14.00

d. 13.40

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

83. 400.0 mL of 1.00 M NaOH

a. 12.42

b. 11.76

c. 11.21

d. 10.00

e. none of these

ANSWER: b

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.6

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

84. The salt AgCl is soluble in strong acid solution than in water.

a. neither more nor less (that is, AgCl is equally soluble in strong acid and in water)

b. more

c. The Ksp value for AgCl must be known to answer this question.

d. less

ANSWER: a

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.6 8.7 8.8

KEYWORDS: effect of pH on solubility | general chemistry | qualitative effect of pH | solubility | solubility equilibria

85. Silver chromate, Ag2CrO4, has a Ksp of 9.0 × 10–12 Calculate the solubility, in moles per liter, of silver chromate.

Chapter 08 - Applications of Aqueous Equilibria

a. 7.8 × 10–5 M

b. 9.5 × 10–7 M

c. 9.8 × 10–5 M

d. 1.9 × 10–12 M

e. 1.3 × 10–4 M

ANSWER: e

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

86. The concentration of OH– in a saturated solution of Mg(OH)2 is 3.6 × 10–4 M What is Ksp for Mg(OH)2?

a. 1.3 × 10–7

b. 4.7 × 10–11

c. 1.2 × 10–11

d. 3.6 × 10–4

e. none of these

ANSWER: e

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

87. The solubility, in moles per liter, of Ag2CrO4 is 1.3 × 10–4 M at 25°C. Calculate Ksp for this compound.

a. 2.3 × 10–13

b. 6.1 × 10–9

c. 8.8 × 10–12

d. 4.7 × 10–13

e. 8.8 × 10–3

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

88. The solubility of AgCl in water is the solubility of AgCl in strong acid at the same temperature.

a. greater than

b. cannot be determined

Chapter 08 - Applications of Aqueous Equilibria

c. less than

d. about the same as

ANSWER: d

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.7

KEYWORDS: effect of pH on solubility | general chemistry | qualitative effect of pH | solubility | solubility equilibria

89. In the titration of 250.0 mL of 0.20 M H3PO4 with 0.10 M NaOH, the pH of the solution after the addition of some NaOH is 4.66. Which of the following phosphate-containing species is present in the largest amount? For H3PO4, Ka1 = 7.5 × 10–3 , Ka2 = 6.2 × 10–8 , and Ka3 = 4.8 × 10–13 .

a. H2PO4 –

b. H3PO4

c. PO4 3–

d. HPO4 2–

ANSWER: a

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.7

KEYWORDS: acid-base titration curve | acids and bases | general chemistry | solutions of a weak acid or base with another solute | titration of polyprotic acids

90. The solubility of Cd(OH)2 in water is 1.7 × 10–5 mol/L at 25°C. What is Ksp for Cd(OH)2?

a. 5.8 × 10–10

b. 2.9 × 10–10

c. 4.9 × 10–15

d. 2.0 × 10–14

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

91. The solubility of La(IO3)3 in a 0.10 M KIO3 solution is 1.0 × 10–7 mol/L. Calculate Ksp for La(IO3)3

a. 2.7 × 10–9

b. 2.7 × 10–27

c. 1.0 × 10–8

Page 32

Chapter 08 - Applications of Aqueous Equilibria

d. 1.0 × 10–10

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

92. In a solution prepared by adding excess PbI2(s) [Ksp = 1.4 × 10–8] to water, [I–] at equilibrium is

a. 2.4 × 10–3 mol/L.

b. 1.5 × 10–3 mol/L.

c. 8.4 × 10–5 mol/L.

d. 3.0 × 10–3 mol/L.

e. 1.2 × 10–3 mol/L.

ANSWER: d

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

93. Calculate the concentration of chromate ion, CrO4 2–, in a saturated solution of CaCrO4 (Ksp = 7.1 × 10–4).

a. 5.0 × 10–7 M

b. 3.5 × 10–2 M

c. 0.027 M

d. 3.5 × 10–4 M

e. 7.1 × 10–4 M

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.7

KEYWORDS: general chemistry | phases | solubility equilibria | solubility product constant

94. Solubility Products (Ksp)

BaSO4 1.5 × 10–9

CoS 5.0 × 10–22

PbSO4 1.3 × 10–8

AgBr 5.0 × 10–13

BaCO3 1.6 × 10–9

Chapter 08 - Applications of Aqueous Equilibria

Which of the following compounds is the most soluble (in moles per liter)?

a. PbSO4

b. BaSO4

c. CoS

d. AgBr

e. BaCO3

ANSWER: a

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.7

KEYWORDS: general chemistry | relative solubilities | solubility | solubility equilibria | solubility product constant

95. Consider the titration of 200.0 mL of a 0.100 M solution of the weak acid H2A with 0.200 M NaOH. The first equivalence point is reached after 100.0 mL of 0.200 M NaOH has been added, and the pH is 6.27. The pH after 65.0 mL of 0.200 M NaOH has been added, the pH is 4.95.

Calculate the value of Ka1 for H2A.

a. 2.10 × 10–5

b. 5.37 × 10–7

c. 1.40 × 10–8

d. 1.12 × 10–5

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: acids and bases | buffer | general chemistry | Henderson-Hasselbalch equation | solutions of a weak acid or base with another solute

96. How many moles of Fe(OH)2 [Ksp = 1.8 × 10–15] will dissolve in 1 L of water buffered at pH = 12.00?

a. 1.8 × 10–11 mol

b. 1.8 × 10–9 mol

c. 5.0 × 10–12 mol

d. 8.0 × 10–6 mol

e. 4.0 × 10–8 mol

ANSWER: a

POINTS: 1

Page

Chapter 08 - Applications of Aqueous Equilibria

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

97. You have two salts, AgX and AgY, with very similar Ksp values. You know that Ka for HX is much greater than Ka for HY. Which salt is more soluble in acidic solution?

a. AgY

b. This cannot be determined from the information given.

c. They are equally soluble in acidic solution.

d. AgX

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: effect of pH on solubility | general chemistry | qualitative effect of pH | solubility | solubility equilibria

98. Which of the following compounds has the lowest solubility, in moles per liter, in water?

a. MgC2O4 Ksp = 8.6 × 10–5

b. PbSO4 Ksp = 1.3 × 10–8

c. CdS Ksp = 1.0 × 10–28

d. Al(OH)3 Ksp = 2 × 10–32

e. Sn(OH)2 Ksp = 3 × 10–27

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | relative solubilities | solubility | solubility equilibria | solubility product constant

99. Which is the correct mathematical expression for the molar solubility (x) in moles per liter of Fe3(PO4)2?

a. 12x 3

b. 108x 5

c. 6x 2

d. 6x 5

e. 5x 6

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

Chapter 08 - Applications of Aqueous Equilibria

TOPICS: 8.7

KEYWORDS: general chemistry | phases | solubility equilibria | solubility product constant

100. The solubility of Fe(OH)2 in water is 7.9 × 10–6 mol/L at 25° C. What is Ksp for Fe(OH)2 at 25° C?

a. 4.9 × 10–16

b. 6.2 × 10–11

c. 2.5 × 10–10

d. 2.0 × 10–15

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

101. It is observed that 7.5 mmol of BaF2 will dissolve in 1.0 L of water. Use these data to calculate the value of Ksp for barium fluoride.

a. 1.7 × 10–6

b. 5.6 × 10–5

c. 2.1 × 10–12

d. 7.5 × 10–3

e. 4.2 × 10–7

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

102. The concentration of Ag+ in a saturated solution of Ag2CrO4 is 1.6 × 10–4 M. Calculate Ksp for Ag2CrO4

a. 2.6 × 10–8

b. 5.1 × 10–8

c. 1.6 × 10–11

d. 4.1 × 10–12

e. 2.0 × 10–12

ANSWER: e

POINTS: 1

DIFFICULTY: moderate

Chapter 08 - Applications of Aqueous Equilibria

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

103. The value of Ksp for AgI is 1.5 × 10–16 . Calculate the solubility, in moles per liter, of AgI in a 0.28 M NaI solution.

a. 1.2 × 10–8 mol/L

b. 5.4 × 10–16 mol/L

c. 2.1 × 10–9 mol/L

d. 4.2 × 10–17 mol/L

e. 2.3 × 10–8 mol/L

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | phases | solubility and the common-ion effect | solubility equilibria

104. What is the molar solubility of AgCl (Ksp = 1.6 × 10–10) in 0.0020 M sodium chloride at 25°C?

a. 8.0 × 10–8

b. 0.0020

c. 1.7 × 10–10

d. 1.3 × 10–5

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

105. Calculate the concentration of Ag+ in a saturated aqueous solution of Ag2CrO4. Ksp for Ag2CrO4 is 9.0 × 10–12 .

a. 9.0 × 10–12

b. 2.6 × 10–4

c. 2.1 × 10–4

d. 1.3 × 10–4

e. 6.5 × 10–5

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

Chapter 08 - Applications of Aqueous Equilibria

TOPICS: 8.8

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

106. The solubility of M(OH)2 in 0.010 M KOH is 1.0 × 10–5 mol/L. What is Ksp for M(OH)2?

a. 1.0 × 10–10

b. 4.0 × 10–15

c. 1.0 × 10–2

d. 1.0 × 10–12

e. 1.0 × 10–9

ANSWER: e

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

107. What is the solubility of Mg(OH)2 (Ksp = 8.9 × 10–12) in 1.0 L of a solution buffered (with large capacity) at pH 10.0?

a. 8.9 × 10–1 mol

b. 8.9 × 10–4 mol

c. 8.9 × 109 mol

d. 8.9 × 10–7 mol

e. none of these

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

108. Calculate the

a. 3.16 × 10–12 mol/L

b. 5.7 × 10–14 mol/L

c. 6.2 × 10–7 mol/L

d. 1.6 × 10–14 mol/L

e. none of these

ANSWER: b

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

2 M Ca(NO3)2 solution.

–

solubility

Ca3(PO4)2(s) (Ksp = 1.3 × 10–32) in a 1.0 × 10

Chapter 08 - Applications of Aqueous Equilibria

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

109. Calculate the solubility of Cu(OH)2 in a solution buffered at pH = 8.50. (Ksp = 1.6 × 10–19)

a. 5.7 × 10–10 M

b. 1.6 × 10–2 M

c. 1.8 × 10–7 M

d. 1.6 × 10–8 M

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

110. Given the following Ksp values Ksp

PbCrO4 2.0 × 10–16 Pb(OH)2 1.2 × 10–15

Zn(OH)2 4.5 × 10–17 MnS 2.3 × 10–13

which statement about solubility, in moles per liter, in water is correct?

a. The solubility of MnS in water will not be pH dependent.

b. MnS has the highest molar solubility in water.

c. PbCrO4 has the lowest solubility in water.

d. A saturated PbCrO4 solution will have a higher [Pb2+] than a saturated Pb(OH)2 solution.

e. PbCrO4, Zn(OH)2, and Pb(OH)2 have equal solubilities in water.

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | relative solubilities | solubility | solubility equilibria | solubility product constant

111. The Ksp value for PbSO4(s) is 1.3 × 10–8 Calculate the solubility, in moles per liter, of PbSO4(s) in a 0.0010 M solution of Na2SO4.

a. 4.5 × 10–6 M

b. 1.3 × 10–11 M

c. 1.3 × 10–5 M

d. 1.3 × 10–8 M

Ksp

Chapter 08 - Applications of Aqueous Equilibria

e. 1.4 × 10–4 M

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

112. Chromate ion is added to a saturated solution of Ag2CrO4 to reach 0.10 M CrO4 2– Calculate the final concentration of silver ion at equilibrium (Ksp for Ag2CrO4 is 9.0 × 10–12).

a. 9.5 × 10–6 M

b. 1.7 × 10–6 M

c. 5.5 × 10–11 M

d. 6.6 × 10–6 M

e. 1.1 × 10–13 M

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

113. Which of the following salts shows the lowest solubility in water? Ksp values are as follows:

Ag2S = 1.6 × 10–49 ; Bi2S3 = 1.0 × 10–72 ; HgS = 1.6 × 10–54 ; Mg(OH)2 = 8.9 × 10–12 ; MnS = 2.3 × 10–13)

a. HgS

b. MnS

c. Bi2S3

d. Ag2S

e. Mg(OH)2

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | relative solubilities | solubility | solubility equilibria | solubility product constant

114. How many moles of CaF2 will dissolve in 3.0 L of 0.050 M NaF solution? (Ksp for CaF2 = 4.0 × 10–11)

a. 4.8 × 10–8 mol

b. 8.9 × 10–9 mol

c. 8.0 × 10–8 mol

Chapter 08 - Applications of Aqueous Equilibria

d. 2.7 × 10–8 mol

e. none of these

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

115. Which of the following compounds has the lowest solubility, in moles per liter, in water at 25°C?

a. CdS Ksp = 1.0 × 10–28

b. CaSO4 Ksp = 6.1 × 10–5

c. Al(OH)3 Ksp = 2.0 × 10–32

d. Sn(OH)2 Ksp = 3.0 × 10–27

e. Ag3PO4 Ksp = 1.8 × 10–18

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | relative solubilities | solubility | solubility equilibria | solubility product constant

116. Calculate the solubility of Ag2CrO4 [Ksp = 9.0 × 10–12] in a 1.0 × 10–2 M AgNO3 solution.

a. 1.3 × 10–4 mol/L

b. 2.3 × 10–8 mol/L

c. 9.0 × 10–8 mol/L

d. 9.0 × 10–10 mol/L

e. none of these

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

117. Calculate the molar solubility of BaCO3 (Ksp = 1.6 × 10–9) in a 0.42M BaCl2 solution.

a. 6.2 × 10–4 M

b. 2.6 × 10–5 M

c. 6.7 × 10–10 M

d. 4.0 × 10–5 M

Chapter 08 - Applications of Aqueous Equilibria

e. 3.8 × 10–9 M

ANSWER: e

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

118. The two salts AgX and AgY have very similar solubilities in water. The salt AgX is much more soluble in acid than is AgY. What can be said about the relative strengths of the acids HX and HY?

a. HY is stronger than HX.

b. Nothing can be said about their relative strengths.

c. HX is stronger than HY.

d. The acids have equal strengths.

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

119. You have a solution of 0.10 M Cl– and . You add 0.10 M silver nitrate dropwise into the solution. Ksp for Ag2CrO4 is 9.0 × 10–12 and for AgCl is1.6 × 10–10 . Which of the following will precipitate first?

a. silver nitrate

b. silver chromate

c. cannot be determined from the information given

d. silver chloride

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: fractional precipitation | general chemistry | precipitation calculations | solubility | solubility equilibria

120. 2.42 × 10–3 g of BaSO4 can dissolve in 1.00 L of water. Calculate Ksp for this salt.

a. 1.08 × 10–10

b. 2.42 × 10–3

c. 5.86 × 10–6

d. 1.05 × 10–5

e. none of these

Chapter 08 - Applications of Aqueous Equilibria

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

121. Consider the titration of 200.0 mL of a 0.100 M solution of the weak acid H2A with 0.200 M NaOH. The first equivalence point is reached after 100.0 mL of 0.200 M NaOH has been added, and the pH is 6.27. The pH after 65.0 mL of 0.200 M NaOH has been added, the pH is 4.95.

Calculate the value of Ka2 for H2A.

a. 5.37 × 10–7

b. 2.10 × 10–5

c. 1.12 × 10–5

d. 1.40 × 10–8

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.7

KEYWORDS: acids and bases | buffer | general chemistry | Henderson-Hasselbalch equation | solutions of a weak acid or base with another solute

122. Titrating 30.00 mL of a saturated calcium iodate solution requires 28.91 mL of a 0.092 M solution of Na2S2O3 according to the equation

Calculate Ksp for Ca(IO3)2.

a. 7.14 × 10–8

b. 6.97 × 10–4

c. 2.79 × 10–3

d. 1.61 × 10–6

e. none of these

ANSWER: d

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.7

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

123. Determine the pH of a solution prepared by mixing the following: 25.0 mL of 0.200 M HCl

Page 43

3 –

6S2O3 2–+ 6H+ → I–+ 2S3O6 2–+ 3H2O

Chapter 08 - Applications of Aqueous Equilibria

75.0 mL of 0.100 M NaOH

50.0 mL of 0.200 M NaH2PO4

For H3PO4 Ka1 = 7.5 × 10–3 , Ka2 = 6.2 × 10–8 , and Ka3 = 4.8 × 10–13 .

a. 6.73

b. 5.18

c. 1.65

d. 4.32

e. 4.25

ANSWER: a

POINTS: 1

DIFFICULTY: difficult

TOPICS: 8.7

KEYWORDS: acids and bases | buffer | general chemistry | pH of a buffer | solutions of a weak acid or base with another solute

124. Sodium chloride is added slowly to a solution that is 0.010 M in Cu+ , Ag+ , and Au+ . The Ksp values for the chloride salts are 1.9 × 10–7 , 1.6 × 10–10 , and 2.0 × 10–13 , respectively. Which compound will precipitate first?

a. AgCl(s)

b. CuCl(s)

c. AuCl(s)

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.8

KEYWORDS: general chemistry | relative solubilities | solubility | solubility equilibria | solubility product constant

125. The concentration of Al3+ in a saturated solution of Al(OH)3 at 25°C is 5.2 × 10–9 M. Calculate the Ksp for Al(OH)3.

a. 7.3 × 10–34

b. 2.9 × 10–33

c. 2.0 × 10–32

d. 5.6 × 10–25

e. none of these

ANSWER: c

POINTS: 1

DIFFICULTY: easy

TOPICS: 8.8

KEYWORDS: general chemistry | solubility | solubility equilibria | solubility product constant

Chapter 08 - Applications of Aqueous Equilibria

126. Calculate the solubility of Ag2SO4 [Ksp = 1.2 × 10–5] in a 2.0 M AgNO3 solution.

a. 6.0 × 10–6 mol/L

b. 1.2 × 10–5 mol/L

c. 3.0 × 10–6 mol/L

d. 1.4 × 10–2 mol/L

e. none of these

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.8

KEYWORDS: general chemistry | solubility | solubility and the common-ion effect | solubility equilibria

127. The Ksp of Al(OH)3 is 2.0 × 10–32 . At what pH will a 0.2 M Al3+ solution begin to show precipitation of Al(OH)3?

a. 8.4

b. 5.6

c. 3.7

d. 10.3

e. 1.0

ANSWER: c

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.8

KEYWORDS: effect of pH on solubility | general chemistry | solubility | solubility equilibria

128. If 30 mL of 5.0 × 10–4 M Ca(NO3)2 is added to 70 mL of 2.0 × 10–4 M NaF, will a precipitate form? (Ksp of CaF2 = 4.0 × 10–11)

a. No, because the ion product is less than Ksp

b. No, because the ion product is greater than Ksp.

c. Not enough information is given.

d. Yes, because the ion product is less than Ksp.

e. Yes, because the ion product is greater than Ksp

ANSWER: a

POINTS: 1

DIFFICULTY: moderate

TOPICS: 8.8

KEYWORDS: criterion for precipitation | general chemistry | precipitation calculations | solubility | solubility equilibria