Exam Questions Ch.16 Applications Of Aqueous Equilibria 774 - Chemistry Canada 4e | Complete Test Bank by John A. Olmsted. DOCX document preview.

Exam Questions Ch.16 Applications Of Aqueous Equilibria 774

CHAPTER 16

APPLICATIONS OF AQUEOUS EQUILIBRIA

CHAPTER STUDY OBJECTIVES

1. Calculate the pH of a buffered solution.

SKILLS TO MASTER: Using the buffer equation to calculate the pH of a buffer solution; calculating the pH of a buffered solution after adding a small amount of acid or base

KEY CONCEPTS: A buffer solution contains both a weak acid and its conjugate base, or a weak base and its conjugate acid, as major species in solution.

2. Explain how to prepare a buffered solution of known pH and capacity.

SKILLS TO MASTER: Calculating the capacity of a buffered solution; preparing a buffered solution

3. Calculate an acid or a base concentration from titration data.

SKILLS TO MASTER: Calculating the concentration of a standard base solution; calculating the pH at the stoichiometric point of a titration; recognizing qualitative features of a titration curve; constructing a titration curve for a polyprotic acid; choosing an appropriate indicator for a particular acid–base titration

KEY CONCEPTS: At the stoichiometric point, just enough hydroxide ions have been added to react with every acidic proton present in the acid solution before the titration was started.

4. Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

SKILLS TO MASTER: Using Ksp to determine solution concentrations and precipitate masses; using the common-ion effect to calculate solution concentrations; calculating the effect of pH on the solubility of a salt

KEY CONCEPTS: Addition of a common ion decreases the solubility of a salt. Lowering the pH increases the solubility of a salt that contains a basic anion.

5. Calculate the concentrations of species involved in complex-ion formation.

SKILLS TO MASTER: Understanding the role of complex formation on the solubility of a solid

KEY CONCEPTS: Electron-donating ligands are Lewis bases. Electron-accepting metals are Lewis acids. The chelate effect results in very high formation constants of complexes.

Multiple Choice QUESTIONS

1. When 0.1 moles of HC2H3O2 and 0.1 moles of NaC2H3O2 are added to make 1 L of aqueous solution, the major species present are

a) Na+, H2O and HC2H3O2.

b) HC2H3O2 and C2H3O2-.

c) H2O, Na+ and C2H3O2-.

d) NaC2H3O2, HC2H3O2 and H2O.

e) Na+, H2O, HC2H3O2, C2H3O2-.

Difficulty: Easy

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

Feedback: This question requires that student recognize the species present in a simple acetic acid buffer solution. a) neglects the acetate anion; b) fails to recognize that Na+ ion is also in solution; c) neglects the acetic acid molecule; d) fails to recognize that sodium acetate salt dissociates; e) correct answer

2. When 0.1 moles of HCl are added to 1 L of solution containing 0.12 moles of aqueous Na2CO3, the major species present are

a) Na+, Cl-, H2O, HCO3-.

b) Na+, Cl-, H2O, CO32-.

c) Na+, Cl-, H2O, HCO3-, CO32-.

d) HCl, Na+, CO32-.

e) Na+, Cl-, H2O, H2CO3, CO32.

Difficulty: Hard

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

3. What are the major species present when 10 ml of 0.1 M NaOH are added to 40 ml of 0.0.05 M CH3COOH?

a) Na+, OH-, H2O and CH3COOH

b) CH3COOH and CH3COO-

c) H2O, Na+ and CH3COO-

d) NaOH, CH3COOH and H2O

e) Na+, H2O, CH3COOH, CH3COO-

Difficulty: Hard

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

Feedback: Recognize major species in buffer prepared from strong base and a weak acid. a) neglects neutralization of strong base by weak acid; b) neglects spectator ion; c) acetic acid is in excess and some will remain; d) neglects all reactions; e) correct answer

4. Which of the following mixtures would make a buffer of pH = 4.74? (pKa of acetic acid is 4.74)

a) 100 mL of 0.10 M HCl and 100 mL of 0.10 M NaC2H3O2

b) 100 mL of 0.10 M HCl and 100 mL of 0.10 M HC23O2

c) 50 mL of 0.10 M HC23O2 and 100 mL of 0.10 M NaC2H3O2

d) 100 mL of 0.10 M NaOH and 100 mL of 0.10 M NaC2H3O2

e) 50 mL of 0.10 M NaOH and 100 mL of 0.10 M HC23O2

Difficulty: Hard

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

5. Which of the following solutions is a buffer?

I. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.05 M NaOH

II. 100 mL of 0.1 M acetic acid mixed with 25 mL of 0.5 M NaOH

III. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.05 M sodium acetate

IV. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.2 M HCl

V. 50 mL of 0.1 M sodium acetate mixed with 25 mL of 0.05 M NaOH

a) I and III

b) III only

c) I, II and V

d) III, IV and V

e) II, IV and V

Difficulty: Hard

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

6. A solution is made by the addition of 0.34 mole of Na2HPO4 and 0.65 mole of NaH2PO4 and sufficient water to give a total volume of 1.2 L. How many grams of NaOH would need to be added to increase the pH by 0.2 pH units?

If needed, use the following equation: pH = pKa + log(A/HA)

a) 0.34

b) 3.4

c) 4.4

d) 5.4

e) 6.5

Difficulty: Hard

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

7. Two buffer solutions of the same pH are prepared. After adding 15 millimoles of acid to 250 mL of one solution, the pH changes by 0.02 pH unit. Adding 15 millimoles of acid to 250 mL of the second solution results in a 0.05 pH unit change. Which statement most accurately describes the differences between the two buffer solutions?

a) The first solution has a lower capacity than the second solution.

b) The first solution has a higher concentration of conjugate acid than conjugate base.

c) The second solution has a lower concentration of conjugate base that the first solution.

d) The second solution has a higher acid to conjugate base ratio than the first solution.

e) The first solution has a higher capacity than the second solution.

Difficulty: Hard

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

8. The pKa’s of the three acid-conjugate base pairs derived from phosphoric acid, H3PO4, are 2.12, 7.21 and 12.32. How many g of sodium hydroxide would have be added to 150 mL of 0.1 M H3PO4 to prepare a buffer of pH = 7.41?

a) 0. 21 g

b) 0.34 g

c) 0.56 g

d) 0.83 g

e) 0.91 g

Difficulty: Hard

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

9. A biochemist wishes to study the activity of an enzyme that produces one mole hydrogen ion for every mole of product produced. However, the enzyme is sensitive to changes in pH and will become inactive at pH less than 7.1. If the initial pH of the solution is 7.21 set with a phosphate buffer (the pKa’s of the three acid-conjugate base pairs derived from phosphoric acid, H3PO4, are 2.12, 7.21 and 12.32) and the reaction is designed to produce 0.15 moles of product in a volume of 250 mL, what is the minimum concentration of the conjugate acid in the buffer solution?

a) 0.46 M

b) 0.53 M

c) 0.82 M

d) 1.12 M

e) 1.25 M

Difficulty: Hard

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

10. Which of the following will have the smallest pH change upon the addition of 0.001 mole strong acid?

a) 50 mL of a total concentration of 0.10 M pH = 5.2 acetate buffer

b) 100 mL of a total concentration of 0.050 M pH = 5.2 acetate buffer

c) 75 mL of a total concentration of 0.067 M pH = 5.2 acetate buffer

d) 250 mL of a total concentration of 0.020 M pH = 5.2 acetate buffer

e) 25 ml of total concentration of 0.20 M pH = 5.2 acetate buffer

Difficulty: Medium

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

11. In which of the following titrations will the pH be less than 7 at the equivalence point?

a) acetic acid titrated with sodium hydroxide

b) perchloric acid titrated with lithium hydroxide

c) sodium hydroxide titrated with hydrochloric acid

d) ammonium bromide titrated with perchloric acid

e) perchloric acid titrated with methyl amine

Difficulty: Easy

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

12. In which of the following titrations will the pH be greater than 7 at the equivalence point?

a) perchloric acid titrated with lithium hydroxide

b) acetic acid titrated with sodium hydroxide

c) sodium hydroxide titrated with hydrochloric acid

d) ammonium bromide titrated with perchloric acid

e) perchloric acid titrated with methyl amine

Difficulty: Easy

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

13. The equivalence point for titration of 50 ml of 0.1 M formic acid, HCO2H (a weak acid) requires what volume of 0.2 M sodium hydroxide?

a) 50 ml

b) 25 ml

c) less than 25 ml as formic acid is a weak acid

d) 100 ml

e) between 25 and 50 ml as formic acid is a weak acid

Difficulty: Easy

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

Feedback: Equivalence point is reached when moles of base added are the same as moles of acid initially present, C1V1 = C2V2.

14. The equivalence point for titration of 50 ml of 0.1 M formic acid, HCO2H (a weak acid) requires what volume of 0.2 M methylamine, CH3NH2 (a weak base)?

a) 50 ml

b) 25 ml

c) impossible to determine unless Ka for formic acid is known

d) impossible to determine unless Kb for methalamine is known

e) impossible to determine unless both Ka for formic acid and Kb for methylamine are known

Difficulty: Easy

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

Feedback: Equivalence point is reached when moles of base added are the same as moles of acid initially present, C1V1 = C2V2.

15. The equivalence point for titration of 50 ml of 0.1 M formic acid, HCO2H (a weak acid) requires what volume of 0.2 M calcium hydroxidem, Ca(OH)2?

a) 50 ml

b) 25 ml

c) less than 12.5 ml as formic acid is a weak acid

d) 12.5 ml

e) between 12.5 and 25 ml as formic acid is a weak acid

Difficulty: Medium

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

Feedback: Each Ca(OH)2 provides 2 OH-; equivalence point is reached when moles of base added (OH-) are the same as moles of acid initially present, C1V1 = C2V2.

16. Sodium carbonate, also called soda ash, is often analyzed by titration with strong acid. Initial concentrations of the carbonate ions are about 0.1 M. What indicator(s) would be suitable for detecting the stoichiometric point in this titration? (for H2CO3: pKa1 = 3.75 pKa2 = 10.33)

Indicator

Color at pH’s

I.

methyl orange

2.8 (red)-3.5 (yellow)

II.

methyl red

4.4 (red)-6.2 (yellow)

III.

phenol red

6.4 (yellow) - 8.0(red)

IV.

thymol blue

8.0(yellow)-9.6(blue)

V.

alizarin yellow R

10.0(yellow)-12.1(red)

a) I only

b) I and II

c) III only

d) IV only

e) I and V

Difficulty: Easy

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

17. Your friend at Podunk U. urgently emails asking if you remember any acid–base chemistry. His problem is that for his final organic lab problem he needs to report the pKa of his acid. He did the pH titration, but only wrote down the following data for the titratant volumes: V = 14.27 mL; pH = 4.73. Vequivalence point = 22.43 mL. What is the pKa of the unknown acid?

a) 3.5

b) 4.0

c) 4.5

d) 4.8

e) 4.9

Difficulty: Hard

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

18. The following graph shows the titration of a 0.150 M benzoic acid solution. At what pH range would benzoic acid be appropriate to make a buffer?

a) 2.2

b) 3.5

c) 4.0

d) 7.5

e) 10.0

Difficulty: Easy

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

19. What is the solubility product expression for barium phosphate?

a) Ksp = [Ba2+][PO43-]

b) Ksp = [Ba2+]2[PO43-]3

c) Ksp = [Ba2+]3[PO43-]2

d) Ksp = [Ba2+]3[PO43-]3

e) Ksp = [Ba2+][PO43-]3

Difficulty: Easy

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

20. What is the solubility product expression for iron (III) hydroxide?

a) Ksp = [Fe3+][OH-]

b) Ksp = [Fe3+][OH-]3

c) Ksp = [Fe3+]2[OH-]3

d) Ksp = [Fe3+]3[OH-]

e) Ksp = [Fe3+]3[OH-]3

Difficulty: Easy

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

21. At what pH will an 0.0010 M of iron (III) nitrate (Ksp (Fe(OH)3)= 4.0 x 10-38) begin to precipitate?

a) 2.55

b) 4.22

c) 5.55

d) 7.41

e) 9.03

Difficulty: Medium

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

22. Which of the following is NOT one of the equilibria in the complexation of Ru(II) by ammonia (coordination number = 6)?

a) Ru2+(aq) + NH3(aq) ⇌Ru(NH3)2+ (aq)

b) Ru2+(NH3)(aq) + NH3(aq) ⇌Ru2+(NH3)2 (aq)

c) Ru2+(NH3)4 (aq) + NH3(aq) ⇌Ru2+(NH3)5 (aq)

d) Ru2+(NH3)3 (aq) + NH3(aq) ⇌Ru2+(NH3)4 (aq)

e) Ru2+(NH3)6 (aq) + NH3(aq) ⇌Ru2+(NH3)7 (aq)

Difficulty: Easy

Learning Objective: Calculate the concentrations of species involved in complex-ion formation.

Section Reference: 16.5 Complexation Equilibria

23. In which solution will copper nitrate (Cu(NO3)2) be the least soluble?

a) 0.1 M NaNO3

b) 0.1 M NH3

c) pure water

d) 0.1 M CuCl2

e) 0.1 M NaOH

Difficulty: Medium

Learning Objective: Calculate the concentrations of species involved in complex-ion formation.

Section Reference: 16.5 Complexation Equilibria

ESSAY QUESTIONS

24. Write the equation showing how Na2HPO4 and NaH2PO4 act as a buffer. Also write an equation showing how this buffer reacts when HCl is added.

Difficulty: Hard

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

25. When is a solution a buffer?

Difficulty: Easy

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

26. What is the pH of a solution made by the addition of 0.34 mole of Na2HPO4 and 0.65 mole of NaH2PO4 and sufficient water to give a total volume of 1.2 L?

If needed, use the following equation: pH = pKa + log(A/HA)

Difficulty: Medium

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

27. What is the pH upon adding 10 mL of 1.0 M NaOH to a) 1.2 L of water and b) a solution made by the addition of 0.34 mole of Na2HPO4 and 0.65 mole of NaH2PO4 and sufficient water to give a total volume of 1.2 L?

If needed, use the following equation: pH = pKa + log(A/HA)

Difficulty: Hard

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

28. What is the pH of the resulting solution when 0.1 moles of HCl are added to 1 L of solution containing 0.12 moles of aqueous Na2CO3?

If needed, use the following equation: pH = pKa + log(A/HA)

Difficulty: Hard

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

29. A solution is prepared by adding 0.14 mole Na2HPO4 and 8.2 grams of NaH2PO4 to sufficient water to prepare 1.00 L of solution. What is the pH of the solution? What is the pH of the solution after 0.041 moles HCl are added?

If needed, use the following equation: pH = pKa + log(A/HA)

Difficulty: Hard

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

30. Determine the ratio of conjugate base to acid for a buffered solution at pH = 4.3 made from propionic acid, (Ka = 1.34 x10-5).

Difficulty: Medium

Learning Objective: Calculate the pH of a buffered solution.

Section Reference: 16.1 Buffer Solutions

31. In the selective precipitation of metal ions as sulphide salts, usable concentrations of the S2- ions can be maintained by bubbling H2S through a solution buffered at a pH of about 10.0. How many mL of 6 M HCl would need to be added to 3.5 mL of a solution 5.6 M in aqueous ammonia to form a solution buffered at pH 10.0?

Difficulty: Hard

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

32. A buffer is made by adding 0.82 moles of monohydrogen phosphate and 0.82 moles of dihydrogen phosphate to sufficient water to give a total volume of 1.0 L. Hydrogen chloride, 0.60 moles, is absorbed into the solution. What is the pH?

Difficulty: Hard

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

33. A buffer solution made from ammonia, NH3, and ammonium chloride, NH4Cl, is prepared with pH 9.0. If Kb of NH3 is 1.78x10-5 and the total concentration of all nitrogen species is 0.25 M, what are the nitrogen concentrations?

Difficulty: Hard

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

Feedback: This is a difficult question as student must determine Ka from given data prior to using Henderson-Hasselbalch equation.

34. What mass of sodium acetate must be added to 250 mL of 0.10 M acetic acid (pKa = 4.75) to give a buffer of pH = 4.90?

Difficulty: Medium

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

35. What mass of ammonium chloride (pKa (NH4+) = 9.25) must be added to 2.0 L of 0.65 M aqueous ammonia to give a buffer with pH = 8.75?

Difficulty: Medium

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

36. How would you make a buffer of pH = 3.75 from the following solutions with a total volume of 300 ml? (Ka = 1.77 x10-4 for Formic Acid, HCO­2H); 0.100M HCOOH; 0.100 M HClO­4; 0.100 M NaOH; 0.100 M NH3

Difficulty: Hard

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

37. How would you make a buffer of pH = 3.75 from the following solutions with a total volume of 450 ml? (Ka = 1.77 x10-4 for Formic Acid, HCO­2H); 0.100M NaHCOO; 0.100 M HClO­4; 0.100 M NaOH, 0.100 M NH3

Difficulty: Hard

Learning Objective: Explain how to prepare a buffered solution of known pH and capacity.

Section Reference: 16.2 Capacity and Preparation of Buffer Solutions

38. Calculate the pH in the titration of a 0.325 g sample of acetylsalicylic acid (see line drawing below) initially in 25.0 mL water with 0.102 M NaOH when (a) 6.23 mL and (b) 20.00 mL of titrant have been added.

HA = acetylsalicylic acid (aspirin)

C9H8O4

Ka = 3.00 x 10-4 M, pKa = 3.523

Difficulty: Medium

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

39. Calculate the pH in the titration of a 0.325 g sample of acetylsalicylic acid (see line drawing below) initially in 25.0 mL water with 0.102 M NaOH when (a) 8.84 mL and (b) 17.68 mL of titrant have been added.

HA = acetylsalicylic acid (aspirin)

C9H8O4

Ka = 3.00 x 10-4 M, pKa = 3.523

Difficulty: Medium

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

40. A qualitative sketch of the titration curve for acetylsalicylic acid is shown at right.

Given the indicators listed, which one would be suitable for titrations of acetylsalicylic acid?

Indicator

Color at pH’s

methyl orange

2.8 (red)-3.5 (yellow)

bromocresol green

(3.5 (yellow)-5.6 (blue)

phenol red

6.4 (yellow) - 8.0(red)

thymolphthalein

9.3(colorless)-10.5(blue)

alizarin yellow R

10.0(yellow)-12.1(red)

Difficulty: Easy

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

41. Consider the titration of 25.0 mL of an 0.11 M lactic acid solution CH3CH(OH)COOH (Ka= 8.4 x10-4) with 0.150 M NaOH.

a) Calculate the pH after 10 mL of the NaOH solution has been added.

b) Calculate the pH of the solution at 10 mL of NaOH past the stoichiometric point.

c) Would phenol red (pKin = 7.9) or phenolphthalein (pKin = 9.4) be a better indicator for this titration?

Difficulty: Hard

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

42. A 5.00 mL sample containing sulphurous acid (Ka1 = 1.5 x 10-2; Ka2 = 1.0 x10-7) is titrated with 0.1324M NaOH. The stoichiometric point is at 25.32 mL. How many g of H2SO3 are present in the sample and what is the pH at the stoichiometric point?

Difficulty: Hard

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

43. At what pH would a solution with a total concentration of 0.20 M phosphate NOT be a buffer? Explain your answer.

Difficulty: Medium

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

44. What is the concentration of phosphate in a solution made from a 0.22 M phosphoric acid whose pH has been adjusted to 7.1 by the addition of solid potassium hydroxide (assume no volume change)?

Difficulty: Medium

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

45. What is the concentration of phosphate in a solution made from a 0.15 M phosphoric acid whose pH has been adjusted to 7.6 by the addition of solid sodium hydroxide (assume no volume change)?

Difficulty: Hard

Learning Objective: Calculate an acid or a base concentration from titration data.

Section Reference: 16.3 Acid–Base Titrations

46. An aqueous solution is 0.2 M in both Mg2+ and Pb2+ ions. You wish to separate the two metal ions by adding oxalate, C2O42-. What is the highest possible oxalate-ion concentration such that only one of the two metal ions will form a solid?

Ksp (MgC2O4) = 8.5 x 10-5 Ksp (PbC2O4) = 2.7 x 10-11

Difficulty: Medium

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

47. An aqueous solution is 0.2 M in both Mg2+ and Pb2+ ions. You wish to separate the two metal ions by adding oxalate, C2O42-. What is the concentration of lead when Mg2+ begins to precipitate?

Ksp (MgC2O4) = 8.5 x 10-5 Ksp (PbC2O4) = 2.7 x 10-11

Difficulty: Hard

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

Feedback: Student must first determine concentration of C2O42- required for 0.2 M Mg2+ to ppt, using this value determine concentration of Pb2+ in solution.

48. 100 ml of aqueous solution is 0.2 M in both Mg2+ and Pb2+ ions. You wish to separate the two metal ions by adding oxalate, C2O42-. What is the mass of PbC2O4 precipitated before Mg2+ begins to precipitate?

Ksp (MgC2O4) = 8.5 x 10-5 Ksp (PbC2O4) = 2.7 x 10-11

Difficulty: Hard

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

Feedback: Pb2+ concentration drops from 0.2 M to 6.3x10-8 M; essentially all of the Pb2+ is precipitated before magnesium oxalate begins to precipitate.

49. At 100°C, 2.1 x 10-2 g of AgCl dissolves in 1.00 L of water. What is the Ksp of AgCl at 100°C?

Difficulty: Medium

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

50. What is the concentration of Ca2+ in a saturated solution of calcium phosphate Ca3(PO4)2 (Ksp = 2.07 x 10-33)?

Difficulty: Easy

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

51. A real world application of solubility products is found in softening “hard water” by removing calcium and magnesium ions present as the sulphate and carbonate salts. We can decrease the amount of Ca2+ in solution by addition of sodium carbonate, a strong electrolyte:

Na2CO3 (s)  2 Na+ (aq) + CO32- (aq)

By increasing the concentration of carbonate, we tend to drive the solubility equilibrium toward the reactant:

CaCO3 (s) Ca2+ (aq) + CO32- (aq) Ksp = 4.7 x 10-9

If the initial [Ca2+] = 5.0 x 10-3 M, what percent of the [Ca2+] will be removed if the carbonate concentration is maintained at 1.0 x 10-3 M?

Difficulty: Medium

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

52. How many grams of AgNO3 will dissolve in 1.32 L of a pH = 12.23 solution (Ksp AgOH = 1.52 x 10-8)?

Difficulty: Medium

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

53. What is the molar concentration of barium in a 0.01 M ammonium sulphate solution saturated with Ba(NO3)2?

Difficulty: Medium

Learning Objective: Use the concepts of Ksp and the common-ion effect to calculate solution concentrations.

Section Reference: 16.4 Solubility Equilibria

54. A 1.32 g sample of solid CuCl2•2H2O is added to 1.21 L of 2.0 M aqueous ammonia. The formation constant for tetra-amminecopper (II) is 1.0 x 1012. What is the concentration of aqueous copper (II) in this solution?

Difficulty: Medium

Learning Objective: Calculate the concentrations of species involved in complex-ion formation.

Section Reference: 16.5 Complexation Equilibria

55. Trisbipyridylruthenium (II), Ru2+(BIPY)3, has been used in many studies, most recently to understand electrical conductivity in DNA molecules. Write all the equilibria in the formation of this complex.

Difficulty: Medium

Learning Objective: Calculate the concentrations of species involved in complex-ion formation.

Section Reference: 16.5 Complexation Equilibria

56. 1.1 mg of HgS (Ksp1.6 x 10-54) is added to 1.0 L of 0.01 M NaCN ((Hg(CN)4+2), Kf = 4 x 1041)? What is the value of the reaction quotient for Qsp and will all of the HgS dissolve?

Difficulty: Hard

Learning Objective: Calculate the concentrations of species involved in complex-ion formation.

Section Reference: 16.5 Complexation Equilibria

57. Write the formation constant expression for the formation of [Zn(OH)4]-2 from Zn(OH)2(s).

Zn(OH)2(s) + OH-(aq) [Zn(OH)3­]-1(aq)

[Zn(OH)3]-1(aq) + OH-(aq) [Zn(OH)4]-2(aq)

Difficulty: Medium

Learning Objective: Calculate the concentrations of species involved in complex-ion formation.

Section Reference: 16.5 Complexation Equilibria

58. Write the equilibrium constant expression and determine the solubility constant for CuS dissolving in an ammonia solution. (CuS, K­sp = 8 x10-37) (Cu(NH3)4+2, Kf = 1.1 x 1013)

Difficulty: Medium

Learning Objective: Calculate the concentrations of species involved in complex-ion formation.

Section Reference: 16.5 Complexation Equilibria

59. What is the concentration of [PbCl3]-prepared from a solution of 0.5 M CaCl2 and solid PbCl2?

Pb2+ + 3Cl- ⇌ [PbCl3]- Kf = 2.4 x 101

pKsp(PbCl2) = 4.77

Difficulty: Hard

Learning Objective: Calculate the concentrations of species involved in complex-ion formation.

Section Reference: 16.5 Complexation Equilibria

Feedback: Multi-step problem. Must use complex-ion equilibrium expression for [PbCl3]- and convert pKsp to Ksp, and recognize that 0.5 M CaCl2 results in initial concentration of 1.0 M Cl-.

60. What is the concentration of [PbCl3]- when PbCl2 solid first begins to precipitate on addition of NaCl salt to a 0.5 M PbNO3 solution?

Pb2+ + 3Cl- ⇌ [PbCl3]- Kf = 2.4 x 101

pKsp(PbCl2) = 4.77

Difficulty: Hard

Learning Objective: Calculate the concentrations of species involved in complex-ion formation.

Section Reference: 16.5 Complexation Equilibria

Feedback: Multi-step problem involving determination of Cl- concentration required for PbCl2 precipitation, then the determination of the concentration of [PbCl3]-.

61. Write the equilibrium constant expression and determine the solubility constant for HgS dissolving in a cyanide solution. (HgS, Ksp = 1.6 x 10-54) ((Hg(CN)2), Kf = 4 x 1041)

Difficulty: Medium

Learning Objective: Calculate the concentrations of species involved in complex-ion formation.

Section Reference: 16.5 Complexation Equilibria

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Document Information

Document Type:
DOCX
Chapter Number:
16
Created Date:
Aug 21, 2025
Chapter Name:
Chapter 16 Applications Of Aqueous Equilibria 774
Author:
John A. Olmsted

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