GENERAL CHEMISTRY II CHEM 1312 SPRING 1999 EXAM 4 STUDY QUESTIONS
1. What is the molar solubility of barium fluoride, BaF₂ in distilled water at 25 ^oC? BaF₂ is a slightly soluble salt. It establishes the following equilibrium when put in water:

BaF₂(s)    <---------->    Ba²⁺(aq)    +    2F^-(aq)

At 25 ^oC, K_sp = 1.0 x 10^-6 for the above equilibrium reaction.

a) 7.1 x 10-4

b) 1.0 x 10-3

c) 6.3 x 10-3

d) 7.9 x 10-3

e) 1.0 x 10-2

2. In the previous problem, you calculated the molar solubility of BaF₂ in distilled water, that is, solubility in units of moles of BaF₂ per liter of solution. What would this solubility be in units of grams per liter?

a) 0.12 g / L

b) 0.18 g / L

c) 1.10 g / L

d) 1.39 g / L

e) 1.75 g / L

3. In the previous two problems, you looked at the solubility of BaF₂ in distilled water. Now consider its solubility in a 0.20 M sodium fluoride (NaF) solution (still at 25 ^oC). Note that NaF is a freely soluble salt. It will not establish an equilibrium, but instead, will completely dissociate when put in water:

NaF(s)    ==========>    Na⁺(aq)    +    F^-(aq)

When the BaF₂ is dissolved in the NaF solution, it establishes its usual equilibrium, that is,

BaF₂(s)    <---------->    Ba²⁺(aq)    +    2F^-(aq)

but there is already some F^- present even before the BaF₂ dissolves, due to the presence of the NaF in the solution. What is the molar solubility of BaF₂ in the 0.20 M NaF solution?

a) 2.5 x 10-6

b) 5.0 x 10-6

c) 2.5 x 10-5

d) 5.0 x 10-3

e) 6.3 x 10-3

4. In the previous problem, you calculated the molar solubility of BaF₂ in a 0.20 M NaF solution, that is, solubility in units of moles of BaF₂ per liter of solution. What would this solubility be in units of grams per liter?

a) 4.4 x 10-4

b) 8.8 x 10-4

c) 4.4 x 10-3

d) 8.8 x 10-1

e) 1.1 x 100

5. What is the molar solubility of cadmium sulfide (CdS) in distilled water at 25 ^oC?. Cadmium sulfide is a salt of VERY low solubility. When CdS is put in water, it establishes the following equilibrium:

CdS(s)    <---------->    Cd²⁺(aq)    +    S^2-(aq)

At 25 ^oC, K_sp = 8.0 x 10^-27 for the above equilibrium reaction.

a) 6.4 x 10-53

b) 4.0 x 10-27

c) 8.0 x 10-27

d) 8.9 x 10-14

e) 4.4 x 10-14

6. What is the molar solubility of CdS in a 1.00 M NH₃ solution? As you have seen in the previous problem, the equilibrium of CdS in water is described by the equation

CdS(s)    <---------->    Cd²⁺(aq)    +    S^2-(aq)       K_sp = 8.0 x 10^-27

However, if NH₃ is present, there is an additional equilibrium reaction taking place:

Cd²⁺(aq)    +    4NH₃    <---------->    Cd(NH₃)₄²⁺(aq)       K_sp = 1.0 x 10⁷

The overall equilibrium reaction is the sum of the above two reactions. After finding this overall reaction, and the numerical value of its associated equilibrium constant, you can do an ICE table calculation to find the molar solubility. It will be necessary to make an approximation to simplify the equation, just as I have done in class. Without this approximation, the mathematics will be horrible!

a) 8.0 x 10-20

b) 4.0 x 10-20

c) 2.0 x 10-10

d) 2.8 x 10-10

e) 3.2 x 103

7. Suppose you mix 22.0 mL of 0.100 M AgNO₃ with 45.0 mL of 0.0260 M NaC₂H₃O₂. The two possible products of the metathesis reaction are NaNO₃ and AgC₂H₃O₂. NaNO₃ is a freely soluble salt, but AgC₂H₃O₂ has limited solubility:

AgC₂H₃O₂(s)    <---------->    Ag⁺(aq)    +    C₂H₃O₂^-(aq)       K_sp = 2.3 x 10^-3

Would a precipitate of AgC₂H₃O₂ form when these solutions are mixed?

a) Yes, it would

b) No, it would not

8. Lead(II) chloride, PbCl₂ and lead(II) iodide, PbI₂ are both salts with limited solubility in water. The chemical equations for their equilibria in distilled water are as follows:

PbCl₂(s)    <---------->    Pb²⁺(aq)    +    2Cl^-(aq)       K_sp = 1.6 x 10^-5

PbI₂(s)    <---------->    Pb²⁺(aq)    +    2I^-(aq)       K_sp = 6.5 x 10^-9

If a solution contained Cl^- and I^- ions at a concentration of 0.100 M each, and you began adding Pb²⁺ ions to the solution, which substance would be the first to precipitate out of the solution?

a) PbCl2

b) PbI2

9. For the anion that precipitated first, what would be its remaining concentration in solution (in units of moles per liter) at the moment the second ion began precipitating? (Assume no change in solution volume as the Pb²⁺ ions are added.)

a) 3.2 X 10-7

b) 1.0 x 10-6

c) 4.1 x 10-5

d) 2.0 x 10-3

e) 6.4 x 10-3

10. What is the hydroxide concentration ( [OH^-] ) in a 0.010 M HCl solution? Note that HCl is a strong acid. That is, it ionizes completely without setting up an equilibrium:

HCl(aq)    ==========>    H⁺(aq)    +    Cl^-(aq)

a) 1.0 x 10-16

b) 1.0 x 10-14

c) 1.0 x 10-12

d) 1.0 x 10-7

e) 1.0 x 10-2

11. What is the pH of a 1.50 M NH₃ solution. Note that NH₃ is a weak base. The equation for its equilibrium in water is

NH₃(aq)    +    H₂O(l)    <---------->    NH₄⁺(aq)    +    OH^-(aq)       K_b = 1.8 x 10^-5

a) 1.5

b) 2.3

c) 4.8

d) 9.2

e) 11.7

12. What is the pOH of a 1.95 M HC₂H₃O₂ solution? Note that HC₂H₃O₂ is a weak acid. The equation for its equilibrium in water is

HC₂H₃O₂(aq)    <---------->    H⁺(aq)    +    C₂H₃O₂-(aq)       K_a = 1.8 x 10^-5

a) 2.2

b) 5.9

c) 8.1

d) 11.8

e) 13.6

13. What is the hydrogen ion concentration ( [H⁺] ) in a 0.50 M HCN solution? Note that HCN is a weak acid. The equation for its equilibrium in water is

HCN(aq)    <---------->    H⁺(aq)    +    CN^-(aq)       K_a = 6.2 x 10^-10

a) 1.0 x 10-14

b) 5.7 x 10-10

c) 2.3 x 10-7

d) 1.8 x 10-5

e) 6.3 x 10-3

14. What is the pH of a 0.025 M NaOH solution? Note that NaOH is a strong base. It completely dissociates when put in water, without establishing an equilibrium:

NaOH(s)    ==========>    Na⁺(aq)    +    OH^-(aq)

a) 1.6

b) 2.5

c) 11.5

d) 12.4

e) 13.8

15. What is the pH of a 2.50 M potassium acetate (KC₂H₃O₂) solution? Note that potassium acetate is a salt that undergoes hydrolysis after being dissolved in water. The salt is freely soluble, and first dissolves without establishing an equilibrium with the solid:

KC₂H₃O₂(s)    ==========>    K⁺(aq)    +    C₂H₃O₂^-(aq)

The K⁺, being the conjugate acid of a strong base (KOH) is too weak an acid to show any reaction with water. However, the C₂H₃O₂^-, being the conjugate base of a weak acid, has enough base strength to react with water:

C₂H₃O₂^-(aq)    +    H₂O(l)    <---------->    HC₂H₃O₂(aq)    +    OH^-(aq)

The numerical value of the equilibrium constant for the above reaction is normally not published in tables, because it can be obtained from the equilibrium constants of two other reactions that ARE published:

HC₂H₃O₂(aq)    <---------->    H⁺(aq)    +    C₂H₃O₂^-(aq)    K_a = 1.8 x 10^-5

H₂O(l)    <---------->    H⁺(aq)    +    OH^-(aq)       K_w = 1.0 x 10^-14

By combining these two reactions in the appropriate way, you can produce the reaction for which the numerical value of the equilibrium constant is desired. Once you have that, you can do an ICE table calculation in the usual way. It should be possible to make an approximation in your solution to the ICE table calculation to avoid having to solve the quadratic formula.

a) 2.8

b) 4.4

c) 7.5

d) 8.4

e) 9.6

16. What is the hydroxide ion concentration ( [OH^-] ) in a 1.25 M amonium chloride (NH₄) solution? Note that ammonium chloride is a salt that undergoes hydrolysis after being dissolved in water. The salt is freely soluble, and dissolves without establishing an equilibrium:

NH₄Cl(aq)    ==========>    NH₄⁺(aq)    +    Cl^-(aq)

The Cl^-, being the conjugate base of a strong acid (HCl) is too weak a base to react with water. The NH₄⁺, being the conjugate acid of a weak base (NH₃) has enough acid strength to react with water:

NH₄⁺(aq)    <---------->    H⁺(aq)    +    NH₃(aq)

The numerical value of the equilibrium constant for the above reaction is normally not published in tables, because it can be obtained from the equilibrium constants of two other reactions that ARE published:

NH₃(aq)    +    H₂O(l)    <---------->    NH₄⁺(aq)    +    OH^-(aq)       K_b = 1.8 x 10^-5

H₂O(l)    <---------->    H⁺(aq)    +    OH^-(aq)       K_w = 1.0 x 10^-14

By combining these two reactions in the appropriate way, you can produce the reaction for which the numerical value of the equilibrium constant is desired. Once you have that, you can do an ICE table calculation in the usual way. It should be possible to make an approximation in your solution to the ICE table calculation to avoid having to solve the quadratic formula.

a) 4.1 x 10-12

b) 3.8 x 10-10

c) 1.3 x 10-7

d) 2.6 x 10-5

e) 7.4 x 10-3

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ANSWERS:

1 C           2 C           3 C           4 C

5 D           6 D           7 B           8 B

9 D           10 C           11 E           12 D

13 D           14 D           15 E           16 B

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