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, Ksp = 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)    +    S2-(aq)

At 25 oC, Ksp = 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)    +    S2-(aq)       Ksp = 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)       Ksp = 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)       Ksp = 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)       Ksp = 1.6 x 10-5

PbI2(s)    <---------->    Pb2+(aq)    +    2I-(aq)       Ksp = 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)       Kb = 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)       Ka = 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)       Ka = 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)    Ka = 1.8 x 10-5

H2O(l)    <---------->    H+(aq)    +    OH-(aq)       Kw = 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)       Kb = 1.8 x 10-5

H2O(l)    <---------->    H+(aq)    +    OH-(aq)       Kw = 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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