5 Steps to a 5 AP Chemistry

29.B—The low value of Kmeans that the equilib-
rium lies to the left. The equilibrium always lies
away from the stronger acid and the stronger
base.

30. B—Nitric acid, being an acid, will react with a
    base. In addition to obvious bases containing
    OH−, the salts of weak acids are also bases. All of
    the anions, except CN−, are from strong acids.
    31. A—The equilibrium constant expression for the
    dissolving of manganese(II) hydroxide is:

Ksp=[Mn^2 +][OH−]^2 =1.6 × 10 −^13

If sis used to indicate the molar solubility, the

equilibrium expression becomes:

Ksp=(s)(2s)^2 = 4 s^3 =1.6 × 10 −^13

This rearranges to:

s=^3 K/ 4

Equilibrium  237

 Free-Response Questions

You have 15 minutes. You may use a calculator and the tables at the back of the book.

An aqueous solution is prepared that is initially 0.100 M in CdI 42 −. After equilibrium is

established, the solution is found to be 0.013 M in Cd^2 +.

a. Derive the expression for the dissociation constant, Kd, and determine the value of the

constant.

b. What will be the cadmium ion concentration arising when 0.400 mol of KI is added

to 1.00 L of the solution in part a?

c. A solution is prepared by mixing 0.500 L of the solution from part band 0.500 L of

2.0 × 10 −^5 M NaOH. Will cadmium hydroxide, Cd(OH) 2 , precipitate? The Kspfor

cadmium hydroxide is 2.2 × 10 −^14.

d. When the initial solution is heated, the cadmium ion concentration increases. Is the

equilibrium an exothermic or an endothermic process? Explain how you arrived at your

conclusion.

 Answers and Explanations

a. Kd+[Cd^2 +][I−]^4 /[CdI 42 −] Give yourself 1 point for this expression. Using the following table:

CdI 42 −(aq) Cd^2 +(aq) I−(aq)

Initial 0.100 M 0 0

Change −x +x + 4 x

Equilibrium 0.100 − xx 4 x

The value of [Cd^2 +] is given (=0.013), and this is x. This changes the last line of the table to:

CdI 42 −(aq) Cd^2 +(aq) I−(aq)

Equilibrium 0.100 −x=0.087 x= 0.013 4 x=0.052

CdI aq^24 −()Cd^2 +−()aq + 4 I aq()