Equilibrium ❮ 237- C—The solubility-product constant expression
is Ksp = [Zn^2 + ] [IO 3 - ]^2 = 4 × 10 -^6. This may be
rearranged to [IO 3 - ]^2 =
[]
× −
+4 10
Zn6
2. Inserting the
desired zinc ion concentration gives [IO 3 - ]^2 =
×
×−
−4 10
1 10
6
6 = 4. Taking the square of each side
leaves a desired IO 3 - concentration of 2 M. Two
moles of KIO 3 must be added to 1.00 L of solu-
tion to produce this concentration. Since you
can estimate the answer, no actual calculations
are necessary.- A—When dealing with gaseous equilibriums,
volume changes are important when there is a
difference in the total number of moles of gas
on opposite sides of the equilibrium arrow. All
the answers, except A, have differing numbers of
moles of gas on opposite sides of the equilibrium
arrow. - C—The hydrolysis of any ion begins with the
interaction of that ion with water. Thus, both
the ion and water must be on the left side of the
equilibrium arrow and hence in the denomi-
nator of the equilibrium-constant expression
(water, like all solvents, will be left out of the
expression). The oxalate ion is the conjugate
base of a weak acid. As a base, it will produce
OH- in solution along with the conjugate acid,
HC 2 O 4 - , of the base. The equilibrium reaction is
HC 2 O 4 - (aq) + H 2 O(l) OH-(aq) + H 2
HC 2 O 4 (aq). - C—The equilibrium given is actually the sum of
the following three equilibriums:
ZnS(s) ^ Zn^2 +(aq) + S^2 - (aq)
Ksp = 1.6 × 10 -^24
S^2 - (aq) + H+(aq) ^ HS-(aq)
K = 1/ Ka2 = 1/1 × 10 -^19
HS-(aq) + H+(aq) ^ H 2 S(aq)
K’ = 1/ Ka1 = 1/9.5 × 10 -^8
Summing these equations means you need to
multiply the equilibrium constants:
Ksum = KspK K’ =K
KK
sp
aa 21 =
()()
×
××
−
−−1.6 10
1 10 9.5 10
24
19 8- B—The addition or removal of some solid, as
long as some remains present, will not change
the equilibrium. An increase in volume will
cause the equilibrium to shift toward the side
with more moles of gas (right). Raising the tem-
perature of an endothermic process will shift the
equilibrium to the right. Any shift to the right
will increase the amounts of the products. - B—Using the following table:
[CH 4 ] [CO 2 ] [CO] [H 2 ]
Initial 0.30 0.40 0 0
Change –x –x +2x +2x
Equilibrium 0.30 – x 0.40 – x 2x 2x
The presence of 0.20 mole of CO (0.20 M) at
equilibrium means that 2x = 0.20 and that x =
0.10. Using this value for x, the bottom line of
the table becomes[CH 4 ] [CO 2 ] [CO] [H 2 ]
Equilibrium 0.20 0.30 0.20 0.20The equilibrium expression is K =CO H
CH CO
2
2242[]
.
Entering the equilibrium values into the equilib-rium expression gives K =[][]
[][]
0.20 0.20
0.20 0.30
22
.- A—The addition of a product will cause the
equilibrium to shift to the left. The amounts of
all the reactants will increase, and the amounts
of all the products will decrease (the O 2 will
not go below its earlier equilibrium value since
excess was added). The value of K is constant,
unless the temperature is changed. The rates of
the forward and reverse reactions are equal at
equilibrium. - C—The only way to change the value of K is
to change the temperature. For an exothermic
process (ΔH < 0), K increases with a decrease in
temperature. - 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
the anions, except CN-, are from strong acids
and, as such, are not bases.