5 Steps to a 5 AP Chemistry

solids, liquids, and solvents are included in the equilibrium constant.) The Kspexpression
for the PbSO 4 system would be:

For this particular salt the numerical value of Kspis 1.6 × 10 −^8 at 25°C. Note that the
Pb^2 +and SO 42 −ions are formed in equal amounts, so the right-hand side of the equation
could be represented as [x]^2. If the numerical value of the solubility product constant is
known, then the concentration of the ions can be determined. And if one of the ion con-
centrations can be determined, then Kspcan be calculated.
For example, the Kspof magnesium fluoride in water is 8 × 10 −^8. How many grams of
magnesium fluoride will dissolve in 0.250 L of water?

If a slightly soluble salt solution is at equilibrium and a solution containing one of the
ions involved in the equilibrium is added, the solubility of the slightly soluble salt is
decreased. For example, let’s again consider the PbSO 4 equilibrium:

Suppose a solution of Na 2 SO 4 is added to this equilibrium system. The additional sul-
fate ion will disrupt the equilibrium, by Le Cha^telier’s principle, and shift it to the left,
decreasing the solubility. The same would be true if you tried to dissolve PbSO 4 in a solu-
tion of Na 2 SO 4 instead of pure water—the solubility would be lower. This application of
Le Cha^telier’s principle to equilibrium systems of slightly soluble salts is called the
common-ion effect. Calculations like the ones above involving finding concentrations and
Kspcan still be done, but the concentration of the additional common ion will have to be
inserted into the solubility product constant expression. Sometimes, if Kspis very small and
the common ion concentration is large, the concentration of the common ion can simply
be approximated by the concentration of the ion added.
For example, calculate the silver ion concentration in each of the following solutions:

a.Ag 2 CrO 4 (s) +water
b.Ag 2 CrO 4 (s) +1.00 MNa 2 CrO 4

Ksp=1.9 × 10 −^12

a.

2 xx

Ksp=(2x)^2 (x) =1.9 × 10 −^12 = 4 x^3

x=7.8 × 10 −^5

[Ag+] = 2 x=1.6 × 10 −^4 M

Ag CrO s 244 2 Ag ag CrO ag

2

() +()+ ()

−

PbSO s 4 ()Pb2+(aq) + SO 42 −(aq)

MgF s) Mg F

Mg F

)

2

2+

sp

2+

2

(

[][]

()(

 +

==×

=

−

−−

2

810

2

K^28

xx===×

=× =

×

−

−

−

4810

310

8

3

x

x

3

32+

2+

310 Mg

mol M g

L

[]

()

()

(00 250

62 3

11

.

)(. )

()(

L)

(1 mol MgF g MgF

mol M g

22

2+ mmol M gF )

g

2

=005.

Ksp=[][ ]Pb2+ SO 4 2–

Equilibrium  229