South Pasadena • AP Chemistry
Name
3 ▪ Chemical Equilibrium
Period
3.3 LESSON
–
Date
COMMON
IONS
Le Châtelier’s Principle and Ksp
Example 1: Consider the dissolution of Ca(OH)2:
Ca(OH)2 (s)  Ca2+ (aq) + 2 OH− (aq)
Ksp = 1.3 × 10−6
 Calculate the molar solubility of Ca(OH)2 when the salt is placed in pure water. Is this solution acidic or
basic? Explain.
Ca(OH)2 (s)  Ca2+ (aq) + 2 OH− (aq)
Initial
0
0
Change
−s
+s
+s
Equil.
s
s
2+
− 2
Ksp = [Ca ][OH ]
1.3 × 10−6 = (s)(2s)2
s = 6.9 × 10−3 M
Solution is basic because Ca(OH)2 releases OH−.


Suppose some acid were added to the saturated solution of Ca(OH)2.
o Write the equation of the second reaction that would take place.
H+ (aq) + OH− (aq)  H2O (ℓ)
o
Would the solubility of Ca(OH)2 increase or decrease? Explain the change in solubility using Le
Châtelier’s Principle.
Solubility increases. Adding acid will remove OH− from solution, so reaction will shift to the right to
restore equilibrium. The solubility increases.
o
Notice that the dissolved ion OH− reacts with added H+ in a side reaction. In general, when there is a
substance that reacts with a dissolved ion, the solubility of the salt [ increases | decreases ]
Suppose Ca(OH)2 were placed in a basic solution.
o Would the solubility of Ca(OH)2 increase or decrease? Explain the change in solubility using Le
Châtelier’s Principle.
Solubility decreases. Adding base will increase OH− in the solution, so reaction will shift to the left,
forming more precipitate. The solubility decreases.
o
Notice that when an ion of the salt is already present in the solution, the solubility of the salt
[ increases | decreases ].
In a solubility equilibrium problem, the equilibrium can be affected by adding or removing the ions in solution.
This will affect the solubility of the compound. If a solution has the same ion as one of the ions in the dissolving
compound, we call that a common ion and the effect it has on the solubility of the compound is the common ion
effect.
Common Ion Effect
Example 2: Ca(OH)2 is placed in a solution that contains 0.15 M NaOH.
 Write the equation for the dissolution of Ca(OH)2.
Ca(OH)2 (s)  Ca2+ (aq) + 2 OH− (aq)

Calculate the molar solubility of Ca(OH)2 in the NaOH solution
Ca(OH)2 (s)  Ca2+ (aq) + 2 OH− (aq)
Initial
0
0.15
Change
−s
+s
+2s
Equil.
s
0.15 + 2s
2+
− 2
Ksp = [Ca ][OH ]
1.3 × 10−6 = (s)(0.15 + 2s)2
Assume 2s << 0.15
−5
s = 5.8 × 10 M

Compare the solubility of Ca(OH)2 in pure water with that in the NaOH solution.
Compare with 6.9 × 10−3 M
Example 3: What is the solubility of PbCl2 in 0.10 M NaCl? The Ksp of PbCl2 is 1.6 × 10−5.
Ksp = [Pb2+][Cl−]2
PbCl2 (s)  Pb2+ (aq) + 2 Cl− (aq)
1.6 × 10−5 = (s)(0.10 + 2s)2 Assume 2s << 0.10
Initial
0
0.10
s = 1.6 × 10−3 M
Change
−s
+s
+2s
Equil.
s
0.10 + 2s

Calculate the solubility of PbCl2 in pure water. Compare the solubilities.
(s)(2s)2 = 1.6 × 10−5
s = 0.016 M
Example 4: Which of the following salts would be more soluble in an acidic solution?
ZnCO3
BaF2
AgCl
Ca3(PO4)2
PbI2
Come up with a “rule” to quickly determine how you can tell.
Salts with anions of weak bases are more soluble in acidic solution.
Mg(OH)2
Selective Precipitation
Example 5: Consider a solution that contains Ca2+ and Pb2+ ions, both at 0.010 M. If solid Na2SO4 is slowly
added to the solution, which precipitates first? What is the concentration of the first ion that precipitates when the
second salt begins to precipitate? The Ksp for CaSO4 and PbSO4 are 6.1 × 10−5 and 1.3 × 10−8. (Ch. 19, #57)
(a) PbSO4 has a lower Ksp (and smaller solubility since they both dissociate in the same ratio), so it will
precipitate before CaSO4.
(b) When precipitate forms, Q = Ksp.
Ksp = [Pb2+][SO42−] = (0.010)[SO42−] = 1.3 × 10−8
[SO42−] = 1.3 × 10−6 M when PbSO4 precipitates
(c) When CaSO4 precipitates:
Ksp = [Ca2+][SO42−] = (0.010)[SO42−] = 6.1 × 10−5
[SO42−] = 6.1 × 10−3 M when CaSO4 precipitates
For PbSO4:
Ksp = [Pb2+][SO42−] = [Pb2+](6.1 ×10−3) = 1.3 × 10−8
[Pb2+] = 2.1 × 10−4