SOLUBILITY EQUILIBRIUM: pp. 743-759
For slightly soluble substances (less than 0.10 M at saturation), we recognize
equilibrium situations. Consider the dissociation of barium carbonate
BaCO3 (s)  Ba2+ (aq) + CO32- (aq)
The equilibrium expression is
Kc = Ksp = [ Ba2+ ] [ CO32- ]
Equation 1
(Notice that the solid barium carbonate is omitted.)
For magnesium phoshate, the dissociation is
Mg3(PO4)2 (s)  3Mg2+ (aq) + 2PO43- (aq)
Notice the rules for equilibrium constant writing are followed; coefficients are
used to define powers.
Equation 2
Ksp = [ Mg2+ ]3 [ PO43- ]2
For barium carbonate (MW = 197.34) only 0.0014 g dissolve in 100 mL of
water. For every one barium carbonate that dissociates, one barium and one
carbonate ion are produced. Convert the mass and volume data to molarity
for the equilibrium expression. The molarity of both ions is 7.09 x 10-5. Using
equation 1 above, Ksp = 5.0 x 10-9.
We can reverse the question and apply it to the magnesium phosphate. The
Ksp is 1.0 x 10-25. What is the solubility of Mg3(PO4)2?
Let "s" be the amount of Mg3(PO4)2 that dissociates. It will produce 3 Mg2+
ions and 2 PO43- ions. Therefore it will produce 3s magnesium and 2s
phosphate. Since there were no ions present initially, 3s and 2s are the
equilibrium concentrations.
Mg3(PO4)2 (s)
Change
-s
 3Mg2+ (aq)
+3s
+
2PO43- (aq)
+2s
Substituting these values into the equilibrium expression (equation 2)
1.0 x 10-25 = [3s]3 [2s]2
This leaves
1.0 x 10-25 = 27s3 • 4s2 = 108s5
Solving for s gives 3.92 x 10-6 M. Multiply by 0.100 L (the volume of water in
solubility problems) and multiply by molecular weight (265.0) gives the
solubility of magnesium phosphate at 0.00010 grams.
Up to this point, a direct comparison of the magnitude of K values gave a
good indication of the extent of reaction; an acid with a Ka of 5.4 x 10-5 is
clearly stronger than an acid with a Ka of 3.2 x 10-8. But here, the issue is
complicated by the fact that one salt dissociating into two particles can have a
Ksp value substantially different than a salt dissociating into 5 particles – even
though, their solubilities may be quite close to each other. Examine the Ksp
and solubility values for the two salts above. The moral of the story is that
you may compare Ksp values of salts that dissociate into the same number of
particles.
COMMON ION EFFECT
What is the solubility of PbI2 in a solution already 0.10 M Pb(NO3)2?
PbI2 (s) 
Pb2+ (aq) +
2I- (aq)
I
blob
0.10M
0
C
-s
+s
+2s
E
blob - s
0.10 + s
2s
The Ksp for lead(II) iodide is 7.1 x 10-9. Write the equilibrium expression.
7.1 x 10-9 = [0.10 + s] [ 2s ]2
How big is "s" compared to 0.10M?
Simplify by omitting s for the Pb2+ ion: 7.1 x 10-9 = 0.10 • 4s2
Solve and molar solubility of Pb2+ (or PbI2; 1 to 1 mole ratio) is 1.3 x 10-4M.
CRITERIA FOR PRECIPITATION
Pour two solutions together allowing ions known to be insoluble to intermix.
Will lead(II) chloride precipitate if 155 mL of 0.016M KCl is added to 245 mL
of 0.175M lead(II) nitrate? Ksp for lead(II) chloride is 1.6 x 10-5.
the ion concentrations after mixing are solved using the classic M1V1 = M2V2
(0.016M)(155mL) = (Cl-)(400mL)
and Cl- = 0.00620M
(0.175M)(245mL) = (Pb2+)(400mL)
and Pb2+ = 0.107M
Insert these values into the correct equilibrium expression for PbCl2.
PbCl2 (s)  Pb2+ (aq) + 2Cl- (aq)
Q = [0.107] [0.00620]2 = 4.1 x 10-6
Notice that Q is less than Ksp, therefore no precipitation will exist in the final
mixed solution. The solution is unsaturated.
SOLUBILITY AND pH
Can the acidity or basicity of the solution affect solubility? Yes, if the ions
involved have either acid or base character. For example, many carbonates
are insoluble. But, if you acidify the solution, the carbonate ions react (they
are basic), shifting the equilibrium allowing more carbonate ions to enter the
solution; the salt dissolves. Quantitative problems are more revealing and
interesting.
Will iron(III) hydroxide precipitate from a 0.013M iron(III) solution buffered with
0.150M HC2H3O2 and 0.250M NaC2H3O2? Ksp for Fe(OH)3 is 4.0 x 10-38.
We have a buffer so the Henderson-Hasselbach equation is in play.
pH  4.74  log
0.250
0.150
The pH is 4.96. Since the question is related to hydroxide concentration, we
convert to pOH which is 9.04 and then find [OH-] from this. This value is
9.16 x 10-10M. Substitute this value into the expression for Q as before.
Fe(OH)3 (s)  Fe3+ (aq) + 3OH- (aq)
Q = [0.013] [9.16 x 10-10]3 = 1.0 x 10-30
In this case, Q exceeds Ksp so precipitation will occur.
You should examine all the sample problems found in the pages assigned at
the top of this handout. Concepts are equally important so please read for
comprehension.