Chapter 10
Solubility (Precipitation Equilibria: The
solubiloity Product- Section 10.5 (p 326 -332)
Precipitation (Insoluble Salts)
• Many metal ions form compounds that are insoluble in water.
• We call them insoluble or slightly soluble salts or precipitates.
• Common precipitates are carbonates, hydroxides, sulfates, and
sulfides.
• An insoluble salt in contact with water maintains an equilibrium
with the ions. In simple cases where there are no common ions
or competing equilibria, the ion concentrations depend only on
the equilibrium constant for the particular precipitate. When we
talk about solubility equilibria we always write the equilibrium
with the solid on the left. For example:
Ba(IO3)2 (s)
Ba2+(aq) + 2 IO3-(aq)
• The equilibrium constant expression for an insoluble salt is
written following the same rules as for any other equilibrium.
• The equilibrium constant is called the solubility product, Ksp.
The Ksp expression for the above equilibrium is:
Ksp = [Ba2+][IO3-]2
Ksp Values for Some Precipitates
Formula
Name
Ksp
AgCl
silver chloride
1.8x10-10
Al(OH)3
aluminum hydroxide
2x10-32
BaCO3
barium carbonate
5x10-9
Ba(IO3)2
barium iodate
1.6x10-9
BaSO4
barium sulfate
1.3x10-10
Fe(OH)2
iron(II) hydroxide
8x10-16
Fe(OH)3
iron(III) hydroxide
4x10-38
FeS
iron sulfide
6x10-18
PbCrO4
lead chromate
1.8x10-14
Pb(OH)2
lead hydroxide
2.5x10-16
PbS
lead sulfide
7x10-28
PbSO4
lead sulfate
1.6x10-8
Solubility (Precipitation Equilibria)
Solubility is defined as moles/L, g/L, or mg/L of the
dissolving species in solution.
Example :
• How is the solubility of lead chloride, PbCl2, related
to the solution concentrations of Pb2+ and Cl-?
• Each formula unit of PbCl2 that dissolves produces
one lead ion, Pb2+, and two chloride ions, Cl-.
• The molar solubility of PbCl2 is equivalent to the
concentration of Pb2+, SPbCl2 = [Pb2+].
• Since two Cl-s are produced per PbCl2 formula unit
that dissolves, the molar solubility of PbCl2 equals
one-half the solution concentration of Cl-,
SPbCl2 = 0.5*[Cl-].
• SPbCl2 = [Pb2+] = 0.5*[Cl-].
Example 2
What is the solubility of barium iodate, Ba(IO3)2, in pure water at 25 oC?
The solubility equilibrium is:
Ba2+(aq) + 2 IO3-(aq)
x
2x
Ksp = [Ba2+][IO3-]2 = 1.5x10-9
Ba(IO3)2 (s)
[Ba2+] = x
[IO3-] = 2x
Ksp = (x) (2x)2 = 1.5x10-9
4*[x]3 = 1.5x10-9
*[Ba2+]3
x3 = [Ba2+]3 = 3.75x10-10 M
X = S (solubility) = [Ba2+] = 7.2x10-4 M
Or in mass terms: S = (7.2x10-4 M)(487 g/mol) = 0.35 g/L
• Why did we choose of Ba(IO3)2 as an example??
• Determining the solubility of Ba(IO3)2 is simple
compared to determining the solubility of many
insoluble salts.
• Ba2+ is a spectator ion so we do not worry about. It
does not react with water.
• What about the iodate anion? It is a base and can
react with water:
IO3-(aq) + H2O
HIO3(aq) + OH-(aq)
• The extent to which this reaction proceeds is given
by the value of Kb. Ka for iodic acid is 0.17
(it is a reasonably strong acid).
Kb = Kw/Ka = 1x10-14/0.17 = 5.9x10-14
• This value of Kb is so small that we can conclude that the
hydrolysis of iodate does not affect the solubility of Ba(IO3)2.
Common –ion effect
Example 3
• What is the solubility of barium iodate,
Ba(IO3)2, at 25 oC in a solution of 0.10 M
barium nitrate, Ba(NO3)2?
• The problem is set up the same as before:
• Ba(IO3)2 (s)
Ba2+(aq) + 2 IO3-(aq)
Initial
After addition
x
0.1 + x
2x
2x
• Ksp = [Ba2+][IO3-]2 = 1.5x10-9
• The difference is that there are two sources
of Ba2+, the 0.10 M Ba(NO3)2 and the
dissolution of Ba(IO3)2:
Common –ion effect
Example 3
Ksp = [Ba2+][IO3-]2 = 1.5x10-9
• Ksp = (0.10 M +x)*[2x-]2 = 1.5x10-9
• We could solve a cubic equation or we could
try neglecting the [Ba2+] that comes from
dissolution of Ba(IO3)2 compared to 0.10 M
barium nitrate.
• Since 0.1>>> x
then 0.1 + x  0.1
• Ksp = (0.1)(2x)2 = 1.5x10-9
Ksp = (0.10)[2x]2 = 1.5x10-9
(2x)2 = 1.5x10-8
(x)2 = 1.5x10-8 / 4 = 3.75x 10-9
X = S = [Ba2+] =
= 6.1x10-5 M
3.75x109
S = 0.5*[IO3-], so
[IO3-] = 2x = 2 x = 6.1x10-5 = 1.2x10-4 M
• Our assumption that S << 0.1 M is justified.
• Compare this result to the solubility in pure water, S
= 7.2x10-4 M. The solubility of barium iodate is much
less when one of the ions involved in the equilibrium
is present in solution.
• This effect is called the common-ion effect, and is
explained qualitatively by LeChatelier's principle.
• Will addition of other ions such as Na+ and Cl- affect
this equilibrium? To a first approximation no.
However, adding spectator ions to a solution
increases the ionic strength and affects the activity
coefficients.
In which of the following examples should we worry about activity?
•solubility of barium iodate in water
•solubility of barium iodate in a solution of 0.10 M barium nitrate
•solubility of barium iodate in a solution of 0.50 M NaCl
Example 4
• Will a precipitate form when 100.0 mL of 0.2 M Fe3+ in
0.5 M H2SO4 is mixed with 100.0 mL of 1.0 M NaOH?
• Before thinking about equilibrium, look at this
problem and decide if anything will happen when the
two solutions are mixed.
• A solution of a strong base is mixed with a solution
containing strong acid.
• The acid and base will react until one of both are
consumed.
• In this case an equal number of moles of H+ and OHare mixed so the resulting solution will be at pH=7.
• The solubility equilibrium is:
Fe(OH)3 (s)
Fe3+(aq) + 3OH-(aq)
Ksp = 2x10-39 = [Fe3+] [OH-]3
• To determine if a precipitate will form:
find Q and compare it to Ksp.
• (this quantity is called the ion product, P, but it is the
same expression as Q.)
• If Q > Ksp the ion concentrations are greater than
their equilibrium concentrations and a precipitate
will form.
• If Q < Ksp the ion concentrations are below their
equilibrium concentrations and no precipitate forms.
• Q = CFe3+COH-3 = (0.100 M)(1x10-7 M)3 = 1.00x10-22
• 1x10-22 > 2x10-39
Q > Ksp
So a precipitate will form.
• What will be the [Fe3+] of the resulting solution?
2x10-39 = [Fe3+](1x10-7 M)3
[Fe3+] = 2x10-18 M
• This is the predicted [Fe3+] in water.
• In real water systems the total amount of iron in
solution can be much higher since the iron can exist
in a variety of forms such as iron complexes and
colloids.