A10.3 Solubility Product
For a metal hydroxide that is only very slightly (sparingly) soluble, when the solid is in a saturated solution, the following equilibrium exists with exchange between ions in the solid and in the solution:
M(OH)n (s)\textsf{M(OH)}_\textsf{n}\textsf{ (s)}
Mnn+(aq) + nOH(aq)\textsf{Mn}{^\textsf{n+}} \textsf{(aq) + nOH}{^{-}} \textsf{(aq)}
At this point, the addition of any more hydroxide ions will cause a little more of the metal hydroxide to precipitate.
Since this is an equilibrium reaction we can write :
Kc =[Mn+][OH]n[M(OH)n]\textsf{Kc =} \frac{\textsf{[M}^\textsf{n+}]\textsf{[OH}^{-}]^\textsf{n}}{\textsf{[M(OH)}_\textsf{n}\textsf{]}}
where [ ] represents the concentration in
mol dm3\textsf {mol dm}^{-3}
Since M(OH)n is a solid, its concentration is a constant. So we can now make use of a new constant, called the Solubility Product,
Ksp\textsf K_\textsf{sp}
Ksp\textsf{K}_\textsf{sp}
=
[Mn+][OH]n\textsf{[M}^\textsf{n+}\textsf{][OH}^{-}\textsf{]}^\textsf{n}
Since the substance is barely soluble.
Ksp\textsf K_{\textsf{sp}}
will have a very low value.

Worked example
🖊

Calculations using the of solubility product
(These are most difficult for a 1:2 like cadmium hydroxide)
The
Ksp\textsf{K}_\textsf{sp}
of cadmium hydroxide =
2.5×10-14mol3dm9 \textsf{2.5}\times \textsf{10} ^\textsf{-14} \textsf{mol}^{3} \textsf{dm}^{-9}
at 298K
  1. 1.
    Calculate the solubility of
    Cd(OH)2\textsf{Cd(OH)}_\textsf{2}
    in
    g dm-3\textsf{g dm}^\textsf{-3}
    in pure water Let the solubility
    Cd(OH)2\textsf{Cd(OH)}_\textsf{2}
    = s g dm-3\textsf{= s g dm}^\textsf{-3}
    this means that in a saturated solution
    [Cd2+] = s\textsf{[Cd}^\textsf{2+} \textsf{] = s}
    and
    [OH-] = 2s\textsf{[OH} ^\textsf{-}\textsf{] = 2s}
    Ksp=[Cd2+] [OH-] 2 = s ×(2s)2 = 4s3\textsf{K}_\textsf{sp} = \textsf{[Cd}^\textsf{2+} \textsf{] }\textsf{[OH} ^\textsf{-}\textsf{] }^\textsf{2}\textsf{ = s }\times\textsf{(2s)}^\textsf{2}\textsf{ = 4s}^{3}
    so
    s3 = 14(2.5×10-14), s = 2.92×105\textsf{s}^\textsf{3}\textsf{ = }\frac{1}{4} \textsf{(2.5}\times\textsf{10}^\textsf{-14}\textsf{), s = 2.92}\times\textsf{10}^{-5}
    So the solubility of Cd(OH)
    Cd(OH)2 = 2.92×10-5mol dm3\textsf{Cd(OH)}_\textsf{2}\textsf{ = 2.92}\times\textsf{10}^\textsf{-5}\textsf{mol dm}^{-3}
    and in
    g dm-3\textsf{g dm}^\textsf{-3}
    = 2.92×10-6×molar mass = 2.92×10-6×[112.4 +2(16.00 + 1.01)]= 2.92×106×146.42 = 4.27×104g dm3\textsf{= 2.92}\times\textsf{10}^{\textsf{-6}}\times\textsf{molar mass } \\ \textsf{= 2.92}\times \textsf{10}^\textsf{-6}\times\textsf{[112.4 +2(16.00 + 1.01)]} \\ \textsf{= 2.92}\times\textsf{10}^{-6}\times\textsf{146.42 = 4.27}\times\textsf{10}^{-4}\textsf{g dm}^{-3}
  2. 2.
    Calculate the solubility of
    Cd(OH)2\textsf{Cd(OH)}_\textsf{2}
    in
    g dm-3\textsf{g dm}^\textsf{-3}
    in a
    0.10 mol dm-3\textsf{0.10 mol dm}^\textsf{-3}
    solution of
    NaOH\textsf{NaOH}
    This time the concentration of hydroxide ions from the NaOH will be affected the way the
    Cd(OH)2\textsf{Cd(OH)}_\textsf{2}
    dissolves As above: Let the solubility of
    Cd(OH)2= s g dm-3\textsf{Cd(OH)}_\textsf{2}\textsf{= s g dm}^\textsf{-3}
    this means that in a saturated solution
    [Cd2+] = s\textsf{[Cd}^\textsf{2+} \textsf{] = s}
    but
    [OH-] = 0.10\textsf{[OH} ^\textsf{-}\textsf{] = 0.10}
    Ksp=[Cd2+ ][OH]2= s×(0.10)2 = 0.01s\textsf{K}_\textsf{sp} = \textsf{[Cd}{^\textsf{2+}}\textsf{ ][OH}{^{-}}\textsf{]}{^\textsf{2}} \textsf{= s} \times \textsf{(0.10)} {^\textsf{2}} \textsf{ = 0.01s}
    so
    2.5×10-14 = 0.01s\textsf{2.5}\times \textsf{10} {^\textsf{-14}} \textsf{ = 0.01s}
    and in
    g dm-3= 2.6×1012×146.42 = 3.80×1010 g dm-3\textsf{g dm} {^\textsf{-3}}\textsf{= 2.6}\times\textsf{10}^{-\textsf{12}}\times \textsf{146.42 = 3.80}\times\textsf{10}^{-\textsf{10}}\textsf{ g dm} {^\textsf{-3}}
    From the calculations above, you can see that the solubility of cadmium ions decreases from 2.6 x10
    6{^{-6}}
    to 2.6 x10
    12{^{-12}}
    by adding the sodium hydroxide. This explains why the metal ions are precipitated by increasing the pH of the solution.
The calculations are much simpler for 1:1 compounds like barium sulphate. You can follow the same steps as in the example above. The
Ksp\textsf K_\textsf{sp}
=BaSO4=\textsf{BaSO}_\textsf4
at
298\textsf{298}
K is
1.1×10-10\textsf{1.1}\times\textsf{10}{^\textsf{-10}}
[Ba2+]\textsf{[Ba}{^\textsf{2+}}\textsf{]}
= SO42-\textsf{= SO}_\textsf4{^\textsf{2-}}
= s =\textsf{= s =}
(1.1×10-10)\sqrt{\textsf{(1.1}\times\textsf{10}}{^\textsf{-10}\textsf)}
and from this s =
1.05×10-5mol dm3\textsf{1.05}\times\textsf{10}{^\textsf{-5}}\textsf{mol dm}^{-\textsf{3}}
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