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:
$\textsf{M(OH)}_\textsf{n}\textsf{ (s)}$
$\textsf{Mn}{^\textsf{n+}} \textsf{(aq) + nOH}{^{-}} \textsf{(aq)}$
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 :
$\textsf{Kc =} \frac{\textsf{[M}^\textsf{n+}]\textsf{[OH}^{-}]^\textsf{n}}{\textsf{[M(OH)}_\textsf{n}\textsf{]}}$
where [ ] represents the concentration in
$\textsf {mol dm}^{-3}$
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,
$\textsf K_\textsf{sp}$
$\textsf{K}_\textsf{sp}$
=
$\textsf{[M}^\textsf{n+}\textsf{][OH}^{-}\textsf{]}^\textsf{n}$
Since the substance is barely soluble.
$\textsf K_{\textsf{sp}}$
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
$\textsf{K}_\textsf{sp}$
of cadmium hydroxide =
$\textsf{2.5}\times \textsf{10} ^\textsf{-14} \textsf{mol}^{3} \textsf{dm}^{-9}$
at 298K
1. 1.
Calculate the solubility of
$\textsf{Cd(OH)}_\textsf{2}$
in
$\textsf{g dm}^\textsf{-3}$
in pure water Let the solubility
$\textsf{Cd(OH)}_\textsf{2}$
$\textsf{= s g dm}^\textsf{-3}$
this means that in a saturated solution
$\textsf{[Cd}^\textsf{2+} \textsf{] = s}$
and
$\textsf{[OH} ^\textsf{-}\textsf{] = 2s}$
$\textsf{K}_\textsf{sp} = \textsf{[Cd}^\textsf{2+} \textsf{] }\textsf{[OH} ^\textsf{-}\textsf{] }^\textsf{2}\textsf{ = s }\times\textsf{(2s)}^\textsf{2}\textsf{ = 4s}^{3}$
so
$\textsf{s}^\textsf{3}\textsf{ = }\frac{1}{4} \textsf{(2.5}\times\textsf{10}^\textsf{-14}\textsf{), s = 2.92}\times\textsf{10}^{-5}$
So the solubility of Cd(OH)
$\textsf{Cd(OH)}_\textsf{2}\textsf{ = 2.92}\times\textsf{10}^\textsf{-5}\textsf{mol dm}^{-3}$
and in
$\textsf{g dm}^\textsf{-3}$
$\textsf{= 2.92}\times\textsf{10}^{\textsf{-6}}\times\textsf{molar mass } \\ \textsf{= 2.92}\times \textsf{10}^\textsf{-6}\times\textsf{[112.4 +2(16.00 + 1.01)]} \\ \textsf{= 2.92}\times\textsf{10}^{-6}\times\textsf{146.42 = 4.27}\times\textsf{10}^{-4}\textsf{g dm}^{-3}$
2. 2.
Calculate the solubility of
$\textsf{Cd(OH)}_\textsf{2}$
in
$\textsf{g dm}^\textsf{-3}$
in a
$\textsf{0.10 mol dm}^\textsf{-3}$
solution of
$\textsf{NaOH}$
This time the concentration of hydroxide ions from the NaOH will be affected the way the
$\textsf{Cd(OH)}_\textsf{2}$
dissolves As above: Let the solubility of
$\textsf{Cd(OH)}_\textsf{2}\textsf{= s g dm}^\textsf{-3}$
this means that in a saturated solution
$\textsf{[Cd}^\textsf{2+} \textsf{] = s}$
but
$\textsf{[OH} ^\textsf{-}\textsf{] = 0.10}$
$\textsf{K}_\textsf{sp} = \textsf{[Cd}{^\textsf{2+}}\textsf{ ][OH}{^{-}}\textsf{]}{^\textsf{2}} \textsf{= s} \times \textsf{(0.10)} {^\textsf{2}} \textsf{ = 0.01s}$
so
$\textsf{2.5}\times \textsf{10} {^\textsf{-14}} \textsf{ = 0.01s}$
and in
$\textsf{g dm} {^\textsf{-3}}\textsf{= 2.6}\times\textsf{10}^{-\textsf{12}}\times \textsf{146.42 = 3.80}\times\textsf{10}^{-\textsf{10}}\textsf{ g dm} {^\textsf{-3}}$
From the calculations above, you can see that the solubility of cadmium ions decreases from 2.6 x10
${^{-6}}$
to 2.6 x10
${^{-12}}$
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
$\textsf K_\textsf{sp}$
$=\textsf{BaSO}_\textsf4$
at
$\textsf{298}$
K is
$\textsf{1.1}\times\textsf{10}{^\textsf{-10}}$
$\textsf{[Ba}{^\textsf{2+}}\textsf{]}$
$\textsf{= SO}_\textsf4{^\textsf{2-}}$
$\textsf{= s =}$
$\sqrt{\textsf{(1.1}\times\textsf{10}}{^\textsf{-10}\textsf)}$
and from this s =
$\textsf{1.05}\times\textsf{10}{^\textsf{-5}}\textsf{mol dm}^{-\textsf{3}}$