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. 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} sos3 = 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. 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}}

Last updated