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)⇌Mnn+(aq) + nOH−(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 =[M(OH)n][Mn+][OH−]nwhere [ ] represents the concentration in 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, KspKsp= [Mn+][OH−]n
Since the substance is barely soluble. Ksp 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 of cadmium hydroxide = 2.5×10-14mol3dm−9 at 298K
Calculate the solubility of Cd(OH)2 in g dm-3 in pure water
Let the solubility Cd(OH)2= s g dm-3
this means that in a saturated solution [Cd2+] = s and [OH-] = 2sKsp=[Cd2+] [OH-] 2 = s ×(2s)2 = 4s3 sos3 = 41(2.5×10-14), s = 2.92×10−5
So the solubility of Cd(OH) Cd(OH)2 = 2.92×10-5mol dm−3 and in g dm-3= 2.92×10-6×molar mass = 2.92×10-6×[112.4 +2(16.00 + 1.01)]= 2.92×10−6×146.42 = 4.27×10−4g dm−3
Calculate the solubility of Cd(OH)2 in g dm-3 in a 0.10 mol dm-3 solution of NaOHThis time the concentration of hydroxide ions from the NaOH will be affected the way the Cd(OH)2dissolves
As above:
Let the solubility of Cd(OH)2= s g dm-3
this means that in a saturated solution [Cd2+] = sbut[OH-] = 0.10Ksp=[Cd2+ ][OH−]2= s×(0.10)2 = 0.01s so 2.5×10-14 = 0.01s
and in g dm-3= 2.6×10−12×146.42 = 3.80×10−10 g dm-3From 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 Ksp=BaSO4 at 298K is 1.1×10-10