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:
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 :
where [ ] represents the concentration in
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,
Since the substance is barely soluble.
will have a very low value.
Calculations using the of solubility product
(These are most difficult for a 1:2 like cadmium hydroxide)
of cadmium hydroxide =
- 1.Calculate the solubility ofinin pure water Let the solubilitythis means that in a saturated solutionandsoSo the solubility of Cd(OH)and in
- 2.Calculate the solubility ofinin asolution ofThis time the concentration of hydroxide ions from the NaOH will be affected the way thedissolves As above: Let the solubility ofthis means that in a saturated solutionbutsoand inFrom the calculations above, you can see that the solubility of cadmium ions decreases from 2.6 x10to 2.6 x10by 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
and from this s =
Last modified 2yr ago