Is it true that the GREATER the solubility product, Ksp, the more soluble the salt?

1 Answer
Oct 12, 2015

The answer is true.

Explanation:

For the sparingly soluble binary salt, MX, we can represent its dissolution in water as:

MX(s) M+(aq) + X(aq).

This is an equilibrium reaction. The (aq) designates the aquated ion, i.e. a metal ion or a negative ion that is aquated or solvated by several water molecules. As for any equilibrium, we can write the equilibrium reaction:

Ksp = [M+(aq)][X(aq)][MX(s)]

Now both [M+] and [X] can be measured in that there are measurable concentrations in gL1 or molL1, but we cannot speak of the concentration of a solid; so [MX(s)] is meaningless.

So now (finally!), we have, Ksp = [M+][X].

This solubility expression (the solubility product!) is dependent solely on temperature (a hot solution can normally hold more solute than a cold one). Ksp constants have been measured for a great number of sparingly soluble salts, and assume standard laboratory conditions.

Because it is a constant, the greater the Ksp, the more soluble the solute. Note that Ksp expressions do not differentiate as to the source of the M+ and X ions. The salt should be less soluble in a solution where X ions were already present.