Is it true that the GREATER the "solubility product," K_"sp", 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) rarr 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:

K_(sp) = ([M^+(aq)][X^(-)(aq)])/[[MX(s)]]

Now both [M^+] and [X^-] can be measured in that there are measurable concentrations in g*L^-1 or mol*L^-1, but we cannot speak of the concentration of a solid; so [MX(s)] is meaningless.

So now (finally!), we have, K_(sp) = [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). K_(sp) 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 K_(sp), the more soluble the solute. Note that K_(sp) 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.