# Is it true that the GREATER the "solubility product," K_"sp", the more soluble the salt?

Oct 12, 2015

#### Explanation:

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

$M X \left(s\right)$ $\rightarrow$ ${M}^{+} \left(a q\right)$ $+$ ${X}^{-} \left(a q\right)$.

This is an equilibrium reaction. The $\left(a q\right)$ 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:

${K}_{s p}$ $=$ $\frac{\left[{M}^{+} \left(a q\right)\right] \left[{X}^{-} \left(a q\right)\right]}{\left[M X \left(s\right)\right]}$

Now both $\left[{M}^{+}\right]$ and $\left[{X}^{-}\right]$ can be measured in that there are measurable concentrations in $g \cdot {L}^{-} 1$ or $m o l \cdot {L}^{-} 1$, but we cannot speak of the concentration of a solid; so $\left[M X \left(s\right)\right]$ is meaningless.

So now (finally!), we have, ${K}_{s p}$ $=$ $\left[{M}^{+}\right] \left[{X}^{-}\right]$.

This solubility expression (the solubility product!) is dependent solely on temperature (a hot solution can normally hold more solute than a cold one). ${K}_{s p}$ 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 ${K}_{s p}$, the more soluble the solute. Note that ${K}_{s p}$ 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.