# What is a saturated solution?

Feb 11, 2016

Neither of the previous answers have defined a saturated solution, and would properly be rejected by an examiner. Saturation defines an EQUILIBRIUM condition.

#### Explanation:

A saturated solution is a solution in which the concentration of the SOLUTE is equal to that concentration that would be in equilibrium with UNDISSOLVED solute. As for any equilibrium we would normally specify a temperature (because at higher temperature, the solvent could probably dissolve more solute).

So for a water soluble salt, ${M}^{+} {X}^{-}$, we could write the following equilibrium reaction:

${M}^{+} {X}^{-} \left(s\right) \stackrel{{H}_{2} O}{r} i g h t \le f t h a r p \infty n s {M}^{+} \left(a q\right) + {X}^{-} \left(a q\right)$.

We would probably also specify a temperature, and we could write the equilibrium condition as : ${K}_{s p} = \left[{M}^{+} \left(a q\right)\right] \left[{X}^{-} \left(a q\right)\right]$ ($s p$ stands for $\text{solubility product}$, and has been measured for many salts). If ${K}_{s p}$ were small, what could you say with respect to the solubility of the salt?

When the ion product (i.e. $\left[{M}^{+} \left(a q\right)\right] \left[{X}^{-} \left(a q\right)\right]$ $=$ ${K}_{s p}$, the solution is said to be $\text{saturated}$ with respect to $M X$; when the ion product $>$ ${K}_{s p}$, the solution is said to be $\text{SUPERSATURATED}$, and when the ion product $<$ ${K}_{s p}$ the solution is $\text{UNSATURATED}$.

These definitions are very poorly understood at undergraduate level, and I urge you to consider them and test your understanding. I reiterate that saturation defines an equilibrium condition, that has been measured for many salts at various temperatures.