# If a crystal added to an aqueous solution causes many particles to come out of solution, what can be said about the original solution?

Apr 12, 2017

That the original solution was $\text{supersaturated}$ with respect to the solute.

#### Explanation:

The definition of $\text{saturation}$ and $\text{supersaturation}$ is recognized as a problem area of A-level, and even of undergraduate chemistry. Many ionic salts and solutes have some solubility in water, and this is commonly used to demonstrate $\text{saturation}$:

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

As we try to show in the equation, saturation defines $\text{AN EQUILIBRIUM CONDITION}$, i.e. the $\text{rate of dissolution}$ is equal to the $\text{rate of precipitation}$, (i.e. solute coming out of solution).

With certain salts, however, sometimes we can drive the equilibrium to the right; i.e. we can take a saturated solution, $\text{WITH UNDISSOLVED SOLUTE}$, heat it up, and bring all the solute into solution. If we slowly cool this solution, so that precipitation does not occur, we might get a $\text{SUPERSATURATED}$ solution, where the solution contains an amount of solute $\text{GREATER}$ than that which would be in equilibrium with undissolved solute. Such solutions are metastable, and the solution might be brought back to saturation, to equilibrium, by scratching the sides of the flask, or by introducing a seed crystal.

I am sorry to labour the point so much, but if you can follow what I say (and certainly, we can elaborate on this answer), you will understand the concepts of $\text{saturation}$, $\text{solubility}$, and $\text{equilibrium}$.

Also see here.