# What are some common mistakes students make with solubility equilibria?

Nov 27, 2015

Many students fail to realize that the precipitate is irrelevant.

#### Explanation:

For an insoluble salt, $M X$, usually a solubility product, ${K}_{s p}$, at some particular temperature, we can write the normal equilibrium expression:

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

As for any equilibrium, we can write the equilibrium expression,

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

Now normally, we have some handle on $\left[{X}^{-}\right]$ or $\left[{M}^{+}\right]$, but the concentration of the solid material $\left[M X \left(s\right)\right]$ is meaningless and irrelevant; it is arbitrarily treated as $1$. So,

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

There may often be a precipitate of $M X \left(s\right)$ in the bottom of the flask, however, this is completely irrelevant to the solubility product, and to the equilibrium. It is out of the game as a precipitate. $\left[{X}^{-}\right]$, may be artificially raised to some extent as well (i.e. by introducing beforehand a soluble salt of ${X}^{-}$; such a procedure is called "salting out"). If $M$ was a precious metal (say gold or rhodium or iridium), you would want to precipitate all this is out as an insoluble salt, as opposed to washing it down the sink.