What are some common mistakes students make with solubility equilibria?

1 Answer
Nov 27, 2015

Answer:

Many students fail to realize that the precipitate is irrelevant.

Explanation:

For an insoluble salt, #MX#, usually a solubility product, #K_(sp)#, at some particular temperature, we can write the normal equilibrium expression:

#MX(s)rightleftharpoons M^+(aq) + X^(-)(aq)#

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

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

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

#[M^(+)(aq)][X^(-)(aq)]# #=# #K_(sp)#.

There may often be a precipitate of #MX(s)# 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. #[X^-]#, 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.