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

1 Answer
Oct 12, 2015

Answer:

The answer is true.

Explanation:

For the sparingly soluble binary salt, #MX#, we can represent its dissolution in water as:

#MX(s)# #rarr# #M^+(aq)# #+# #X^(-)(aq)#.

This is an equilibrium reaction. The #(aq)# 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_(sp)# #=# #([M^+(aq)][X^(-)(aq)])/[[MX(s)]]#

Now both #[M^+]# and #[X^-]# can be measured in that there are measurable concentrations in #g*L^-1# or #mol*L^-1#, but we cannot speak of the concentration of a solid; so #[MX(s)]# is meaningless.

So now (finally!), we have, #K_(sp)# #=# #[M^+][X^-]#.

This solubility expression (the solubility product!) is dependent solely on temperature (a hot solution can normally hold more solute than a cold one). #K_(sp)# 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_(sp)#, the more soluble the solute. Note that #K_(sp)# 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.