What is #K_"sp"#?

1 Answer
Apr 21, 2017

Answer:

Consider the so-called #"solubility product"# of an ionic compound.

Explanation:

Let's suppose we have a sparingly soluble ionic salt, #MX_2#. Its dissolution in water is governed by the following equilibrium expression:

#MX_2(s) rightleftharpoonsM^(2+) + 2X^-#

We write the #K_"solubility product"# expression in this way:

#K_"sp"=[M^(2+)][X^-]^2#.

The larger the #K_"sp"#, clearly the GREATER the solubility of the salt #MX_2#. #K_"sp"# values have been measured for many inorganic salts. Why? Well, if you are in the precious metal caper, and want to isolate gold or platinum or rhodium salts, then you would want to make sure that your experiments precipitate out all of the precious metal that is in solution. If you are working with #Hg^(2+)# or #Cd^(2+)#, you would also want to make sure that you are NOT releasing these toxic ions into water waste.

Given the expression, #K_"sp"=[M^(2+)][X^-]^2#, then if we represent the solubility of #MX_2# as #S#, then, clearly, #K_"sp"=[S][2S]^2=4S^(3)#. Now the equilibrium responds to ALL the #X^-# and #M^(2+)# ions in solution. It does not differentiate the source of the ion. So suppose, you have artificially raised #[X^-]# by adding a soluble source of this ion, which acts as a #"common ion"#. Because now added #X^-# contributes to the ion product, you are at the same time REDUCING the solubility of #MX_2#.

In your text you can read of #"salting out"#, and #"salting in"# which these solubility equilibria are manipulated.

Just to add, it may seem that using #K_"sp"=[M^(2+)][X^-]^2#, you could reduce the solubility of #MX_2# TO ANY LEVEL, simply by loading the solution with a soluble form of #X^-#. At very high anion concentrations, complex ion formation MAY occur:

#M^(2+) + 4X^(-) rightleftharpoons[MX_4]^(2-)#

And this equilibrium would bring the metal ion into solution.