# Solubility Equilibria

### In-class Problem Set #5

Calculate the solubility of lead(II)phosphate under the following constraints.

SOLUBILITY:  For our purposes, the solubility of a substance is defined as the moles of the solid that will dissolve in one liter of solution.

a)  No other simultaneous equilibria occur.

The first step in a problem like this is to write the relevant reaction that describes the process.  This involves the solubility of a sparingly soluble substance.  Reactions of sparingly soluble substances are always written with the solid on the reactant side and the dissolved ions on the product side.

$\ce{Pb3(PO4)2(s) \leftrightarrow 3Pb^2+(aq) + 2PO4^3- (aq)} \hspace{60px} \mathrm{K_{sp} = 8.1\times 10^{-47}}$

The equilibrium expression for this reaction is written as follows:

$\mathrm{K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2}$

and is known as the solubility product.  Note that the solid does not appear in the equilibrium constant expression.

The way to solve this problem is to write two expressions for the solubility (S), one in terms of lead ion, the other in terms of phosphate ion.  What we need to consider is that the only way we get lead or phosphate ions in solution is to have some of the lead phosphate dissolve.  Remember, solubility refers to the moles of solid that dissolve in a liter of solution.

If we consider the equation, one thing we would see is that for every one molecule of solid lead phosphate that dissolves, we get three lead ions.  This leads to the following expression for solubility:

$\mathrm{S =\dfrac{[Pb^{2+}]}{3} \hspace{30px} or \hspace{30px} [Pb^{2+}] = 3S}$

Before we continue, we need to make sure that this makes sense.  Remember, S is a measure of the number of lead phosphate molecules that dissolve, and if we have three lead ions, only one lead phosphate has dissolved.  If [Pb2+] = 3, S = 1 in the above equation.

We can write a similar equation for phosphate ion, keeping in mind that for every one molecule of solid lead phosphate that dissolves, we get two phosphate ions.

$\mathrm{S =\dfrac{[PO_4^{3-}]}{2} \hspace{30px} or \hspace{30px} [PO_4^{3-}] = 2S}$

We can now substitute these two solubility expressions into the Ksp expression:

$\mathrm{K_{sp} = [Pb^{2+}]^3[PO_4^{3-}]^2 = (3S)^3(2S)^2 = 108S^5}$

$\mathrm{S = 2.37\times 10^{-10}}$

So this is a sparingly soluble material and we have an exceptionally low solubility.