Question

For the following equilibrium, if the concentration of phosphate ion is X, what will be the molar solubility of copper phosphate? Cu3(PO4)2(s)↽−−⇀3Cu2+(aq)+2PO3−4(aq)

Answer 1

Answer: The molar solubility of copper phosphate is $\frac{X}{2}$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation is given as:

$Cu_3(PO_4)_2(s)\rightarrow 3Cu^{2+}(aq)+2PO_4^{3-}$

By stoichiometry of the reaction:

1 mole of  $Cu_3(PO_4)_2$ gives 3 moles of $Cu^{2+}$ and 2 moles of $PO_4^{3-}$

When the solubility of  $Cu_3(PO_4)_2$ is S moles/liter, then the solubility of $Cu^{2+}$  will be 3S moles\liter and solubility of $PO_4^{3-}$ will be 2S moles/liter.

Molar concentration of $PO_4^{3-}$ = X

Given : 2S = X

Thus S =$\frac{X}{2}$

Thus the molar solubility of copper phosphate is $\frac{X}{2}$