Question

A saturated solution of pbcl2 is found to contain 2.88 x 10-2 m chloride ions. what is the correct value of the solubility product, ksp

Answer 1

The solubility equilibrium of PbCl$_{2}$:

$PbCl_{2}(aq) Pb^{2+}(aq) + 2 Cl^{-}(aq)$

$K_{sp}=[Pb^{2+}][Cl^{-}]^{2}$

$[Cl^{-}] = 2.88 * 10^{-2} M$

$[Pb^{2+}]=\frac{[Cl^{-}]}{2} = \frac{2.88 * 10^{-2}}{2}=1.44 *10^{-2}$

$K_{sp}=[Pb^{2+}][Cl^{-}]^{2}$

= $(1.44 * 10^{-2})(2.88*10^{-2})^{2}$

= $1.19 * 10^{-5}$

So, the corrected solubility product will be $1.19 * 10^{-5}$