Question

A saturated solution of $PbC{l}_2$ contains $2\times {10}^{-3}mol$ of $PbC{l}_2$ per litre. What is the $K_{sp}$ of $PbC{l}_2$?

Answer 1

$PbC{l}_2\left(s\right)\to P{b}^{2+}+C{l}^{-}$

Solubility$=2\times {10}^{-2}$ mol/L

So, $\left[P{b}^{2+}\right]=2\times {10}^{-2}$ mol/L

$\left[C{l}^{-}\right]=2\times 2\times {10}^{-2}$ mol/L

$K_{sp}=\left[P{b}^{2+}\right]\left[C{l}^{-}\right]$

$K_{sp}=(2\times {10}^{-2})\times (4\times {10}^{-2})$

$K_{sp}=32\times {10}^{-6}$

$K_{sp}=3.2\times {10}^{-6}$ or $3.2\times {10}^{-5}$.