Question

Find the minimum concentration of $P{b}^{2+}$ ions that will initiate the precipitation of $PbS{O}_4$ from a solution containing $0.005mol{L}^{-1}$ of $S{O}_4^{2-}$ ions?
(Given $K_{sp}$ for $BaS{O}_4=1.8\times {10}^{-8}$ )

• A. $3.6\times {10}^{-6}mol{L}^{-1}$
• B. $4.5\times {10}^{-5}mol{L}^{-1}$
• C. $0.9\times {10}^{-4}mol{L}^{-1}$
• D. $2.7\times {10}^{-9}mol{L}^{-1}$

Answer 1

The correct option is A $3.6\times {10}^{-6}mol{L}^{-1}$

If ionic product $\left[P{b}^{2+}\right]\left[S{O}_4^{2-}\right]$ is greater than $K_{sp}$ for $PbS{O}_4,$ then the precipitation of $PbS{O}_4$ will start.
So,
$Q_{sp}>K_{sp}$

$\left[P{b}^{2+}\right]\left[S{O}_4^{2-}\right]>1.8\times {10}^{-8}$
For the limiting case, i.e. when precipitation just starts
$\left[P{b}^{2+}\right]\left[S{O}_4^{2-}\right]=1.8\times {10}^{-8}$
$\left[P{b}^{2+}\right]\times \left(0.005\right)=1.8\times {10}^{-8}$
$\left[P{b}^{2+}\right]=3.6\times {10}^{-6}mol{L}^{-1}$
Thus, minimum concentration of $P{b}^{2+}$ ions required for the precipitation of $PbS{O}_4$ is $3.6\times {10}^{-6}M$.