Question

The solubility of $Pb(OH{)}_2$ in water is $6.7\times {10}^{-6}$M. Calculate the solubility of $Pb(OH{)}_2$ in a buffer solution of $pH=8$.

• A. $1.0\times {10}^{-2}M$
• B. $1.3\times {10}^{-5}M$
• C. $1.2\times {10}^{-3}M$
• D. $2.1\times {10}^{-3}M$

Answer 1

Answer: (C)

$1.2\times {10}^{-3}M$

The solubility of $Pb(OH{)}_2$ in water is $6.7\times {10}^{-6}$ M.

The expression for the solubility product is $K_{sp}=\left[P{b}^{2+}\right][O{H}^{-}{]}^2$.

$K_{sp}=4S^3$

Substitute values in the above expression.
$K_{sp}=(6.7\times {10}^{-6}{)}^3\times 4=1.2\times {10}^{-15}$

In buffer solution of pH 8, $pOH=14-8=6$ or $\left[O{H}^{-}\right]={10}^{-6}$.

Substitute values in the expression for the solubility product.

$1.2\times {10}^{-15}=\left[P{b}^{2+}\right]({10}^{-6}{)}^2$

$\left[P{b}^{2+}\right]=1.2\times {10}^{-3}$ M

So, the solubility of $Pb(OH{)}_2$ in a buffer solution of $pH=8$ is $1.2\times {10}^{-3}$ M.