Question

The $K_{sp}$ of $PbCr{O}_4$ is $1.0\times {10}^{-16}.$ The the molar solubility of $PbCr{O}_4$ is :

• A. $1.0\times {10}^{-6}$
• B. $1.0\times {10}^{-4}$
• C. $1.0\times {10}^{16}$
• D. $1.0\times {10}^{-8}$

Answer 1

The correct option is D $1.0\times {10}^{-8}$

$PbCr{O}_4\leftrightharpoons P{b}^2+Cr{O}_4^{2-}$

$K_{sp}=\left(S\right)\left(S\right)$

or $S^2=K_{sp}$

or $S=\sqrt{K_{sp}}$

$=\sqrt{1.0\times {10}^{-16}}$

or $S=1.0\times {10}^{-8}M$

Hence, option D is correct.