Question

The $K_{sp}$ of $PbCrO_4$ is $1.0 \times 10^{-16}$. Then the molar solubility of $PbCrO_4$ is

• A. $1.0 \times 10^{-6}$
• B. $1.0 \times 10^ {-4} $
• C. $1.0 \times 10^{-16}$
• D. $1.0 \times 10^8$
• E. $1.0 \times 10^{-12}$

Answer 1

Answer: (D)

$PbCrO _{4} \leftrightharpoons Pb ^{2}+ CrO _{4}^{2-}$
$K _{ sp }=( S )( S )$
or $S ^{2}= K _{ sp }$
or $S =\sqrt{ K _{ sp }}$
$=\sqrt{1.0 \times 10^{-16}}$
or $S =1.0 \times 10^{-8} M$