Question

If the solubility product of lead iodide $(PbI_2)$ is $3.2 \times 10^{-8}$, its solubility will be

• A. $2 \times 10^{-3} \,M$
• B. $4 \times 10^{-4} \,M$
• C. $1.6 \times 10^{-5} \,M$
• D. $1.8 \times 10^{-5} \,M$

Answer 1

Answer: (A)

$PbI_{2} {\rightleftharpoons} \underset{\text{S}}{{Pb^{2+}}}+\underset{\text{ 2S}}{{2I^{-}}} ↑ ; K_{sp}=4S^{3}$
$\therefore \, 4S^{3}=3.2\times 10^{-8}$
$\Rightarrow S=2\times10^{-3}\,M$