Question

If the solubility product of lead iodide $(Pb{I}_2)$ 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)

$Pb{I}_2\rightleftharpoons \underset{\text{S}}{P{b}^{2+}}+\underset{\text{ 2S}}{2I^{-}}\uparrow ;K_{sp}=4S^3$
$\therefore 4S^3=3.2\times 10^{-8}$
$\Rightarrow S=2\times 10^{-3}M$