Question

The molar solubility of $PbI_{2}$ in $0.2MPb\left(\left(NO\right)_{3}\right)_{2}$ solution in terms of solubility product $\left(K_{sp}\right)$ of $PbI_{2}$ is :-

• A. $\sqrt{\left(K_{SP} / 0 . 2\right)}$
• B. $\sqrt{K_{SP} / 0 . 4}$
• C. $\sqrt[3]{K_{SP} / 0 . 8}$
• D. $\sqrt{K_{SP} / 0 . 8}$

Answer 1

Answer: (D)

$Pb\left(\left(NO\right)_{3}\right)_{2} \rightarrow \left(Pb\right)^{2 +}+2NO_{3}^{-}0.2M0.4M$
$\left(PbI\right)_{2}\left(\right.s\left.\right)\rightleftharpoons\left(Pb\right)^{+ 2}\left(\right.aq\left.\right)+2I^{-}\left(\right.aq\left.\right)S+0.2M2S$
$K_{sp}=\left[Pb^{2 +}\right]\left[I^{-}\right]^{2}$
$K_{sp}=\left[\right.S+0.2\left]\right.\left[\right.2s\left]\right.^{2}$
$\left(\right.2s\left(\left.\right)^{2}=\frac{K_{sp}}{0 . 2}\Rightarrow S=\sqrt{\frac{K_{sp}}{0 . 8}}$