Question

In equilibrium analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $A{g}^{+}$ and $P{b}^{2+}$ at a concentration of $0.10$M. Aqueous HCl is added to this solution until the $C{l}^{-}$ concentration is $0.10$M. What will the concentrations of $A{g}^{+}$ and $P{b}^{2+}$ be at equilibrium?
($K_{sp}$ for $AgCl=1.8\times {10}^{-10}$, $K_{sp}$ for $PbC{l}_2=1.7\times {10}^{-5}$).

• A. $\left[A{g}^{+}\right]=1.8\times {10}^{-11}$M; $\left[P{b}^{2+}\right]=1.7\times {10}^{-4}$M
• B. $\left[A{g}^{+}\right]=1.8\times {10}^{-7}$M; $\left[P{b}^{2+}\right]=1.7\times {10}^{-6}$M
• C. $\left[A{g}^{+}\right]=1.8\times {10}^{-11}$M; $\left[P{b}^{2+}\right]=8.5\times {10}^{-5}$M
• D. $\left[A{g}^{+}\right]=1.8\times {10}^{-9}$M; $\left[P{b}^{2+}\right]=1.7\times {10}^{-3}$M

Answer 1

The correct option is D $\left[A{g}^{+}\right]=1.8\times {10}^{-9}$M; $\left[P{b}^{2+}\right]=1.7\times {10}^{-3}$M

According to the problem:

$K_{sp}$ for $AgCl=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

Since, $\left[A{g}^{+}\right]=\frac{1.8\times {10}^{-10}}{{10}^{-1}}$

=$=1.8\times {10}^{-9}M$

$K_{sp}$ for $PbC{l}_2=\left[P{b}^2\right]{\left[C{l}^{-}\right]}^2$

Since, $\left[P{b}^2\right]=\frac{1.7\times {10}^{-5}}{{10}^{-1}\times {10}^{-1}}$

=$1.7\times {10}^{-3}$