Question

In qualitative analysis, the metals of Group I can be separated form other ions by precipitating them as chloride salts. A solution initially contains ${Ag}^{+}$ and ${Pb}^{2+}$ at a concentration of $0.10M$. Aqueous $HCl$ is added to this solution until the ${Cl}^{-}$ concentration is $0.10M$. What will the concentrations of ${Ag}^{+}$ and ${Pb}^{2+}$ be at equilibrium ?

$(K_{sp}forAgCl=1.8\times {10}^{-10},K_{sp}forPb{Cl}_2=1.7\times {10}^{-5})$

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

Answer 1

The correct option is C $\left[{Ag}^{+}\right]=1.8\times {10}^{-9}M;\left[{Pb}^{2+}\right]=1.7\times {10}^{-3}M$

The solubility product expression for silver chloride is

$K_{sp}=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]$

Substitute values in the above expression.

$1.8\times {10}^{-10}=\left[{Ag}^{+}\right]\left[0.1\right]$

$\left[{Ag}^{+}\right]=1.8\times {10}^{-9}M$

The solubility product expression for lead (II) chloride is

$K_{sp}=\left[{Pb}^{+2}\right]{\left[{Cl}^{-}\right]}^2$

Substitute values in the above expression.

$1.7\times {10}^{-5}=\left[{Pb}^{+2}\right]{\left[0.1\right]}^2$

$\left[{Pb}^{+2}\right]=1.7\times {10}^{-3}M$

Thus, $\left[{Ag}^{+}\right]=1.8\times {10}^{-9}M;\left[{Pb}^{2+}\right]=1.7\times {10}^{-3}M$.