Question

$K_{sp}$ of $Pb{Cl}_2$ is ${10}^{-13}$. $\left[{Pb}^{2+}\right]$ in solution prepared by mixing $100mL$ of $0.1M$ $Pb{\left({NO}_3\right)}_2$ and $1mL$ of $1M$ $HCl$ will be:

• A. $\left[{Pb}^{2+}\right]=9.4\times {10}^{-2}mol$ ${litre}^{-1}$
• B. $\left[{Pb}^{2+}\right]=8.4\times {10}^{-2}mol$ ${litre}^{-1}$
• C. $\left[{Pb}^{2+}\right]=9.0\times {10}^{-2}mol$ ${litre}^{-1}$
• D. $\left[{Pb}^{2+}\right]=9.2\times {10}^{-2}mol$ ${litre}^{-1}$

Answer 1

Answer: (A)

$\left[{Pb}^{2+}\right]=9.4\times {10}^{-2}mol$ ${litre}^{-1}$

$\underset{Millimoleadded100\times 0.1=10Millimoleleft9.5}{Pb{\left({NO}_3\right)}_2}+\underset{1\times 1=10}{2HCl}⟶\underset{00.5}{Pb{Cl}_2}+\underset{01}{2H{NO}_3}$

$\therefore$ $\left[{Pb}^{2+}\right]=\frac{Millimolesof{Pb}^{2+}insolution}{TotalvolumeofsolutioninmL}$

$\left[{Pb}^{2+}\right]=\frac{9.5+0.5}{101}$

If $Pb{Cl}_2$ gets precipitated then contribution $0.5$ of $\left[{Pb}^{2+}\right]$ from $Pb{Cl}_2$ should be left, i.e,

$\left[{Pb}^{2+}\right]=\frac{9.5}{101}$

To see precipitation $Pb{Cl}_2$

$\left[{Pb}^{2+}\right]{\left[{Cl}^{-}\right]}^2=\left[\frac{10}{101}\right]{\left[\frac{1}{101}\right]}^2=9.70\times {10}^{-6}$

The value is higher than $K_{sp}$ of $Pb{Cl}_2$ and therefore, $Pb{Cl}_2$ gets precipitated.

$\therefore$ $\left[{Pb}^{2+}\right]=\frac{9.5}{101}=9.4\times {10}^{-2}$ $mol$ ${litre}^{-1}$