Question

The concentration of ${NH}_3$ solution whose one litre can dissolve $0.10$ mole $AgCl$ is:
[$K_{sp}$ of $AgCl$ and $K_f$ of $Ag{\left({NH}_3\right)}_2^{+}$ are ${10}^{-10}{M}^2$ and $1.6\times {10}^7{M}^{-2}$ respectively].

• A. $2.5M$
• B. $3.0M$
• C. $2.8M$
• D. none of these

Answer 1

Answer: (A)

$2.5M$

$AgCl+2{NH}_3\rightleftharpoons Ag{\left({NH}_3\right)}_2^{+}+{Cl}^{-}$

$AgCl\left(s\right)\rightleftharpoons {Ag}^{+}+{Cl}^{-}$

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

Also, ${Ag}^{+}+2{NH}_3\rightleftharpoons \left[Ag{\left({NH}_3\right)}_2^{+}\right]$

$K_f=\frac{\left[Ag{\left({NH}_3\right)}_2^{+}\right]}{\left[{Ag}^{+}\right]{\left[{NH}_3\right]}^2}.....\left(ii\right)$

Since $AgCl$ is present with ${NH}_3$ thus, $\left[{Ag}^{+}\right]$ in solution remains some, thus form Eqn, $\left(i\right)$ and $\left(ii\right)$

$\frac{\left[Ag{\left({NH}_3\right)}_2^{+}\right]\left[{Cl}^{-}\right]}{{\left[{NH}_3\right]}^2}=K_{sp}\times K_f$

$\frac{\left[Ag{\left({NH}_3\right)}_2^{+}\right]\left[{Cl}^{-}\right]}{{\left[{NH}_3\right]}^2}=1\times {10}^{-10}\times 1.6\times {10}^7=1.6\times {10}^{-3}$

Given solubility of $AgCl=0.1M$

$\therefore$ $\left[Ag{\left({NH}_3\right)}_2^{+}\right]=\left[{Cl}^{-}\right]=0.1M$

$\therefore$ ${\left[{NH}_3\right]}^2=\frac{0.1\times 0.1}{1.6\times {10}^{-3}}=6.25$

$\therefore$ $\left[{NH}_3\right]=2.5M$