Question

Which is/are the correct statement(s) about the solubility of $AgCl\left(s\right)$ ?
Given : $K_{sp\left(AgCl\right)}={10}^{-10};K_f\left[Ag\right(N{H}_3{)}_2^{+}]={10}^8$

• A. Solubility of $AgCl$ in pure water is ${10}^{-5}{\text{gL}}^{-1}$
• B. Solubility of $AgCl$ in $2\text{M}$ $KN{O}_3$ is ${10}^{-5}{\text{molL}}^{-1}$
• C. Solubility of $AgCl$ in $2\text{M}$ $AgN{O}_3\text{is}5\times {10}^{-11}\text{M}$
• D. Solubility of $AgCl$ in $2\text{M}$ $N{H}_3$ is $0.166\text{M}$

Answer 1

Answer: (B), (C), (D)

Solubility of $AgCl$ in $2\text{M}$ $N{H}_3$ is $0.166\text{M}$

Let the solubility of $AgCl$ in pure water be $s$.
So.
$$\begin{array}{ccccc}& AgCl\rightleftharpoons & A{g}^{+}& +& C{l}^{-}\\ & & s& & s\end{array}$$
Again, $K_{sp}={10}^{-10}$
$\therefore {10}^{-10}=s^2$
$\Rightarrow s={10}^{-5}{\text{molL}}^{-1}$
Thus option (A) is incorrect but option (B) is correct.
Again, when $2\text{M}$ $AgN{O}_3$ is present,
$$\begin{array}{ccccc}& AgCl\rightleftharpoons & A{g}^{+}& +& C{l}^{-}\\ & & 2+s& & s\end{array}$$
So, $K_{\text{sp}}=(2+s)\left(s\right)$
But since $s$ is very small compared to $2$,
thus, we can write, $K_{\text{sp}}=2s$
$\Rightarrow s=\frac{K_{\text{sp}}}{2}=\frac{{10}^{-10}}{2}=5\times {10}^{-11}\text{M}$
Let,
$AgCl\left(s\right)\rightleftharpoons A{g}^{+}\left(\text{aq}\right)+C{l}^{-}\left(\text{aq}\right)K_{sp}={10}^{-10}\left(i\right)$
$A{g}^{+}2N{H}_3\rightleftharpoons Ag(N{H}_3{)}_2^{+}{K}_f={10}^8(ii)$
Now adding $\left(i\right)$ and $\left(ii\right)$,
$$\begin{array}{cccccccc}& AgCl\left(s\right)& +& 2N{H}_3\left(s\right)\left(\text{aq}\right)& \rightleftharpoons & Ag\left(N{H}_3{)}_2^{+}\right(\text{aq})& +& C{l}^{-}\left(\text{aq}\right)\\ \text{at equm}& & & 2-2s& & s& & s\end{array}$$
Therefore,
$K_{\text{eq}}=K_{\text{sp}}\times K_f={10}^{-2}$
$\Rightarrow \frac{s^2}{(2-2s{)}^2}={10}^{-2}$
$\Rightarrow s=0.2-0.2s$
$\Rightarrow s=0.166\text{M}$