Question

One liter of $N{H}_3$ solution dissolves 0.10 mole of AgCl. Its concentration is: $(K_{sp}$ of AgCl and $K_f$ of $\left[Ag\right(N{H}_3{)}_2{]}^{+}$ are ${10}^{-10}\text{and}1.6\times {10}^7$ respectively)

• A. $2.5M$
• B. $2.7M$
• C. $2.5\times {10}^{3}M$
• D. $6.25M$

Answer 1

The correct option is B $2.7M$

To dissolve 0.1 mole of $AgCl$ in 1 L, let conc of $N{H}_3$ be C and when simultaneous equilibria will be established, conc of $C{l}^{-}$ will be 0.1 M and of complex will be also almost 0.1 M as $K_f$ is large. Let conc $A{g}^{+}$ left over be $x$.
$AgCl\left(s\right)\stackrel{K_{sp}={10}^{-10}}{\rightleftharpoons }A{g}_{aq}^{+}+C{l}_{aq}^{-}x0.1$
$A{g}_{aq}^{+}+2N{H}_3\stackrel{K_f=1.6\times {10}^7}{\rightleftharpoons }\left[Ag\right(N{H}_3{)}_2{]}^{+}$
$xC-0.1\times 20.1$

So, $K_{sp}={10}^{-10}=x\times 0.1\Rightarrow x={10}^{-9}M$
And $K_f=1.6\times {10}^7=\frac{0.1}{x\times (C-0.1\times 2{)}^2}=\frac{0.1}{{10}^{-9}\times (C-0.1\times 2{)}^2}$
$\Rightarrow (C-0.1\times 2{)}^2=\frac{1}{0.16}\Rightarrow C-0.1\times 2=\frac{1}{0.4}\Rightarrow C=2.7M$