Question

Equal volumes of $0.02M$ $AgN{o}_3$ and $0.02MHCN$ are mixed. Calculate $\left[A{g}^{\oplus }\right]$ in solution after attaining equilibrium.

$K_a\left(HCN\right)=6.2\times {10}^{-10}$ and $K_{sp}$ $\left(AgCN\right)=2.2\times {10}^{-16}$

• A. $6\times {10}^{-5}M$
• B. $5.6\times {10}^{-6}M$
• C. $3.3\times {10}^{-5}M$
• D. $9\times {10}^{-5}M$

Answer 1

The correct option is A $6\times {10}^{-5}M$

$\begin{array}{cccccccc}& A{g}^{\oplus }& +& HCN& ⟶& AgCN& +& H^{\oplus }\\ Initial& \frac{0.02\times V}{2V}& & \frac{0.02\times V}{2V}& & 0& & 0\\ Final& 0& & 0& & 0.01& & 0.01\end{array}$

$AgCN\rightleftharpoons A{g}^{\oplus }+C{N}^{⊝}$

$\therefore K_{sp}=2.2\times {10}^{-16}=\left[A{g}^{\oplus }\right]\left[C{N}^{⊝}\right]$

$HCN\rightleftharpoons H^{\oplus }+C{N}^{⊝}{K}_a=6.2\times {10}^{10}=\frac{\left[H^{\oplus }\right]\left[C{N}^{⊝}\right]}{\left[HCN\right]}$

Now, soluble $C{N}^{⊝}$ formed will hydrolyse as follows:

$C{N}^{⊝}+H_{2}O\rightleftharpoons HCN+\stackrel{⊝}{O}H$

$\therefore \left[A{g}^{\oplus }\right]$ of soluble $AgCN=C{N}^{⊝}$ left after hydrolysis + $HCN$ formed after hydrolysis

$\therefore \left[\frac{2.2\times {10}^{-10}}{\left[C{N}^{⊝}\right]}\right]=\left[C{N}^{⊝}\right]+\frac{\left[H^{\oplus }\right]\left[C{N}^{⊝}\right]}{6.2\times {10}^{-10}}$

$=\left[C{N}^{⊝}\right]+\frac{{10}^{-2}\left[C{N}^{⊝}\right]}{6.2\times {10}^{-10}}[\because C{N}^{⊝}<<<\frac{{10}^{-2}\left[C{N}^{⊝}\right]}{6.2\times {10}^{-10}}]$

$or\left[C{N}^{⊝}\right]=\frac{2.2\times {10}^{-16}\times 6.2\times {10}^{-10}}{{10}^{-2}}$

$\therefore \left[C{N}^{⊝}\right]=3.7\times {10}^{-12}$

$\therefore \left[A{g}^{\oplus }\right]=\frac{K_{sp}}{\left[C{N}^{⊝}\right]}=\frac{22.2\times 106-16}{3.7\times 106-12}=5.96\times {10}^{-5}M$