Question

Equal volumes of $0.02M$ $Ag{NO}_3$ and $0.02M$ $HCN$ were mixed. If $\left[{Ag}^{+}\right]=6.66\times {10}^{-x}M$ at equilibrium, find the value of $x$.

Take $K_a\left(HCN\right)=9\times {10}^{-10}$; $K_{sp}\left(AgCN\right)=4\times {10}^{-16}$

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

Answer: (B)

$5$

After mixing with equal volume
$\left[{Ag}^{+}\right]=0.01M$, $HCN=0.01M$
$HCN\rightleftharpoons H^{+}+{CN}^{-}{K}_a$
${Ag}^{+}+{CN}^{-}\rightleftharpoons Ag{CN}_{\left(s\right)}$ $\frac{1}{K_{sp}}$
----------------------------------------------------------
$HCN+{Ag}^{+}\rightleftharpoons H^{+}+Ag{CN}_{\left(s\right)}$
$K=\frac{K_a}{K_{sp}}=2.25\times {10}^6$
$0.01$ $0.01$
$x$ $x$ $0.01$
since $K$ value is very high almost all of reactant will convert into product.
$\frac{0.01}{x^2}=2.25\times {10}^6$ $X=6.6\times {10}^{-5}$
$\left[{Ag}^{+}\right]=6.66\times {10}^{-5}M$