Question

For the reaction $\left[ Ag ( CN )_{2}\right]^{-}( aq ) \rightleftharpoons Ag ^{+}( aq )+2 CN ^{-} (aq)$, the equilibrium constant at $25^{\circ} C$ is $4.0 \times 10^{-19}$. Calculate the silver ion concentration in a solution which was originally $0.10\, M$ in $KCN$ and $0.03\, M$ in $AgNO _{3}$.

• A. $2.5 \times 10^{-18} M$
• B. $1.5 \times 10^{-18} M$
• C. $5.5 \times 10^{-18} M$
• D. $7.5 \times 10^{-18} M$

Answer 1

Answer: (D)

$K =\frac{\left[ Ag ^{+}\right]\left[ CN ^{-}\right]^{2}}{\left[ Ag ( CN )^{2}\right]^{-}} $
$=4 \times 10^{-19} $
$=\frac{ a (0.04+2 a )^{2}}{\left[ Ag ( CN )_{2}\right]^{-}} $
On solving $4 \times 10^{-19} \approx \frac{ a (0.04)^{2}}{(0.03- a )}$
i.e., $a =7.5 \times 10^{-18} M$