Chemical and Ionic Equilibrium - Result Question 63-1

Solution:

63. $\quad \quad\quad \quad \mathrm{Ag}^{+} \quad + \quad 2 \mathrm{CN}^{-} \quad \rightleftharpoons \quad \quad \mathrm{Ag}(\mathrm{CN})_2^{-}$

$\text {Initial }: \quad \quad \quad 0.03 \quad \quad \quad 0.10 \quad \quad \quad \quad 0 $

$\text { Equilibrium }: \quad x \quad \quad 0.10-0.06 \quad \quad 0.03$

$\begin{aligned} K & =\frac{1}{4 \times 10^{-19}}=2.5 \times 10^{18} \\ \Rightarrow \quad K & =2.5 \times 10^{18}=\frac{0.03}{(0.04)^2 x} \\ x & =7.50 \times 10^{-18} \mathrm{M} \mathrm{Ag}^{+} \end{aligned} $