Question

In a given reaction; $Ag(CN{)}_2^{-}\rightleftharpoons A{g}^{+}+2C{N}^{-}$, the equilibrium constant at ${25}^{\circ }C$ is $4.0\times {10}^{-19}$, then the silver ion concentration in a solution which was originally $0.1M$ in $KCN$ and $0.03M$ in $AgN{O}_3$ is:

• A. $7.5\times {10}^{18}$
• B. $7.5\times {10}^{-18}$
• C. $7.5\times {10}^{19}$
• D. $7.5\times {10}^{-19}$

Answer 1

The correct option is B $7.5\times {10}^{-18}$

$2KCN+AgN{O}_3\rightleftharpoons Ag(CN{)}_2^{-}+KN{O}_3+K^{+}$
0.1 0.03 0 0 0
(0.1-0.06) 0 (0.03) (0.03) (0.03)

$Ag(CN{)}_2^{-}=0.03$
Now,
$\therefore Ag(CN{)}_2^{-}\rightleftharpoons A{g}^{+}+2C{N}^{-}$
0.03 a 0.04( left from $KCN$) (0.03–a) a (0.04+a)
$0.04+a\approx 0.04$
$\therefore K_c=4\times {10}^{-19}=\frac{(0.04{)}^2\times a}{0.03}$
$a=7.5\times {10}^{-18}$