Question

For the reaction, ${\left[Ag{\left(CN\right)}_2\right]}^{-}\rightleftharpoons {Ag}^{+}+2{CN}^{-}$, the equilibrium constant $K_c$ at ${25}^{o}C$ is $4\times {10}^{-19}$. Then ${Ag}^{+}$ concentration in a solution which has $0.1M$ $KCN$ and $0.03M$ $Ag{NO}_3$ 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

Answer: (B)

$7.5\times {10}^{-18}$

$Equation1:$ $2KCN+AgN{O}_3\rightleftharpoons {\left[Ag{\left(CN\right)}_2\right]}^{-}+KN{O}_3+K^{+}$

concentration : $0.1$ $0.03$ $0$ $0$ $0$

$(0.1-0.06)$ $0$ $0.03$ $0.03$ $0.03$

$Equation2:$ ${\left[Ag{\left(CN\right)}_2\right]}^{-}\rightleftharpoons A{g}^{+}+2C{N}^{-}$

$0.03$ $0$ $0.04$

$(0.03-a)$ $a$ $(0.04+2a)$

From the above concentration, we get

$\left[C{N}^{-}\right]=(0.04+2a)M\approx 0.04M$

$\left[Ag\right(CN{)}_2^{-}]=(0.03-a)M\approx 0.03M$

Since $a$ is very small.

$K_{eq}=\frac{\left[A{g}^{+}\right]{\left[C{N}^{-}\right]}^2}{\left[Ag{\left(CN\right)}_2^{-}\right]}=4\times {10}^{-19}$

$K_{eq}=4\times {10}^{-19}=\frac{\left(a\right)(0.04{)}^2}{0.03}$

$a=\frac{(4\times {10}^{-19})\left(0.03\right)}{(0.04{)}^2}$

$a=7.5\times {10}^{-18}$

Hence the correct answer is option $B$.