Question

For the reaction ; ${\left[Ag{\left(CN\right)}_2\right]}^{-}\to A{g}^{+}+2C{N}^{-}$

The equilibrium constant for the above reaction at ${25}^{\circ }C$ is $4.0\times {10}^{-19}$. Calculate the silver ion concentration in a solution which was originally 0.10 molar in $KCN$ and 0.03 molar in $AgN{O}_3$.

• A. $7.50\times {10}^{18}M$
• B. $7.50\times {10}^{-18}M$
• C. $7.50\times {10}^{10}M$
• D. $7.50\times {10}^{-10}M$

Answer 1

The correct option is B $7.50\times {10}^{-18}M$

0.03 moles of $AgN{O}_3$ will react with $2\times 0.03=0.06$ moles of KCN to form 0.03 moles of $\left[Ag\right(CN{)}_2{]}^{-}$.

Let $x$ M be the silver ion concentration obtained due to dissociation of $\left[Ag\right(CN{)}_2{]}^{-}$.

$\underset{0.03-x}{\left[Ag\right(CN{)}_2{]}^{-}}\to \underset{x}{A{g}^{+}}+\underset{0.04+x}{2C{N}^{-}}$

Here, $0.04>>x,0.03>>x$
The expression for the equilibrium constant is as shown below:

$K=\frac{\left[Ag\right(CN{)}_2^{-}]}{\left[A{g}^{+}\right][C{N}^{-}{]}^2}$

The equilibrium constant is equal to the reciprocal of the dissociation constant as the equilibrium reaction is the reverse of the dissociation reaction.

Substitute values in the above expression

$\frac{1}{4\times {10}^{-19}}=\frac{0.03}{x\times (0.04{)}^2}$

Hence, $x=7.5\times {10}^{-18}$.

The correct option is 'B'