Question

Calculate the solubility of$\mathbf{AgCN}\left(s\right)$$\left(K_{\mathrm{SP}}=2.2x{10}^{-12}\right)$ in a solution containing$\mathbf{1}\mathbf{.}\mathbf{0}{\mathbf{MH}}^{\mathbf{+}}$ (${\mathbf{K}}_{\mathbf{a}}\mathbf{}\mathbf{for}\mathbf{}\mathbf{HCN}\mathbf{}\mathbf{is}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{2}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{10}}$).

Short answer

The solubility of$\mathbf{AgCN}\left(s\right)$ in a solution containing$1.0{\mathrm{MH}}^{+}$ is $5.7\times {10}^{-2}\text{mol/L}$.

Step-by-step solution

Introduction to the Concept

The solubility product constant is a constant that determines the solubility equilibrium that occurs between a substance in its solid-state and its ions in the aqueous state. It has the following general expression.

${\mathbf{A}}_{\mathbf{x}}{\mathbf{B}}_{\mathbf{y}}\left(S\right)\mathbf{\rightleftharpoons }{\mathbf{xA}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{yB}}^{\mathbf{-}}\left(\mathrm{aq}\right)$

${\mathrm{K}}_{\mathrm{SP}}=\left[{\mathrm{A}}^{+}\right]\times {\left[{\mathrm{B}}^{-}\right]}^{\mathrm{Y}}$

The solubility product constant is ${\mathrm{K}}_{\mathrm{sp}}$, and the concentration is represented by square brackets.

Determination of the solubility equilibrium.

Given,

${\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{AgCN}\left(\mathrm{s}\right)\right)=2.2\mathrm{x}{10}^{-12}$

${\mathrm{K}}_{\mathrm{a}}\left(\mathrm{HCN}\right)=6.2\mathrm{x}{10}^{-10}$

The solubility equilibrium of$\mathrm{AgCN}\left(\mathrm{s}\right)$is,

$\mathrm{AgCN}\left(\mathrm{s}\right)\to {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CN}}^{-}\left(\mathrm{aq}\right)$

${\mathrm{K}}_{\mathrm{SP}}=2.2\mathrm{x}{10}^{-12}-----\left(1\right)$

Dissociation reaction,

$\mathrm{HCN}\left(\mathrm{aq}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CN}}^{-}\left(\mathrm{aq}\right)$

${\mathrm{K}}_{\mathrm{a}}=6.2\mathrm{x}{10}^{-10}-----\left(2\right)$

By reversing the equation (2)

${\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CN}}^{-}\left(\mathrm{aq}\right)\to \mathrm{HCN}\left(\mathrm{aq}\right)$

$\mathrm{K}=\frac{1}{{\mathrm{K}}_{\mathrm{r}}}-----\left(3\right)$

By adding equation (1) and (2),

$\mathrm{AgCN}\left(\mathrm{s}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+\mathrm{HCN}\left(\mathrm{aq}\right)-----\left(4\right)$

${\mathrm{K}}_1=3.5\mathrm{x}{10}^{-3}$

Determination of the solubility of AgCN(s).

The reaction's equilibrium constant will be:

${\mathrm{K}}_1=\frac{\left[{\mathrm{Ag}}^{+}\right]\times \left[\mathrm{HCN}\right]}{\left[{\mathrm{H}}^{+}\right]}$

The ICE table for the equation (4),

Initial	s	$1.0$	0	0
Change	-s	$1.0-s$	$+s$	$+\mathrm{s}$
Equilibrium	0	$1.0-\mathrm{s}$	$\mathrm{s}$	$\mathrm{s}$

By substituting the values,

$3.5\mathrm{x}{10}^{-3}=\frac{{\mathrm{s}}^2}{\left(1.00-\mathrm{s}\right)}$

$\mathrm{s}=5.7\times {\text{10}}^{-2}\text{mol/L}$

Therefore, the solubility of $\mathrm{AgCN}\left(\mathrm{s}\right)$is $5.7\times {10}^{-2}\text{mol/L}$.