Question

A mixture contains $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{MCu}}^{\mathbf{2}\mathbf{+}}$and $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{MMn}}^{\mathbf{2}\mathbf{+}}$and is saturated with $\mathbf{0}\mathbf{.}\mathbf{10}{\mathbf{MH}}_{\mathbf{2}}\mathbf{S}$. Determine a pHwhere $\mathbf{CuS}$precipitates but $\mathbf{MnS}$does not precipitate.${\mathbf{K}}_{\mathbf{sp}}$for$\mathit{C}\mathit{u}\mathit{S}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{45}}$and${\mathbf{K}}_{\mathbf{sp}}$for $\mathbf{MnS}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{13}}$.

Short answer

The pHwhere CuS precipitates but MnS does not precipitate is$8.7$.

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.

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

${\mathbf{K}}_{\mathbf{sp}}\mathbf{=}{\left[A^{+}\right]}^{\mathbf{x}}{\left[B^{-}\right]}^{\mathbf{Y}}$

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

Determination of the concentration of S2-

From the given$1.0\times {10}^{-3}{\mathrm{MCu}}^{2+}$and$1.0\times {10}^{-3}{\mathrm{MMn}}^{2+}$in a mixture comprising saturated with$0.10{\mathrm{MH}}_2\mathrm{S}$.

${\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{CuS}\right)=8.5\times {10}^{-45}$and${\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{MnS}\right)=2.3\times {10}^{-13}$

$\mathbf{CuS}$is a sparingly soluble salt with weak dissociation in an aqueous solution.

$\mathrm{CuS}\left(\mathrm{s}\right)\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{S}}^{2-}\left(\mathrm{aq}\right)$

The formula for the solubility constant,

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Cu}}^{2+}\right]\times \left[{\mathrm{S}}^{2-}\right]$

Substituting the provided${\mathrm{K}}_{\mathrm{sp}}$and$\left[{\mathrm{Cu}}^{2+}\right]$ values as:

$8.5\times {10}^{-45}=\left(1\times {10}^{-3}\text{M}\right)\times \left[{\mathrm{S}}^{2-}\right]$

$\left[{\mathrm{S}}^{2-}\right]=\frac{8.5\times {10}^{-45}}{1\times {10}^{-3}\text{M}}$

$\left[{\mathrm{S}}^{2-}\right]=8.5\times {10}^{-45}\text{M}$

$\mathrm{CuS}$will precipitate $\left(\mathrm{As}{\mathrm{Q}}_{\mathrm{sp}}>{\mathrm{K}}_{\mathrm{sp}}\right)$when the concentration

$\left[{\mathrm{S}}^{2-}\right]$of is greater than $8.5\times {10}^{-42}\text{M}$.

In aqueous solution, dissociates as follows:

$\mathrm{MnS}\left(\mathrm{s}\right)\to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{S}}^{2-}\left(\mathrm{aq}\right)$

Determination of the pH

The formula for the solubility constant,

${\mathrm{K}}_{\mathrm{SP}}=\left[{\mathrm{Mn}}^{2+}\right]\times \left[{\mathrm{S}}^{2-}\right]$

Substituting${\mathrm{K}}_{\mathrm{sp}}$and $\left[{\mathrm{Mn}}^{2+}\right]$for the given values:

$2.3\times {10}^{-13}=\left(1\times {10}^{-3}\mathrm{M}\right)\left[{\mathrm{S}}^{2-}\right]$

$\left[{\mathrm{S}}^{2-}\right]=2.3\times {10}^{-10}\text{M}$

$\mathrm{MnS}$will precipitate when the concentration of$\left[{\mathrm{S}}^{2-}\right]$ is more than$2.3\times {10}^{-10}\text{M}$$\left(\mathrm{As}{\mathrm{Q}}_{\mathrm{SP}}>{\mathrm{K}}_{\mathrm{SP}}\right)$. Thus, the concentration of$\left[{\mathrm{S}}^{2-}\right]$ must be larger than$8.5\times {10}^{-42}\text{M}$ and less than$2.3\times {10}^{-10}\text{M}$in order to$\mathrm{CuS}$ precipitate$\mathrm{CuS}$ rather than$\mathrm{MnS}$.

At$\left[{\mathrm{S}}^{2-}\right]=2.3\times {10}^{-10}\text{M}$, the pH is determined.

The acid utilized to achieve saturation is ${\mathrm{H}}_2\mathrm{S}$, which is diprotic in nature, meaning it splits into two parts:

$\begin{array}{l}{\mathrm{H}}_2\mathrm{S}\left(\mathrm{aq}\right)\to {\text{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{HS}}^{-}\left(\mathrm{aq}\right)\\ {\mathrm{K}}_{\mathrm{a}1}=1.0\times {10}^{-7}\end{array}$

$\begin{array}{l}{\mathrm{HS}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{S}}^{-}\left(\mathrm{aq}\right)\\ {\mathrm{K}}_{\mathrm{a}2}=1.0\times {10}^{-19}\end{array}$

By adding both the equations,

$\begin{array}{l}{\mathrm{H}}_2\mathrm{S}\left(\mathrm{aq}\right)\to \text{2}{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{S}}^{-}\left(\mathrm{aq}\right)\\ {\mathrm{K}}_{\mathrm{a}}=1.0\times {10}^{-26}\end{array}$

The reaction's equilibrium constant will be:

$\begin{array}{l}\mathrm{K}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2\times \left[{\mathrm{s}}^{-}\right]}{\left[{\mathrm{H}}_2\mathrm{S}\right]}\\ \left[{\mathrm{H}}^{+}\right]=2.1\times {10}^{-9}\text{M}\end{array}$

By using pH formula,

$\begin{array}{l}\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\\ \mathrm{pH}=8.7\end{array}$

Therefore, at any pH lower than the $8.7$, the$\mathrm{CuS}$ precipitates but not$\mathrm{MnS}$