Chapter 22, Problem 22.66P

Interpretation Introduction

Interpretation:

The solubility of $\text{FeS}\left(\text{s}\right)\text{ and CuS}\left(\text{s}\right)$ in a solution buffered at $pH=2.00\&4$ and saturated with hydrogen sulfide should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

  ${\text{M}}_{\text{n}}{\text{X}}_{\text{m}}\text{(s)}\stackrel{}{\rightleftharpoons }\text{n}{\text{M}}^{\text{m+}}\text{(aq)+m}{\text{X}}^{\text{n+}}\text{(aq)}$

The relation between solubility product and molar solubility is as follows:

  ${\text{K}}_{\text{sp}}\text{=}{\left[{\text{M}}^{\text{m+}}\right]}^{\text{n}}{\left[{\text{X}}^{\text{n-}}\right]}^{\text{m}}$

Here

The solubility product of salt is ${\text{K}}_{\text{sp}}$.

The molar solubility of ${\text{M}}^{\text{m+}}$ ion is $\left[{\text{M}}^{\text{m+}}\right]$

The molar solubility of ${\text{X}}^{\text{n-}}$ ion is $\left[{\text{X}}^{\text{n-}}\right]$

Answer to Problem 22.66P

The solubility of $\text{FeS}\left(\text{s}\right)\text{ and CuS}\left(\text{s}\right)$ in a solution buffered at $pH=2.00$ is given as $5.7\times {10}^{-1}M$ and . $7.3\times {10}^{-10}M$ respectively.

The solubility of $\text{FeS}\left(\text{s}\right)\text{ and CuS}\left(\text{s}\right)$ in a solution buffered at $pH=4$ is given as $5.7\times {10}^{-5}M$ and . $7.3\times {10}^{-14}M$ respectively.

Explanation of Solution

Given data is as follows:

  $\begin{array}{c}pH=2.00\\ \\ pH=2.00\\ \\ \left[H_{2}S\right]=0.10M\end{array}$

The two applicable equations are given below:

  $\begin{array}{c}{\text{H}}_2\text{S}(aq)+{\text{H}}_2\text{O}(l)\leftrightharpoons {\text{H}}_3{\text{O}}^{+}(aq)+{\text{HS}}^{-}(aq)K_{{\text{a}}_1}=8.9\times {10}^{-8}\text{M}\left(1\right)\\ \\ {\text{HS}}^{-}(aq)+{\text{H}}_2\text{O}(l)\leftrightharpoons {\text{H}}_3{\text{O}}^{+}(aq)+{\text{S}}^{2-}(aq)K_{{\text{a}}_2}=1.2\times {10}^{-13}\text{M}\left(2\right)\end{array}$

Addition of the both equation will give,

  $\begin{array}{c}{\text{H}}_2\text{S}(aq)+2{\text{H}}_2\text{O}(l)\leftrightharpoons 2{\text{H}}_3{\text{O}}^{+}(aq)+{\text{S}}^{2-}(aq)\left(3\right)\\ \\ K_c=K_{a1}.K_{a2}\\ =8.9\times {10}^{-8}M\times 1.2\times {10}^{-13}M\\ =1.1\times {10}^{-20}\end{array}$

The expression for sulphide ion concentration can be get by rearranging equilibrium constant expression,

  $\begin{array}{l}{K}_{\text{c}}=\frac{{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2\left[{\text{S}}^{2-}\right]}{\left[{\text{H}}_2\text{S}\right]}\\ \\ \left[{\text{S}}^{2-}\right]=\frac{K_{\text{c}}\left[{\text{H}}_2\text{S}\right]}{{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}=\frac{\left(1.1\times {10}^{-20}\right)\left[{\text{H}}_2\text{S}\right]}{{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}\end{array}$

Given that the solution is saturated with hydrogen sulphide which has a molar concentration of $\text{0}\text{.10 M}$.

  $\begin{array}{l}\left[{\text{S}}^{2-}\right]=\frac{K_{\text{c}}\left[{\text{H}}_2\text{S}\right]}{{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}=\frac{\left(1.1\times {10}^{-20}\right)\left[{\text{H}}_2\text{S}\right]}{{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}\\ \\ \left[{\text{S}}^{2-}\right]=\frac{\left(1.1\times {10}^{-20}\right)\left[0.10\right]}{{\left[{\text{H}}_3{\text{O}}^{+}\right]}^2}\end{array}$

The hydronium ion concentration at $pH=2.00$ will be $\text{0}\text{.01 M}$.

  $\left[{\text{S}}^{2-}\right]=\frac{\left(1.1\times {10}^{-20}\right)\left[0.10\right]}{{\left[0.01\right]}^2}=1.1\times {10}^{-17}M$

Using solubility product constant value of $FeS\left(s\right)$,

  $\begin{array}{l}\left[F{e}^{2+}\right]\left[{\text{S}}^{2-}\right]=6.3\times {10}^{-18}{\text{M}}^2\\ \\ \left[F{e}^{2+}\right]=\frac{6.3\times {10}^{-18}{\text{M}}^2}{\left[{\text{S}}^{2-}\right]}\\ \\ \left[F{e}^{2+}\right]=\frac{6.3\times {10}^{-18}{\text{M}}^2}{1.1\times {10}^{-17}M}=5.7\times {10}^{-1}M\end{array}$

The hydronium ion concentration at $pH=4.00$ will be $\text{1}\times {\text{10}}^{-4}\text{ M}$.

  $\left[{\text{S}}^{2-}\right]=\frac{\left(1.1\times {10}^{-20}\right)\left[0.10\right]}{{\left[\text{1}\times {\text{10}}^{-4}\text{ M}\right]}^2}=1.1\times {10}^{-13}M$

Using solubility product constant value of $FeS\left(s\right)$,

  $\begin{array}{l}\left[F{e}^{2+}\right]\left[{\text{S}}^{2-}\right]=6.3\times {10}^{-18}{\text{M}}^2\\ \\ \left[F{e}^{2+}\right]=\frac{6.3\times {10}^{-18}{\text{M}}^2}{\left[{\text{S}}^{2-}\right]}\\ \\ \left[F{e}^{2+}\right]=\frac{6.3\times {10}^{-18}{\text{M}}^2}{1.1\times {10}^{-13}M}=5.7\times {10}^{-5}M\end{array}$

The hydronium ion concentration at $pH=2.00$ will be $\text{0}\text{.01 M}$.

  $\left[{\text{S}}^{2-}\right]=\frac{\left(1.1\times {10}^{-20}\right)\left[0.10\right]}{{\left[0.01\right]}^2}=1.1\times {10}^{-17}M$

Using solubility product constant value of $CdS\left(s\right)$,

  $\begin{array}{l}\left[C{d}^{2+}\right]\left[{\text{S}}^{2-}\right]=8.0\times {10}^{-27}{\text{M}}^2\\ \\ \left[C{d}^{2+}\right]=\frac{8.0\times {10}^{-27}{\text{M}}^2}{\left[{\text{S}}^{2-}\right]}\\ \\ \left[C{d}^{2+}\right]=\frac{8.0\times {10}^{-27}{\text{M}}^2}{1.1\times {10}^{-17}M}=7.3\times {10}^{-10}M\end{array}$

The hydronium ion concentration at $pH=4.00$ will be $\text{1}\times {\text{10}}^{-4}\text{ M}$.

  $\left[{\text{S}}^{2-}\right]=\frac{\left(1.1\times {10}^{-20}\right)\left[0.10\right]}{{\left[\text{1}\times {\text{10}}^{-4}\text{ M}\right]}^2}=1.1\times {10}^{-13}M$

Using solubility product constant value of $FeS\left(s\right)$,

  $\begin{array}{l}\left[C{d}^{2+}\right]\left[{\text{S}}^{2-}\right]=8.0\times {10}^{-27}{\text{M}}^2\\ \\ \left[C{d}^{2+}\right]=\frac{8.0\times {10}^{-27}{\text{M}}^2}{\left[{\text{S}}^{2-}\right]}\\ \\ \left[C{d}^{2+}\right]=\frac{8.0\times {10}^{-27}{\text{M}}^2}{1.1\times {10}^{-13}M}=7.3\times {10}^{-14}M\end{array}$

At $pH=2.00$, both compounds can be separated since there is a huge difference in the solubility’s and at $pH=4.00$ separation is not possible.