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What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide ? (For iron sulphide, ${{K}_{sp}}=6.3\times {{10}^{-18}}$).

Last updated date: 22nd Jun 2024
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Hint: If the product of concentration of ions in a solution is equal to the value of solubility product of a sparingly soluble salt, then no precipitation occurs.

According to the question, we have,
- Ferrous sulphate ($FeS{{O}_{4}}$) and sodium sulphide ($N{{a}_{2}}S$) are equimolar solutions, i.e. their molarity is equal.
- Suppose the maximum concentration of equimolar solutions of $FeS{{O}_{4}}$ and $N{{a}_{2}}S$ required so that there is no precipitation of iron sulphide is ‘$x$’. Concentration is usually expressed in $mol{{L}^{-1}}$.
- Now, $FeS{{O}_{4}}$ and $N{{a}_{2}}S$ are mixed in equal volume. So, the concentrations of $FeS{{O}_{4}}$ and $N{{a}_{2}}S$ after mixing will be half of their initial concentration, i.e. $\dfrac{x}{2}mol{{L}^{-1}}$. This can be calculated as $\dfrac{v\,\times \,x}{v+v}=\dfrac{v\times x}{2v\,}=\dfrac{x}{2}$. Here $v$ is the volume of $FeS{{O}_{4}}$ or $N{{a}_{2}}S$ mixed.
$\left[ FeS{{O}_{4}} \right]=\left[ N{{a}_{2}}S \right]=\dfrac{x}{2}mol{{L}^{-1}}$
- $FeS{{O}_{4}}$ ionizes in the solution into $F{{e}^{2+}}$ and $S{{O}_{4}}^{2-}$.
- If the concentration of $FeS{{O}_{4}}$ is $\dfrac{x}{2}$ and it completely ionizes into $F{{e}^{2+}}$ and $S{{O}_{4}}^{2-}$, then the concentration of $F{{e}^{2+}}$ and $S{{O}_{4}}^{2-}$ will be equal to $\dfrac{x}{2}mol{{L}^{-1}}$.
\begin{align} & FeS{{O}_{4}}\to F{{e}^{2+}}+SO_{4}^{2-} \\ & \,\,\,\,\dfrac{x}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{x}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{x}{2} \\ \end{align}
- Similarly, $N{{a}_{2}}S$ ionizes into $N{{a}^{+}}$ and ${{S}^{2-}}$ completely in the solution.
\begin{align} & N{{a}_{2}}S\to N{{a}^{+}}+{{S}^{2-}} \\ & \,\,\,\,\dfrac{x}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{x}{2}\,\,\,\,\,\,\,\,\,\,\,\dfrac{x}{2} \\ \end{align}
- We now have the concentration of $F{{e}^{2+}}$ and ${{S}^{2-}}$.
$\left[ F{{e}^{2+}} \right]=\left[ {{S}^{2-}} \right]=\dfrac{x}{2}$
According to the question, no precipitates of $FeS$ are formed. It means that the solubility product (${{K}_{sp}}$) for $FeS$should be equal to the product of the concentration of $F{{e}^{2+}}$ and ${{S}^{2-}}$.
${{K}_{sp}}=\left[ F{{e}^{2+}} \right]\times \left[ {{S}^{2-}} \right]$

Given, the solubility product for $FeS$, ${{K}_{sp}}=6.3\times {{10}^{-18}}$ .
\begin{align} & \left[ F{{e}^{2+}} \right]\left[ {{S}^{2-}} \right]={{K}_{sp}}\,of\,FeS \\ & \dfrac{x}{2}\times \dfrac{x}{2}=6.3\times {{10}^{-18}} \\ & {{x}^{2}}=6.3\times {{10}^{-18}}\times 4=25.2\times {{10}^{-18}} \\ & x=\sqrt{25.2\times {{10}^{-18}}}=5.02\times {{10}^{-9}}mol{{L}^{-1}} \\ \end{align}
Therefore, the the maximum concentration of equimolar solutions of $FeS{{O}_{4}}$ and $N{{a}_{2}}S$ so that when mixed in equal volumes, there is no precipitation of $FeS$ is $5.02\times {{10}^{-9}}mol{{L}^{-1}}$

Note:
When the product of concentration of ions of a salt in the solution (ionic product) is more than the solubility product of the salt, then precipitation takes place.