Q19.139 CP - Chemistry: Molecular Nature Of Matter And Change

Question

$\mathbf{A}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{ }\mathbf{M}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{S}$ solution contains  $\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{ }\mathbf{M}\mathbf{ }{\mathbf{NiCl}}_{\mathbf{2}}$ and $\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{ }\mathbf{M}\mathbf{ }\mathbf{Hg}{\left({\mathrm{NO}}_3\right)}_{\mathbf{2}}$. What data-custom-editor="chemistry" $\mathbf{pH}$ is required to precipitate the maximum amount of $\mathbf{HgS}$ but none of the $\mathbf{Ni}\mathbf{ }\mathbf{S}$? (See Appendix C.)

Step-by-Step Solution

Answer

The pH required to precipitate maximum amount $\text{HgS}$ of is $3.8$.

Step 1: Definition of Concept.

Capacity: The amount of heat energy required to raise the temperature of a body by a certain amount is known as heat capacity. The amount of heat in joules required to raise the temperature 1 Kelvin is known as heat capacity (symbol: C) in SI units.

Step 2: Find the pH required to precipitate maximum amount of HgS .

The expression for is used to create the balanced equation,

${\mathrm{NiS}}_{\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\rightleftharpoons {\mathrm{Ni}}^{2+}{\left(\mathrm{aq}\right)+\mathrm{HS}}_{\left(4\mathrm{q}\right)}^{-}+\mathrm{O}{{\mathrm{H}}^{-}}_{\left(\mathrm{aq}\right)}$

The expression for the ion-product is as follows:

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ni}}^{2+}\right]\left[{\mathrm{HS}}^{-}\right]\left[{\mathrm{OH}}^{-}\right]=3\times {10}^{-16}$

From the concentration of and from appendix,${\text{H}}_2\text{S}$ and ${\text{K}}_{a1}$ can be calculated as,

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{S}}_{\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{t}\right)}\rightleftharpoons {+\mathrm{HS}}_{\left(\mathrm{a}\right)}^{-}+{\mathrm{H}}_3{\mathrm{O}}_{\left(\mathrm{aq}\right)}^{+} \\ \begin{array}{c}{\mathrm{K}}_{\mathrm{a}1}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{HS}}^{-}\right]}{\left[{\mathrm{H}}_2\mathrm{S}\right]}\\ 9\times {10}^{-8}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{HS}}^{-}\right]}{[0.050-\mathrm{x}]}\\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{HS}}^{-}\right]=4.5\times {10}^{-9}\\ \left[{\mathrm{HS}}^{-}\right]=\frac{4.5\times {10}^{-9}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}\end{array} \end{gathered}$$

The concentration of OH is calculated as follows:

$\begin{array}{c}\left[{\mathrm{OH}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}=\frac{1.4\times {10}^{-14}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}\\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ni}}^{2+}\right]\left[{\mathrm{HS}}^{-}\right]\left[{\mathrm{OH}}^{-}\right]=3\times {10}^{-16}\\ 3\times {10}^{-16}=(0.15)\times \left(\frac{4.5\times {10}^{-9}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}\right)\left(\frac{4.5\times {10}^{-9}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}\right)\\ 3\times {10}^{-16}=\frac{(0.15)\times \left(4.5\times {10}^{-9}\right)\left(4.5\times {10}^{-9}\right)}{{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^2}\\ {\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^2=\frac{(0.15)\times \left(4.5\times {10}^{-9}\right)\left(4.5\times {10}^{-9}\right)}{3\times {10}^{-16}}\\ =2.25\times {10}^{-8}\\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.5\times {10}^{-4}\mathrm{M}\end{array}$

The buffer solution's pH can be calculated as follows:

$\begin{array}{c}\mathrm{pH} = -\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\\ \mathrm{pH} = -\log \left(1.5\times {10}^{-4}\mathrm{M}\right)\\ = 3.83\end{array}$

Therefore, the pH required to precipitate maximum amount $\text{HgS}$of is $3.8$.