91Ó°ÊÓ

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.

The expression for is used to create the balanced equation,

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

The expression for the ion-product is as follows:

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

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

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

The concentration of OH is calculated as follows:

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

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

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

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