Question

What is the maximum possible concentration of ${\mathbf{Ni}}^{\mathbf{2}\mathbf{+}}$ ion in water at$\mathbf{25}\mathbf{°}\mathbf{C}$ that is saturated with$\mathbf{0}\mathbf{.}\mathbf{10}{\mathbf{MH}}_{\mathbf{2}}\mathbf{S}$ and maintained at$\mathbf{pH}\mathbf{3}\mathbf{.}\mathbf{0}$ with HCl?

Short answer

The maximum possible concentration of ${\mathbf{Ni}}^{\mathbf{2}\mathbf{+}}$ion in water at$\text{25}°\text{C}$that is saturated with$0.10{\mathrm{MH}}_2\mathrm{S}$ and maintained at$\mathrm{pH}3.0$ with$\mathrm{HCI}$ is$3\mathrm{M}$.

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.

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

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

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

Determination of disassociation of H2S

Let take the temperature of 25°C from the given,

Using $0.10{\mathrm{MH}}_2\mathrm{S}$, the${\mathrm{Ni}}^{2+}$ion in water is saturated, and the pH is maintained at 3.0 with.

Due to the presence of two protons, the dissociation of${\text{H}}_{\text{2}}\text{S}$occurs in two phases.

The following are the dissociation equations:

${\mathrm{H}}_2\mathrm{S}\to {\mathrm{H}}^{+}+{\mathrm{HS}}^{-}$ .…..(1)

${\mathrm{K}}_{\mathrm{a}1}=1.0\times {10}^{-7}$

${\mathrm{HS}}^{-}\to {\mathrm{H}}^{+}+{\mathrm{S}}^{2-}$ ……(2)

${\mathrm{K}}_{\mathrm{a}2}=1.0\times {10}^{-19}$

The net equation is found by multiplying both equations together as follows:

${\mathrm{H}}_2\mathrm{S}\to \text{2}{\mathrm{H}}^{+}+{\mathrm{S}}^{2-}$

$\mathrm{K}=1.0\times {10}^{-26}$

Determination of solubility constant expression of NiS.

The reaction's equilibrium constant will be:

$\mathrm{K}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2\times \left[{\mathrm{s}}^{2-}\right]}{\left[{\mathrm{H}}_2\mathrm{S}\right]}$

$\left[{\mathrm{H}}_2\mathrm{S}\right]=0.10\text{M}$

As the ,$\mathrm{pH}=3$

$3=-\log \left[{\mathrm{H}}^{+}\right]$

$\left[{\mathrm{H}}^{+}\right]=1.0\times {10}^{-3}$

Let’s substitute the obtained values,

$1.0\times {10}^{-26}=\frac{{\left[1.0\times {10}^{-3}\right]}^2\times \left[{\mathrm{s}}^{2-}\right]}{\left[0.1\right]}$

$\left[{\mathrm{s}}^{2-}\right]=\frac{\left[1.0\times {10}^{-26}\right]\times \left[0.10\right]}{{\left[1\times {10}^{-3}\right]}^2}$

$=1.0\times {10}^{-21}\text{M}$

Let’s calculate the concentration of ${\mathrm{Ni}}^{2+}$,

$\mathrm{NiS}\left(\mathrm{s}\right)\to {\text{Ni}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{S}}^{2-}\left(\mathrm{aq}\right)$and${\mathrm{K}}_{\mathrm{SP}}=3.0\times {10}^{-21}$

Thus, the solubility constant expression,

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

By substituting all values,

$3.0\times {10}^{-21}=\left[{\mathrm{Ni}}^{2+}\right]\times \left(1\times {10}^{-21}\right)$

$\left[{\mathrm{Ni}}^{2+}\right]=3\text{M}$

Therefore, maximum possible concentration of${\mathrm{Ni}}^{2+}$ is $3\mathrm{M}$.