A problem in solubility of salts with multistep hydrolysis of an ion possible

Question

Calculate the solubility of $\ce{MnS}$ in pure water, $K_\mathrm{sp}(\ce{MnS}) = \pu{2.5e-10}$. Assume hydrolysis of $\ce{S^2-}$ ion. $K_1$ and $K_2$ for $\ce{H2S}$ are $10^{-7}$ and $10^{-14}$, respectively.

Let unknown amount of $\ce{MnS}$ be dissolved in $\pu{1 L}$ of water, so $\ce{Mn^2+} = s~\mathrm{moles}$ and $\ce{S^2-} = s~\mathrm{moles}$ where $s$ is the solubility of $\ce{MnS}$ in $\pu{mol L-1}$.

Now let $\ce{S^2-}$ participate in hydrolysis by dissociating its $x$ moles and combining with $\ce{H2O}$ to form $x$ moles of $\ce{HS-}$ and $x$ moles of $\ce{OH-}$, so final moles of $\ce{S^2-}$ is $(s-x)$.

I considered only $K_\mathrm{h1}$ as it is $10^7$ times larger than $K_\mathrm{h2}$, $K_\mathrm{h1}$ is $K_\mathrm{w}/K_\mathrm{a2}$ which is equal to $1$. So I got $s^2 = (s-x)$ and we know that $K_\mathrm{sp} = s(s-x)$. On substituting I got the following equation:

$$K_\mathrm{sp} = s^2 - s^{1/2} \cdot K_\mathrm{sp}^{1/2}$$

I am stuck here. I know that $K_\mathrm{sp} = \pu{2.5e-10}$, so I tried solving for $s$, but could not find the correct answer.

Can you please point out my conceptual mistake?

Sorry, if this question is of homework type, I really tried solving this before posting.

Answer 1

Let's denote all processes of interest which are occuring or might occur in the system:

\begin{align} \ce{MnS &<=>[$K_\mathrm{sp}$] Mn^2+ + S^2-} && K_\mathrm{sp} = [\ce{Mn^2+}][\ce{S^2-}] \tag{R1} \\ \ce{MnS + H2O &<=>[$K_\mathrm{h1}$] Mn^2+ + HS- + OH-} && K_\mathrm{h1} = [\ce{Mn^2+}] [\ce{HS-}] [\ce{OH-}] \tag{R2} \\ \ce{MnS + 2 H2O &<=>[$K_\mathrm{h2}$] Mn^2+ + H2S + 2 OH-} && K_\mathrm{h2} = [\ce{Mn^2+}][\ce{H2S}][\ce{OH-}]^2 \tag{R3} \\ \ce{H2S &<=>[$K_\mathrm{a1}$] HS- + H+} && K_\mathrm{a1} = \frac{[\ce{HS-}][\ce{H+}]}{[\ce{H2S}]} \tag {R4} \\ \ce{HS- &<=>[$K_\mathrm{a2}$] S^2- + H+} && K_\mathrm{a2} = \frac{[\ce{S^2-}][\ce{H+}]}{[\ce{HS-}]} \tag {R5} \\ \ce{H2S &<=>[K_\mathrm{d}] 2 H+ + S^2-} && K_\mathrm{d} = K_\mathrm{a1} K_\mathrm{a2} \tag{R6} \end{align}

and also agree on labeling $C_\mathrm{A}$ total concentration of sulfide anions in solution in form of $\ce{S^2-}$, $\ce{HS-}$ and $\ce{H2S}$; $\alpha_2$ – fraction of $\ce{S^2-}$ among those; $s$ – solubility. Then from (R1)

$$K_\mathrm{sp} = [\ce{Mn^2+}] C_\mathrm{A} \alpha_2 \implies s = \sqrt{\frac{K_\mathrm{sp}}{\alpha_2}} \tag{1}$$

$\alpha_2$ can be expressed via known equilibrium constants and $\mathrm{pH}$:

\begin{align} \alpha_2 &= \frac{[\ce{S^2-}]}{[\ce{S^2-}] + [\ce{HS-}] + [\ce{H2S}]} &&\Bigg| \cdot \frac{[\ce{H+}][\ce{HS-}]}{[\ce{H+}][\ce{HS-}]}\\ &= \frac{K_\mathrm{a2}}{K_\mathrm{a2} + [\ce{H+}] + \frac{[\ce{H2A}][\ce{H+}]}{[\ce{HA-}]}} &&\Bigg| \cdot \frac{K_\mathrm{a1}}{K_\mathrm{a1}} \\ &= \frac{K_\mathrm{a1}K_\mathrm{a2}}{K_\mathrm{a1}K_\mathrm{a2} + K_\mathrm{a1}[\ce{H+}] + [\ce{H+}]^2} \tag{2} \end{align}

Since hydrolysis might occur partially (R2) or completely (R3), in order to find$[\ce{H+}]$ from

$$[\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} \tag{3}$$

let's estimate whether the equilibrium in (R3) is going to be shifted left or right. Solubility of sulfide anions in the absence of hydrolysis is

$$[\ce{S^2-}] = \sqrt{K_\mathrm{sp}} = \sqrt{\pu{2.5e-10}} = \pu{1.6e-5 M} \tag{4}$$

whereas solubility when $\ce{H2S}$ is forming (from (R6)) would be

$$[\ce{S^2-}] = \sqrt[3]{\frac{K_\mathrm{a1} K_\mathrm{a1}}{4}} = \sqrt[3]{\frac{10^{-7} \cdot 10^{-10}}{4}} = \pu{1.4e-6 M} \tag{5}$$

so that formation of less soluble product would indeed favor (R3). Hence, in order to estimate the increase in $\mathrm{pH}$, it's better to use $K_\mathrm{h2}$:

\begin{align} K_\mathrm{h2} &= [\ce{Mn^2+}][\ce{OH-}]^2[\ce{H2S}] && \Big| \cdot \frac{[\ce{H+}]^2[\ce{S^2-}]}{[\ce{H+}]^2[\ce{S^2-}]} \\ &= \frac{K_\mathrm{sp}K_\mathrm{w}^2}{K_\mathrm{a1}K_\mathrm{a2}} \tag{6} \end{align}

On the other hand,

$$K_\mathrm{h2} = [\ce{Mn^2+}][\ce{OH-}]^2[\ce{H2S}] = 4[\ce{OH-}]^4 \tag{7}$$

Equating (4) and (5):

$$[\ce{OH-}] = \sqrt[4]{\frac{K_\mathrm{sp}K_\mathrm{w}^2}{4 K_\mathrm{a1}K_\mathrm{a2}}} = \sqrt[4]{\frac{\pu{2.5e-10} \cdot (10^{-14})^2}{4 \cdot 10^{-7} \cdot 10^{-10}}} = \pu{5e-6 M} \tag{8}$$

Now back to (3):

$$[\ce{H+}] = \frac{10^{-14}}{\pu{5e-6}} = \pu{2e-9 M} \tag{9}$$

Knowing how the $\mathrm{pH}$ of the system has been changed, let's find what fraction of sulfide anions are in solution:

$$\alpha_2 = \frac{10^{-7} \cdot 10^{-10}}{10^{-7} \cdot 10^{-10} + 10^{-7} \cdot \pu{2e-9} + (\pu{2e-9})^2} = 0.0467 \tag{8}$$

Finally, solubility can be determined:

$$s = \sqrt{\frac{\pu{2.5e-10}}{0.0467}} = \pu{7.3e-5 M} \tag{10}$$

e.g. taking hydrolysis into account, solubility of $\ce{MnS}$ is approx. $4.5$ times higher (see equations (4) and (10)).