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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.

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  • $\begingroup$ I edited the question a bit according to local guidelines (if you are interested in formatting on Chemistry.SE, check out this page, this page and this one). Have you tried to solve the last equation for $s$, and if so, what did you get and what was the answer? $\endgroup$ – andselisk Dec 15 '17 at 11:19
  • $\begingroup$ I think this page might hold your answer : socratic.org/questions/… $\endgroup$ – Shoubhik Raj Maiti Dec 15 '17 at 12:49
  • $\begingroup$ I entered the equation in wolfram, it showed pretty complex solution, the correct answer is $6.3*10^{-4}$ $\endgroup$ – drake01 Dec 16 '17 at 10:25
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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)).

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    $\begingroup$ And yes, quantitative analysis can be tedious at times. $\endgroup$ – andselisk Dec 15 '17 at 16:05
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    $\begingroup$ Thanks @andselisk your answer is correct I think, as I made a typo in the question, K2 was actually 10^(-14). Should I correct it? $\endgroup$ – drake01 Dec 16 '17 at 10:10
  • $\begingroup$ @drake01 No prob, thank you for letting me know. Sure, go ahead and edit your question, I'll edit my answer at some point today:) $\endgroup$ – andselisk Dec 16 '17 at 10:15
  • $\begingroup$ How did you obtain equation (5) and kh2 is net kh right? $\endgroup$ – drake01 Dec 16 '17 at 10:20

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