The problem I have been asked to solve is:

$\ce{Cd^2+}$ and $\ce{Fe^2+}$ (starting concentration $\pu{E-3 mol/L}$) should be separated by precipitation using $\ce{H2S}$. Does a pH range exist allowing to completely precipitate one metal (i.e. the remaining concentration of the precipitated ion has to be $\pu{E-5 mol/L}$ or less) whereas the other stays in solution at its initial concentration? Please calculate this $\pu{pH}$ range!

Solubility product: $\ce{(CdS)}$: $\pu{E-27}$; $\ce{(FeS)}$: $\pu{E-19}$; Acid constant $\ce{(H2S)}$: $\pu{E-20}$; concentration of $\ce{H2S}$ = $\pu{0.1M}$.

My approach to solving the problem was the following:

$\pu{K_{sp}}$ = [$\ce{M+}$][$\ce{S^2-}$] $\ce{->}$ $\pu{K_{sp}}$ for either metal can be defined in the following way:

$\ce{[S^2-]}$ = $\pu{K_{a}}$ $\frac{[\ce{H2S}]}{[\ce{H+}]}$ $\ce{->}$ the concentration of sulfide ions is given by this expression

[$\ce{M+}$] = $\frac{\pu{K_{sp}}\ce{[H+]}}{\pu{K_{a}[\ce{H2S}]}}$ $\ce{->}$ substituting the second equation into the first we get the concentration of metal ions in terms of $\pu{pH}$.

After this, we must find the $\pu{pH}$ range where the concentration of $\ce{Cd^2+}$ is less than $\pu{E-5 M}$ and the concentration of $\ce{Fe^2+}$ is greater than $\pu{E-3 M}$.

For $\ce{Cd^2+}$:

$\pu{E-5} =\frac{\pu{E-27} \times [\ce{H+}]}{\pu{E-20} \times 0.1}$

Solving for $\ce{H+}$ gives $\ce{H+}$ = $\pu{E-3 M}$, or $\pu{pH = 3}$.

For $\ce{Fe^2+}$:

$\pu{E-3} =\frac{\pu{E-19}\times[\ce{H+}]}{\pu{E-20} \times 0.1}$

Solving for $\ce{H+}$ gives $\ce{H+}$ = $\pu{E-5 M}$, or $\pu{pH = 5}$

Thus the $\pu{pH}$ range is $\pu{3-5}$

I realise that this is a homework question, but I hope I have shown that I have given this problem some thought. I am really uncertain, whether my approach is correct, especially the step where I found the $\pu{pH}$ at which the concentration of $\ce{Cd^2+}$ ions is less than $\pu{E-5 M}$ (because I could have also calculated the $\pu{pH}$ at which the concentration of $\ce{Fe^2+}$ ions is less than $\pu{E-5 M}$). Some help would be greatly appreciated.

  • 1
    $\begingroup$ Formatting guides for texts and formulas/equations/expressions. $\endgroup$
    – Poutnik
    Dec 15, 2023 at 13:51
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    $\begingroup$ I'll edit the question. $\endgroup$ Dec 15, 2023 at 13:57
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    $\begingroup$ @HarikrishnanM +1 for your heroic effort! $\endgroup$ Dec 15, 2023 at 20:02
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    $\begingroup$ H2S had two constants. They probably mean the product of both constants, skipping HS-. So K=Ka1.Ka2=[H+]^2[S^2-]/[H2S] . Be aware of the square. $\endgroup$
    – Poutnik
    Dec 16, 2023 at 8:10
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    $\begingroup$ @MaxW I often read Ka2 is near 13.5-14., so I am not sure But if it does not exist, all sulfide solubility products are questionable how they are defined. $\endgroup$
    – Poutnik
    Dec 16, 2023 at 8:23


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