# How do you calculate the pH range for which one metal precipitates but the other stays in solution?

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.

• Formatting guides for texts and formulas/equations/expressions. Dec 15, 2023 at 13:51
• I'll edit the question. Dec 15, 2023 at 13:57
• @HarikrishnanM +1 for your heroic effort! Dec 15, 2023 at 20:02
• 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. Dec 16, 2023 at 8:10
• @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. Dec 16, 2023 at 8:23