# Why can acidification of solution lead to selective precipitation of ions such as Fe(II) and Zn(II)?

Given $$K_\textrm{sp}(\ce{ZnS})= \pu{2.5E-22}$$ and $$K_\textrm{sp}(\ce{FeS})= \pu{3.7E-19}$$, why would $$\ce{Fe^{2+}}$$ ions remain in solution but not $$\ce{Zn^{2+}}$$ with acidification of the solution?

I have little understanding of this, so I don't have a great attempt at the solution.

• Are you adding an acid to $\ce{FeS}$ and $\ce{ZnS}$? If so, what acid? That might make a difference. Apr 16 '14 at 10:31

FeS has 1500 times the solubility of ZnS. Given the slow addition of a small molar fraction of acid that solubilizes both Zn(II) and Fe(II), that fraction will retain at equilibrium. It gets complicated if soluble polysufides exist. Iron gives pyrite and marcasite ($\ce{FeS2}$) via mackinawite ($\ce{FeS}$) and greigite ($\ce{Fe3S4}$), but they are insoluble.

When you acidify a solution, you increase the concentration of hydronium ions and while doing so decrease the concentration of hydroxide ions.

This decrease in the concentration of hydroxide ions cause the equilibrium of the hydrolysis(both complete and partial) of the sulphide ion to shift forward.

$$\ce{S^2- + 2H2O <=> H2S + 2OH-}$$

This forward shift in equilibrium decreases the concentration of sulphide ions in the solution and thus helps in increasing the solubility of the sulphide salts.

The solubility is increased because all the sulphide added to the solution doesn't stay in the form of sulphide ions, a major portion of it is converted to $$\ce{HS-}$$ and $$\ce{H2S}$$. This drives the solubility equilibrium of the salt forward.

$$\ce{MS <=> M^2+ + S^2-}$$

This method saves $$\ce{FeS}$$ from precipitating in acidic conditions but for $$\ce{ZnS}$$ this is not sufficient. $$\ce{ZnS}$$ is just too insoluble in water to salvage it by acidification, a fact which is also evident from it's very low $$K_{\mathrm{sp}}$$. Chemists have also used this fact to seperate mixtures of zinc and ferrous sulphide.

To precipitate ferrous sulphide, you will need to add an alkali to the solution which will alter the aforementioned equilibriums in just the reverse way.

Let's admit that some $$\ce{H2S}$$ is added into a solution containing both ions $$\ce{Zn^2+}$$ and $$\ce{Fe^2+}$$ with a concentration $$\pu{0.1 M}$$. In my table, the solubility products of $$\ce{FeS}$$ is $$\pu{6.3 \times 10^{-18}}$$, which is $$10$$ times higher than your value : $$\pu{6.3 \times 10^{-19}}$$. With your value for $$\ce{ZnS}$$ and mine for $$\ce{FeS}$$, the maximum amount of $$\ce{S^{2-}}$$ for the ions remaining in solution is for $$\ce{Fe^{2+}, \ce{[S^{2-}]_{Fe}} = 6.3 10^{-17} M}$$, and for $$\ce{Zn^{2+}, \ce{[S^{2-}]}_{Zn} = 10^{-21} M}$$.

Now the amount of $$\ce{S^{2-}}$$ ions in a solution of $$\ce{H2S}$$ is dependent of the two $$\mathrm p K_\mathrm a$$ values of $$\ce{H2S}$$, which are $$7$$ and $$13$$. Combining these data gives the following maximum value for $$\ce{S^{2-}}$$ versus $$\ce{[H+]}$$ : $$\ce{[S^{2-}] = 10^{-20} / [H+]^2 }$$ In an acidic solution of $$\ce{H2S}$$ where $$\ce{[H+] = 0.1 M}$$, this formula gives $$\ce{[S^{2-}] = 10^{-18} M}$$, and this is just between the two limits $$\ce{[S^{2-}]_{Fe}}$$ and $$\ce{[S^{2-}]_{Zn}}$$ described above. So, it is enough for dissolving $$\ce{FeS}$$ and precipitating $$\ce{ZnS}$$.

Reference:

(1) Wallace, H. G.; Stark, J. G.; McGlashan, M. L. Chemistry Data Book; Hodder Murray: London, England, 1989.