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I was wondering if it would be possible to use $\ce{HCl}$ acid to dissolve $\ce{PbS}$.

I know that I can dissolve it in $\ce{HNO_3}$ since, the nitric acid oxidises the sulphide ion to elemental sulphur, and some sulphides like $\ce{ZnS}$ do dissolve in $\ce{HCl}$. But what about $\ce{PbS}$?

What is the extent of $\ce{pH}$ sensitivity of lead sulphide precipitate.

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Hydrogen sulfide is a weak acid: $$\begin{align} \mathrm{p}K_{\text{a}1} &\approx 7\\ \mathrm{p}K_{\text{a}2} &\approx 13\\ \end{align}$$

Therefore, the concentration of sulfide $c_{\ce{S^2-}}$ is very low in acidic solutions:

$$c_{\ce{S^2-}} = \frac{K_{\text{a}1}\cdot K_{\text{a}2}\cdot c_{\ce{H2S}}}{c_{\ce{H+}}^2}\approx 10^{-20}\frac{c_{\ce{H2S}}}{c_{\ce{H+}}^2}$$

Besides, hydrogen sulfide is a gas. Its solubility in water is approximately $c_{\ce{H2S}}\approx 0.1\ \mathrm{mol\ l^{-1}}$.

For example, in a solution of a strong acid with $c_{\ce{H+}}=1\ \mathrm{mol\ l^{-1}}$, the concentration of sulfide may be estimated as

$$c_{\ce{S^2-}} = \frac{K_{\text{a}1}\cdot K_{\text{a}2}\cdot c_{\ce{H2S}}}{c_{\ce{H+}}^2}\approx 10^{-20}\frac{0.1\ \mathrm{mol\ l^{-1}}}{\left(1\ \mathrm{mol\ l^{-1}}\right)^2}=10^{-21}\ \mathrm{mol\ l^{-1}}$$

On the other hand, the solubility product of lead(II) sulfide is extremely low:

$$K_\text{sp} = c_{\ce{Pb^2+}} \cdot c_{\ce{S^2-}} \approx 10^{-28}\ \mathrm{mol^2\ l^{-2}}$$

Thus, the solubility of lead(II) sulfide (i.e. the maximum concentration of $\ce{Pb^2+}$) in a solution of a strong acid with $c_{\ce{H+}}=1\ \mathrm{mol\ l^{-1}}$ is

$$c_{\ce{Pb^2+}} = \frac{K_\text{sp}}{c_{\ce{S^2-}}} \approx \frac{10^{-28}\ \mathrm{mol^2\ l^{-2}}}{10^{-21}\ \mathrm{mol\ l^{-1}}} = 10^{-7}\ \mathrm{mol\ l^{-1}}$$

Therefore, lead(II) sulfide precipitates from acidic solutions and will not significantly dissolve in dilute hydrochloric acid.

However, lead(II) sulfide can be dissolved in hot hydrochloric acid with $c_{\ce{HCl}}=8\ \mathrm{mol\ l^{-1}}$ because of the formation of anionic complexes $\ce{[PbCl3]^-}$ and $\ce{[PbCl4]^2-}$.

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  • $\begingroup$ Excellent answer. Makes perfect sense to me. Thank you so much. $\endgroup$ – getafix May 31 '15 at 0:22

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