While looking at aqeuous $\ce{K_{sp}}$ values in the back of my textbook, I noticed that there is a asterisk by all of the sulfide compounds (such as $\ce{HgS}$), and the asterisk says, "For a solubility equilibrium of the type $\ce{MS_{(s)} + H_2O_{(l)} <=> M^{2+}_{(aq)} + HS^{-}_{(aq)} + OH^{-}_{(aq)}}$." Now, why is it this expression, instead of the expected $\ce{MS_{(s)} <=> M^{2+}_{(aq)} + S^{2-}_{(aq)}}$?

Upon examination, I see that the given expression is just the expected expression and then tacking on the base equilibrium of the $\ce{S^{2-}}$ ion, which is $\ce{S^{2-}_{(aq)} + H_2O_{(l)} <=>HS^{-}_{(aq)} + OH^{-}_{(aq)}}$. But why do we involve the base equilibrium of $\ce{S^{2-}}$ when we don't involve other quite basic anions, such as $\ce{F^-}$ or $\ce{CO_3^{2-}}$?

  • $\begingroup$ $\ce{F^-}$ is not that basic. $\ce{CO3^2-}$ is, but still less so than $\ce{S^2-}$. $\endgroup$ Apr 18, 2016 at 23:49
  • $\begingroup$ Why is it a new category in and of itself? $\endgroup$
    – Yunfei Ma
    Apr 18, 2016 at 23:52
  • $\begingroup$ From Wikipedia article on hydrogen sulfide The sulfide dianion $\ce{S^{2−}}$ exists only in strongly alkaline aqueous solutions; it is exceptionally basic with a pKa > 14. en.wikipedia.org/wiki/Hydrogen_sulfide $\endgroup$
    – MaxW
    Apr 19, 2016 at 0:13

1 Answer 1


To summarize the comments, the sulfide anion is a strong base. This site (sourced from the CRC Handbook of Chemistry and Physics) lists the second $K_a$ of $\ce{H2S}$ as $1\times 10^{-19}$, which is less acidic than water ($1\times 10^{-15.7}$).

Thus the $K_b$ of $\ce{S^2-}$, which is the equilibrium constant for the reaction with water, is $1\times 10^5$.

$$\ce{S^2- + H2O -> HS- + OH-}\ K_b= 1\times 10^5$$

After wrangling the algebra, if you start with a $1\ \mathrm{M}$ solution of $\ce{S^2-}$, at equilibrium $[\ce{S^2-}]=3.16\times 10^{-10}\ \mathrm{M}$ - almost all of it is converted to $\ce{HS-}$

Compare for some other anions (assume starting with total concentration $[\ce{A-}]+[\ce{HA}]=1\ \mathrm{M}$

$$\begin{array}{c|c|c|c|c} \ce{A-} & K_b & [\ce{A-}]_{eq} & [\ce{HA}]_{eq} & \text{% protonated}\\ \hline \ce{S^2-} & 1\times 10^5 & 3.16 \times 10^{-10}\ \mathrm{M} & \sim1\ \mathrm{M} & \sim 100\% \\ \ce{CO3^2-} & 2.1\times 10^{-4} & 0.99\ \mathrm{M} & 1.45\times 10^{-2}\ \mathrm{M} & \sim 1\% \\ \ce{F-} & 3.13\times 10^{-11} & \sim1\ \mathrm{M} & 5.59\times 10^{-6}\ \mathrm{M} & \sim 0\% \end{array}$$

  • $\begingroup$ Very well presented. $\endgroup$
    – ringo
    Apr 19, 2016 at 1:03

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