Unique Solubility Expression for Sulfide Salts
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 $1\times 10^5$. 

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

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 $[\ce{A-}]+[\ce{HA}]=1\ \mathrm{M}$

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