$$\(\mathrm{ZnS}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \leftrightarrows \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})\)$$ What is the equilibrium constant for the above reaction? The successive acid dissociation constants for $$\(\mathrm{H}_{2} \mathrm{~S}\)$$ are $$\(9.5 \times 10^{-8}\)$$ $$\(\left(K_{a 1}\right)\)$$ and $$\(1 \times 10^{-19}\left(K_{a 2}\right)\)$$. The $$\(K_{\mathrm{sp}}\)$$, the solubility product constant, for ZnS equals $$\(1.6 \times 10^{-24}\)$$.

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