An aqueous solution contains 0.10 M H$$_2$$S and 0.20 M HCl. If the equilibrium constant for the formation of HS$$^-$$ from H$$_2$$S is $$1.0 \times 10^{-7}$$ and that of S$$^{2-}$$ from HS$$^-$$ ions is $$1.2 \times 10^{-13}$$, then the concentration of S$$^{2-}$$ ions in the aqueous solution is:

We have a solution that already contains a strong acid, HCl, at a concentration of 0.20 M. Because HCl is a strong acid it is completely dissociated, so the hydrogen-ion concentration is fixed at

$$[H^+] = 0.20\ \text{M}$$

The weak diprotic acid H$$_2$$S dissociates in two steps.

First dissociation: $$\mathrm{H_2S \;\rightleftharpoons\; H^+ + HS^-}$$

The equilibrium (acid-dissociation) constant for this step is given as

$$K_{a1} = 1.0 \times 10^{-7}$$

By definition, for this equilibrium we write the relation

$$K_{a1} = \dfrac{[H^+][HS^-]}{[H_2S]}$$

Substituting the known quantities, we let the initial concentration of undissociated H$$_2$$S remain essentially equal to the analytical concentration 0.10 M, because only a very small fraction will dissociate in the strongly acidic medium.

So we have

$$1.0 \times 10^{-7} = \dfrac{(0.20)\,[HS^-]}{0.10}$$

Now we solve for the bisulfide-ion concentration:

$$[HS^-] = \dfrac{(1.0 \times 10^{-7})(0.10)}{0.20}$$

$$[HS^-] = \dfrac{1.0 \times 10^{-8}}{0.20}$$

$$[HS^-] = 5.0 \times 10^{-8}\ \text{M}$$

Second dissociation: $$\mathrm{HS^- \;\rightleftharpoons\; H^+ + S^{2-}}$$

The equilibrium constant for this step is

$$K_{a2} = 1.2 \times 10^{-13}$$

Again we write the defining relation

$$K_{a2} = \dfrac{[H^+][S^{2-}]}{[HS^-]}$$

Rearranging for the required sulfide-ion concentration gives

$$[S^{2-}] = \dfrac{K_{a2}\,[HS^-]}{[H^+]}$$

Substituting the numerical values obtained:

$$[S^{2-}] = \dfrac{(1.2 \times 10^{-13})(5.0 \times 10^{-8})}{0.20}$$

First multiply the numerators:

$$1.2 \times 5.0 = 6.0 \quad\text{and}\quad 10^{-13}\times10^{-8}=10^{-21}$$

So the numerator equals $$6.0 \times 10^{-21}$$.

Now divide by 0.20:

$$[S^{2-}] = \dfrac{6.0 \times 10^{-21}}{0.20} = 30 \times 10^{-21}$$

Finally we adjust the power of ten:

$$[S^{2-}] = 3.0 \times 10^{-20}\ \text{M}$$

Hence, the concentration of sulfide ions in the solution is $$3 \times 10^{-20}\ \text{M}$$.

Hence, the correct answer is Option C.