Hydrogen sulfide ( H_2S ) is a weak diprotic acid that ionizes in...

The ionization steps of $H_2S$ are:

$H_2S \rightleftharpoons H^+ + HS^-$ with ionization constant $K_{a_1}$

$HS^- \rightleftharpoons H^+ + S^{2-}$ with ionization constant $K_{a_2}$

The overall reaction is:

$H_2S \rightleftharpoons 2H^+ + S^{2-}$ with overall ionization constant $K_{eq}$

From the equations, $K_{eq} = K_{a_1} \times K_{a_2}$.

Taking the negative logarithm on both sides:

$-\log K_{eq} = -\log K_{a_1} - \log K_{a_2}$

$pK_{eq} = pK_{a_1} + pK_{a_2}$

Thus, statement A is correct and statement D is incorrect.

For a polyprotic acid, the first ionization constant is always greater than the second because it is more difficult to remove a positively charged proton from a negatively charged ion ($HS^-$) than from a neutral molecule ($H_2S$).

$K_{a_1} > K_{a_2}$

Since $pK_a = -\log K_a$, a larger $K_a$ corresponds to a smaller $pK_a$.

$pK_{a_1} < pK_{a_2}$

Thus, statement B is incorrect.

Since $K_{a_1} > K_{a_2}$, $H_2S$ is a stronger acid than $HS^-$.

Thus, statement C is correct.

Therefore, only statements A and C are correct.

Answer: A and C only