Question

$K_{a_1}, K_{a_2}$ and $K_{a_3}$ are the respective ionisation constants for the following reactions:
$ H _2 S \rightleftharpoons H ^{+}+ HS ^{-}$
$ H _2 S \rightleftharpoons H ^{+}+ S ^{2-}$
$ H _2 S \rightleftharpoons 2 H ^{+}+ S ^{2-}$
The correct relationship between $K_{a_1}, K_{a_2}$ and $K_{a_3}$ is

• A. $K_{a_3}= K_{a_1} \times K_{a_2} $
• B. $K_{a_3}= K_{a_1} + K_{a_2} $
• C. $K_{a_3}= K_{a_1} - K_{a_2} $
• D. $K_{a_3}= K_{a_1} / K_{a_2} $

Answer 1

Answer: (A)

For the reaction:
$ H _2 S \rightleftharpoons H ^{+}+ HS ^{-} ; K_{a_1} $
$K_{a_1}=\frac{\left[H^{+}\right]\left[H S^{-}\right]}{\left[H_2 S\right]} \ldots(i)$
For the reaction:
$ HS ^{-} \rightleftharpoons H ^{+}+ S ^{2-} ; K_{a_2} $
$K_{a_2}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H S^{-}\right]} \ldots(i i)$
For the reaction:
$ H _2 S \rightleftharpoons 2 H ^{+}+ S ^{2-} ; K_{a_3}$
$K_{a_3}=\frac{\left[H^{+}\right]^2\left[S^{2-}\right]}{\left[H_2 S\right]} \ldots(i i i)$
By multiplying eqn. (i) and eqn. (ii), we get
$K_{a_1} \times K_{a_2}=\frac{\left[H^{+}\right]\left[H S^{-}\right]}{\left[H_2 S\right]} \times \frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H S^{-}\right]}$
$=\frac{\left[H^{+}\right]^2\left[S^{2-}\right]}{\left[H_2 S\right]}=K_{a_3}$