Question

$K_{a_1},K_{a_2}$ and $K_{a_3}$ are the respective ionisation constants for the following reactions:
$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}$
$H_{2}S\rightleftharpoons H^{+}+S^{2-}$
$H_{2}S\rightleftharpoons 2H^{+}+S^{2-}$
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:
$H_{2}S\rightleftharpoons H^{+}+H{S}^{-};K_{a_1}$
$K_{a_1}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}\dots (i)$
For the reaction:
$H{S}^{-}\rightleftharpoons H^{+}+S^{2-};K_{a_2}$
$K_{a_2}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}\dots (ii)$
For the reaction:
$H_{2}S\rightleftharpoons 2H^{+}+S^{2-};K_{a_3}$
$K_{a_3}=\frac{{\left[H^{+}\right]}^2\left[S^{2-}\right]}{\left[H_{2}S\right]}\dots (iii)$
By multiplying eqn. (i) and eqn. (ii), we get
$K_{a_1}\times K_{a_2}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}\times \frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}$
$=\frac{{\left[H^{+}\right]}^2\left[S^{2-}\right]}{\left[H_{2}S\right]}=K_{a_3}$