Question

$K_{a_1},K_{a_2}$ and $K_{a_3}$ are respective ionization constants for the following reactions.

$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}$

$H{S}^{-}\rightleftharpoons H^{+}+S^{2-}$

$H_{2}S\rightleftharpoons 2H^{+}+S^{2-}$

The correct relationship between $K_{a_1},K_{a_2}$ and $K_{a_3}$ is

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

Answer 1

The correct option is D $K_{a_3}=K_{a_1}\times K_{a_2}$

We have following equations:

$H_{2}S\rightleftharpoons H^{+}+H{S}^{-}....\left(1\right),K_{a_1}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}$

$H{S}^{-}\rightleftharpoons H^{+}+S^{-}....\left(2\right),K_{a_2}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}$

$H_{2}S\rightleftharpoons 2H^{+}+S^{2-}....\left(3\right),K_{a_3}=\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}$

Adding equations $\left(1\right)$ and $\left(2\right)$ gives equation $\left(3\right)$

So, their equilibrium constant gets multiplied.

Thus,

$K_{a_3}=K_{a_1}\times K_{a_2}$

$=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}\times \frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}$

$\frac{\left[H^{+}{]}^2\right[S^{2-}]}{\left[H_{2}S\right]}=K_{a_3}$

Therefore, the correct option is $\left(A\right)$ i.e.,

$K_{a_3}=K_{a_1}\times K_{a_2}$