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$K_{a_1}, K_{a_2}$ and $K_{a_3}$ are the respective ionisation constants for the following reactions.
$\begin{aligned}
&\mathrm{H}_2 \mathrm{S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-} \\
&\mathrm{HS}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-} \\
&\mathrm{H}_2 \mathrm{S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}
\end{aligned}$
The correct relationship between $K_{a_1}, K_{a_2}$ and $K_{a_3}$ is
Options:
$\begin{aligned}
&\mathrm{H}_2 \mathrm{S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-} \\
&\mathrm{HS}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-} \\
&\mathrm{H}_2 \mathrm{S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}
\end{aligned}$
The correct relationship between $K_{a_1}, K_{a_2}$ and $K_{a_3}$ is
Solution:
1182 Upvotes
Verified Answer
The correct answer is:
$K_{a_3}=K_{a_1} \times K_{a_2}$
$K_{a_3}=K_{a_1} \times K_{a_2}$
For the reaction,
$\mathrm{H}_2 \mathrm{S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-}$
$K_{a_1}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HS}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{S}\right]}\quad\quad...(i)$
For the reaction,
$\begin{aligned}
\mathrm{HS}^{-} & \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-} \\
K_{a_2} &=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{HS}^{-}\right]}\quad\quad...(ii)
\end{aligned}$
$K_{a_3}=\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~S}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{S}\right]}\quad\quad...(iii)$
Hence, $K_{a_3}=K_{a_1} \times K_{a_2}$
$\mathrm{H}_2 \mathrm{S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-}$
$K_{a_1}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HS}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{S}\right]}\quad\quad...(i)$
For the reaction,
$\begin{aligned}
\mathrm{HS}^{-} & \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-} \\
K_{a_2} &=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{HS}^{-}\right]}\quad\quad...(ii)
\end{aligned}$
$K_{a_3}=\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~S}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{S}\right]}\quad\quad...(iii)$
Hence, $K_{a_3}=K_{a_1} \times K_{a_2}$
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