Ncert Exercises & Solutions

$K_{a_{1}},$ $K_{a_{2}}$ $and$ $K_{a_{3}}$ are the respective ionisation constants for the following reactions.
$\mathrm{H}_2 \mathrm{~S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-}$
$\mathrm{HS}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-}$
$\mathrm{H}_2 \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}$
The correct relationship between $K_{a_{1}},$ $K_{a_{2}}$ $and$ $K_{a_{3}}$ is:
1. $\mathrm{K}_{\mathrm{a}_3}=\mathrm{K}_{\mathrm{a}_1} \times \mathrm{K}_{\mathrm{a}_2} $
2. $\mathrm{K}_{\mathrm{a}_3}=\mathrm{K}_{\mathrm{a}_1}+\mathrm{K}_{\mathrm{a}_2} $
3. $K_{a_3}=K_{a_1}-K_{a_2} $
4. $\mathrm{K}_{\mathrm{a}_3}=\mathrm{K}_{\mathrm{a}_1} / \mathrm{K}_{\mathrm{a}_2}$

HINT: Use basic properties of equilibrium constant.

For the reaction,

H_{2} S ⇌ H^{+} + {HS}^{-}

K_{a_{1}} = \frac{[H^{+}] [{HS}^{-}]}{[H_{2} S]}

For the reaction,

{HS}^{-} ⇌ H^{+} + S^{2 -}

K_{a_{2}} = \frac{[H^{+}] [S^{2 -}]}{[{HS}^{-}]}

When, the above two reactions are added, their equilibrium constants are multiplied, thus

K_{a_{3}} = \frac{{[H^{+}]}^{2} [S^{2 -}]}{[H_{2} S]} = K_{a_{1}} \times K_{a_{2}}

Hence,

K_{a_{3}} = K_{a_{1}} \times K_{a_{2}}

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