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  • K_{a1}, , K_{a2}, K_{a3} are the respective ionisation constants for the following reactions. H_{2}S \Leftrightarrow H^{+} + HS^{-} HS^{-} \Leftrightarrow H^{+} + S^{2-} H_{2}S \Leftrightarrow 2H^{+} + S^{2-}

K_{a1}, , K_{a2}, K_{a3}  are the respective ionisation constants for the following reactions.

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

 The correct relationship between K_{a1}, , K_{a2}, K_{a3}  is

(i) K_{a3} = K_{a1} \times K_{a2}

(ii) K_{a3} = K_{a1} + K_{a2}

(iii) K_{a3} = K_{a1} - K_{a2}

(iv) K_{a3} = \frac{K_{a1} }{K_{a2}}

Answers (1)

The answer is the option (i) K_{a3} = K_{a1} \times K_{a2}

Explanation:

K_{a1}=\frac{[H^{+}][HS^{-}]}{[H_{2}S]}.....(1)

K_{a2}=\frac{[H^{+}]+[S^{2-}]}{[HS]}.....(2)

On adding equation (i.e. multiplication of equilibrium constants) 1 and 2, we get,

K_{a3}=\frac{[H^{+}]^{2}[S^{2-}]}{[H_{2}S]}

Which proves that for a dibasic acid K_{a3} = K_{a1} \times K_{a2}

Posted by

infoexpert22

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