Why is $K a_{2} << K a_{1}$ for $H_{2} S O_{4}$ in water?

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Solution

$H_{2} S O_{4}$ is dibasic acid. It ionises in two stages and hence has two dissociation constants as given below:

(i) $H_{2} S O_{4 (a q)} + H_{2} O_{(l)} \to H_{3} O_{(a q)}^{+} + H S O_{4 (a q)}^{-}; K a_{1} > 10$

(ii) $H S O_{4 (a q)}^{-} + H_{2} O_{(l)} \to H_{3} O_{(a q)}^{+} + S O_{4 (a q)}^{2 -}; K a_{2} = 1.2 \times 10^{- 2}$

$K a_{1}$ is greater than $K a_{2}$ , i.e., tendency to move towards the products is greater in (i) than in (ii).

This is because the negatively charged $H S O_{4}^{-}$ ion has much less tendancy to donate a proton to $H_{2} O$ as compared to neutral $H_{2} S O_{4}$ .

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