Why is $K_{a 2}$ $≪$ $K_{a 1}$ for $H_{2} S O_{4}$

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Solution

$H_{2} S O_{4}$ ionizes in two steps:
$H_{2} S O_{4} \to H^{+} + H S O_{4}^{-}, K_{a 1}$
$H S O_{4}^{-} \to H^{+} + S O_{4}^{2 -}, K_{a 2}$

It is seen that Ka1 >> Ka2

This is because a neutral $H_{2} S O_{4}$ has a much higher tendency to lose a proton than the negatively charged $H S O_{4}^{-}$ . Thus, the former is a much stronger acid than the latter.

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Why is for H₂SO₄ in water?

K a_{2} << K a_{1}

H_{2} S O_{4}

H_{3} A ⇌ H^{+} + H_{2} A^{-}; K = K_{a_{1}}

H_{2} A^{-} ⇌ H^{+} + H A^{2 -}; K = K_{a_{2}}

H A^{2 -} ⇌ H^{+} + A^{3 -}; K = K_{a_{3}}

100 mL

0.1 M NaOH

100 mL

0.05 M H_{2} S O_{4}

p H

(for H_{2} S O_{4}, K a_{1} = \infty, K a_{2} = 10^{- 2})

K_{a_{1}}

K_{a_{2}}

H_{2} S O_{4}

10^{- 2}

10^{- 6}

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