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Question
Why is `K_(a_2) "<<" K_(a_1)` for `H_2SO_4` in water?
Solution
`H_2SO_(4(aq)) + H_2O_(l) -> H_3O_(aq)^(+) + HSO_(4(aq))^(-); K_(a1) > 10`
`HSO_(4(aq))^(-) + H_2O_(l) -> H_3O_(aq)^(+) + SO_(4(aq))^(-); K_(a_2) = 1.2 xx10^(-2)`
It can be noticed that `K_(a_2) ">>" K_(a_1)`
This is because a neutral H2SO4 has a much higher tendency to lose a proton than the negatively charged `HSO_4^(-)`. Thus, the former is a much stronger acid than the latter.
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