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For a dibasic acid, H(2)A hArr HA^(-) + ...

For a dibasic acid, `H_(2)A hArr HA^(-) + H^(+) (K_(1))`
`HA^(-) hArr A^(2-) + H^(+)(K_(2))`
`H_(2)A hArr 2H^(+) + A^(-2)(K)` then

A

`K = K_(1) + K_(2)`

B

`K = K_(1) - K_(2)`

C

`K = K_(1) //K_(2)`

D

`K = K_(1).K_(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`K = K_(1)K_(2)`
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