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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)` thenA. `K = K_(1) + K_(2)`B. `K = K_(1) - K_(2)`C. `K = K_(1) //K_(2)`D. `K = K_(1).K_(2)` |
Answer» Correct Answer - D `K = K_(1)K_(2)` |
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