For a diprotic organic acid in aqueous soluti | vedclass

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Question

(D) For a diprotic acid $H_2A$,the first dissociation is $H_2A \rightleftharpoons H^+ + HA^-$ and the second is $HA^- \rightleftharpoons H^+ + A^{2-}$

Vedclass · Accepted Answer

$[H^+]$ and $[HCO_3^-]$ are approximately equal.

Vedclass · Answer

(D) For a diprotic acid $H_2A$,the first dissociation is $H_2A ightleftharpoons H^+ + HA^-$ and the second is $HA^- ightleftharpoons H^+ + A^{2-}$.Since $K_1 \gg K_2$,the concentration of $H^+$ is primarily determined by the first dissociation.For a weak acid,$[H^+] \approx [HA^-] \approx \sqrt{K_1 	imes C} = \sqrt{4.2 	imes 10^{-7} 	imes 0.034} \approx 1.19 	imes 10^{-4} \ M$.The concentration of the second dissociation product $[A^{2-}]$ is approximately equal to $K_2$,which is $4.8 	imes 10^{-11} \ M$.Comparing the species,$[H^+] \approx [HCO_3^-]$ (or $[HA^-]$) because the contribution of $H^+$ from the second step is negligible.

For a diprotic organic acid in aqueous solution,the ionization constants are $K_1 = 4.2 \times 10^{-7}$ and $K_2 = 4.8 \times 10^{-11}$. For a $0.034 \ M$ solution of this acid,select the correct statement.

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