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(B) For a weak diprotic acid like $H_2CO_3$,the first dissociation step is $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$.Since $K_1 \gg K_2$,the concentr

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$[H^+] > [HCO_3^-]$

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(B) For a weak diprotic acid like $H_2CO_3$,the first dissociation step is $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$.Since $K_1 \gg K_2$,the concentration of $H^+$ ions is primarily determined by the first dissociation step.In the first step,one molecule of $H_2CO_3$ produces one $H^+$ ion and one $HCO_3^-$ ion.However,a small amount of $HCO_3^-$ further dissociates into $H^+$ and $CO_3^{2-}$ according to the second dissociation step: $HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}$.This second step produces additional $H^+$ ions,making the total $[H^+]$ slightly greater than $[HCO_3^-]$.

For an aqueous solution of $H_2CO_3$,the ionization constants are $K_1 = 4.2 \times 10^{-7}$ and $K_2 = 4.8 \times 10^{-11}$. Which statement is correct for a $0.034 \ M$ saturated solution of carbonic acid?

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