Simultaneous Acid-base Equilibria

For the stepwise dissociation of a diprotic acid H_2CO_3: First dissociation: H_2CO_3 \rightleftharpoons H^+ + HCO_3^-, K_{a1} = [H^+][HCO_3^-]/[H_2CO_3]. Second dissociation: HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}, K_{a2} = [H^+][CO_3^{2-}]/[HCO_3^-]. Rearranging the second equilibrium expression: [CO_3^{2-}] = K_{a2} \times [HCO_3^-] / [H^+]. This is the exact equilibrium expression and is always correct. Option A ([CO_3^{2-}] = K_{a2}) is dimensionally incorrect since K_{a2} has units of mol/L but this equality requires [HCO_3^-]/[H^+] = 1, which is not generally true. Option B includes an extra factor of C/[H^+] and does not correctly follow from the K_{a2} expression. Option D ([CO_3^{2-}] = K_{a1}K_{a2}/[H^+]^2) is derived by combining both equilibria: [CO_3^{2-}] = K_{a1}K_{a2}[H_2CO_3]/[H^+]^2, which reduces to K_{a1}K_{a2}C/[H^+]^2 only when [H_2CO_3] \approx C; this is an approximate form, not the exact equilibrium expression for the second step alone. The exact expression from K_{a2} is the most fundamental and reliable. This multiprotic acid equilibrium treatment is from NCERT and is characteristic of JEE Advanced multi-step reasoning questions. Plausibility check: [CO_3^{2-}] is very small compared to [HCO_3^-] because the second dissociation is much weaker (K_{a2} << K_{a1}), consistent with [CO_3^{2-}] = K_{a2} \times [HCO_3^-]/[H^+] being a small number.