Question

$\mathrm{Ka}_{1}, \mathrm{Ka}_{2}$ and $\mathrm{Ka}_{3}$ are the respective ionization constants for the following reactions (a),(b), and (c).
(a) $\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HC}_{2} \mathrm{O}_{4}^{-}$
(b) $\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HC}_{2} \mathrm{O}_{4}^{2-}$
(c) $\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$

The relationship between $\mathrm{K}_{\mathrm{a}_{1}}, \mathrm{~K}_{\mathrm{a}_{2}}$ and $\mathrm{K}_{\mathrm{a}_{3}}$ is given as

Accepted Answer

(D) $\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}} \times \mathrm{K}_{\mathrm{a}_{2}}$

Answer

(A) $\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}}+\mathrm{K}_{\mathrm{a}_{2}}$

Answer

(B) $\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}}-\mathrm{K}_{\mathrm{a}_{2}}$

Answer

(C) $\mathrm{K}_{\mathrm{a}_{3}}=\mathrm{K}_{\mathrm{a}_{1}} / \mathrm{K}_{\mathrm{a}_{2}}$

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Equilibrium

Solution

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