JEE Main · 2022 · Shift-IImediumIEQ-021

Ka1, Ka2 and Ka3 are the respective ionization constants for the following reactions: (a) H2C2O4 H+ + HC2O4- (b) HC2O4-…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

Ka1\mathrm{Ka_1}, Ka2\mathrm{Ka_2} and Ka3\mathrm{Ka_3} are the respective ionization constants for the following reactions:

(a) H2C2O4⇌H++HC2O4−\mathrm{H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^-} (b) HC2O4−⇌H++C2O42−\mathrm{HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-}} (c) H2C2O4⇌2H++C2O42−\mathrm{H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-}}

The relationship between Ka1\mathrm{Ka_1}, Ka2\mathrm{Ka_2} and Ka3\mathrm{Ka_3} is:

Options
  1. a

    Ka3=Ka1+Ka2K_{a3} = K_{a1} + K_{a2}

  2. b

    Ka3=Ka1Ka2K_{a3} = \frac{K_{a1}}{K_{a2}}

  3. c

    Ka3=Ka1−Ka2K_{a3} = K_{a1} - K_{a2}

  4. d

    Ka3=Ka1×Ka2K_{a3} = K_{a1} \times K_{a2}

    ✓
Correct Answerd

Ka3=Ka1×Ka2K_{a3} = K_{a1} \times K_{a2}

Detailed Solution

Step 1 — Reaction (c) = Reaction (a) + Reaction (b):

Adding reactions (a) and (b): H2C2O4⇌H++HC2O4−Ka1HC2O4−⇌H++C2O42−Ka2\mathrm{H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^-} \quad K_{a1} \mathrm{HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-}} \quad K_{a2}

Net: H2C2O4⇌2H++C2O42−\mathrm{H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-}} which is reaction (c).

Step 2 — Rule for adding reactions: When reactions are added, equilibrium constants are multiplied: Ka3=Ka1×Ka2K_{a3} = K_{a1} \times K_{a2}

Verification: Ka3=[H+]2[C2O42−][H2C2O4]=[H+][HC2O4−][H2C2O4]×[H+][C2O42−][HC2O4−]=Ka1×Ka2✓K_{a3} = \frac{[\mathrm{H^+}]^2[\mathrm{C_2O_4^{2-}}]}{[\mathrm{H_2C_2O_4}]} = \frac{[\mathrm{H^+}][\mathrm{HC_2O_4^-}]}{[\mathrm{H_2C_2O_4}]} \times \frac{[\mathrm{H^+}][\mathrm{C_2O_4^{2-}}]}{[\mathrm{HC_2O_4^-}]} = K_{a1} \times K_{a2} \checkmark

Answer: Option (4) — Ka3=Ka1×Ka2K_{a3} = K_{a1} \times K_{a2}

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