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JEE Main202225 Jul 2022Evening ShiftChemistryIonic EquilibriumActual

Ka 1 , Ka 2 and Ka 3 are the respective ionization constants for the following reactions a , b and c . (a) H 2 C 2 O 4 ⇌ H + + HC 2 O 4 - (b) HC 2 O 4 - ⇌ H + + HC 2 O 4 2 - (c) H 2 C 2 O 4 ⇌ 2 H + + C 2 O 4 2 - The relationship between K a 1 , K a 2 and K a 3 is given as

Options

  1. AK a 3 = K a 1 + K a 2
  2. BK a 3 = K a 1 K a 2
  3. CK a 3 = K a 1 - K a 2
  4. DK a 3 = K a 1 × K a 2

Correct answer

D. K a 3 = K a 1 × K a 2

Step-by-step solution

H 2 C 2 O 4 ⇌ H + + HC 2 O 4 - K a 1 H 2 C 2 O 4 - ⇌ H + + C 2 O 4 2 - K a 2 H 2 C 2 O 4 ⇌ 2 H + + C 2 O 4 2 -       K a 3 On adding equation 1 and equation 2 we get equation 3 . So, K a 3 = K a 1 × K a 2 .

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