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