K_(a₁), K_(a₂) and K_(a₃) are the respective ionization constants for the following reactions (a), (b) and (c).
(a) H₂C₂O₄ ⇌ H + + HC₂O₄ -
(b)HC₂O₄ - ⇌ H + + C₂O₄²⁻
(c)H₂C₂O₄ ⇌ 2H + + C₂O₄²⁻
The relationship betweenK_(a₁),K_(a₂)andK_(a₃) is given as :

Solution

Chemistry ma’am · 3 steps

Reaction (c) is just (a) plus (b), and when you add reactions you multiply their K values.

1. Add the steps. Add (a) and (b). The HC₂O₄⁻ formed in (a) is used up in (b), so it cancels. What is left is H₂C₂O₄ ⇌ 2H⁺ + C₂O₄²⁻, which is exactly reaction (c).
2. Write each K. K_(a₁) = [H⁺][HC₂O₄⁻]/[H₂C₂O₄] and K_(a₂) = [H⁺][C₂O₄²⁻]/[HC₂O₄⁻]. Multiply them and [HC₂O₄⁻] cancels between the top and the bottom.

   K_(a₁) × K_(a₂) = [H⁺]²[C₂O₄²⁻]/[H₂C₂O₄] = K_(a₃)

3. So. The product of the two step constants is the overall constant: K_(a₃) = K_(a₁) × K_(a₂), which is option (d).

The idea to keep. Adding reactions means multiplying their K values. Reversing a reaction gives 1/K, and multiplying a reaction by n gives Kⁿ. In log form, adding reactions adds the pK values.

Answer from the NTA final answer key. Solution by PrepWiser.