Question

$Ka _1, Ka _2$ and $Ka _3$ are the respective ionization constants for the following reactions (a), (b) and (c)
(a) $H _2 C _2 O _4 \rightleftharpoons H ^{+}+ HC _2 O _4^{-}$
(b) $HC _2 O _4^{-} \rightleftharpoons H ^{+}+ HC _2 O _4^{2-}$
(c) $H _2 C _2 O _4 \rightleftharpoons 2 H ^{+}+ C _2 O _4^{2-}$
The relationship between $K _{ a _1}, K _{ a _2}$ and $K _{ a _3}$ is given as

• A. $K _{ a _3}= K _{ a _1}+ K _{ a _2}$
• B. $K _{ a _3}= K _{ a _1}- K _{ a _2}$
• C. $K _{ a _3}= K _{ a _1} / K _{ a _2}$
• D. $K _{ a _3}= K _{ a _1} \times K _{ a _2}$

Answer 1

The Correct Option is D

To find the relationship among the ionization constants \( K_{a1} \), \( K_{a2} \), and \( K_{a3} \), we start by considering the ionization reactions in sequence and their respective equilibrium expressions.

1. First Reaction: \( H_2C_2O_4 \rightleftharpoons H^+ + HC_2O_4^- \)

For this reaction, the ionization constant \( K_{a1} \) is expressed as:

\(K_{a1} = \frac{[H^+][HC_2O_4^-]}{[H_2C_2O_4]}\)

1. Second Reaction: \( HC_2O_4^- \rightleftharpoons H^+ + C_2O_4^{2-} \)

For this reaction, the ionization constant \( K_{a2} \) is expressed as:

\(K_{a2} = \frac{[H^+][C_2O_4^{2-}]}{[HC_2O_4^-]}\)

1. Overall Reaction: \( H_2C_2O_4 \rightleftharpoons 2H^+ + C_2O_4^{2-} \)

For this reaction, the ionization constant \( K_{a3} \) is expressed as:

\(K_{a3} = \frac{[H^+]^2[C_2O_4^{2-}]}{[H_2C_2O_4]}\)

To find the relationship between \( K_{a3} \), \( K_{a1} \), and \( K_{a2} \), note that the overall reaction is the sum of the first two reactions.

Thus, we multiply the equilibrium expressions of the first two reactions:

\(K_{a1} \times K_{a2} = \left(\frac{[H^+][HC_2O_4^-]}{[H_2C_2O_4]}\right) \times \left(\frac{[H^+][C_2O_4^{2-}]}{[HC_2O_4^-]}\right)\)

Upon simplification, the intermediate species \( [HC_2O_4^-] \) cancels out:

\(K_{a1} \times K_{a2} = \frac{[H^+]^2[C_2O_4^{2-}]}{[H_2C_2O_4]}\)

This result is identical to the expression for \( K_{a3} \), confirming that:

\(K_{a3} = K_{a1} \times K_{a2}\)

Therefore, the correct answer is \( K_{a3} = K_{a1} \times K_{a2} \).