The ionization constant of $H_{2} C O_{3}$ as an acid in aqueous solution at room temperature is X. If the first and second ionization constants of $H_{2} C O_{3}$ are $X_{1}$ and $X_{2}$ respectively then

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$X = \frac{X_{1}}{X_{2}}$

$X = \frac{X_{2}}{X_{1}}$

$X = X_{1} X_{2}$

$X = \frac{X_{1} X_{2}}{2}$

answer is C.

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Detailed Solution

onization of $H_{2} C O_{3}$ results in the following equation,

$H_{2} C O_{3} + H_{2} O \to H C O^{- 3} + H_{3} O +, K a_{1} = X_{1}$

$H C O^{- 3} + H_{2} O \to C O_{3}^{2 -} + H_{3} O +, K a_{2} = X_{2}$

In the case of $H_{2} C O_{3}, X = \frac{[C O_{3}^{2 -}] [H^{3} O +]}{[H^{2} C O^{3}] [H 2^{O}]}$

$X = X_{1} \times X_{2}$

It is evident from the for mentioned equations that $X = X_{1} \times X_{2}$

Therefore, the correct answer is option (C) $X = X_{1} \times X_{2}$ .

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