In a solution of $N a H C O_{3}$ , the amphiprotic anion undergoes ionization to form $H^{+}$ ion and hydrolysis to form $O H^{-} i o n$ .
$H C O_{3}^{-} + H_{2} O i o n i z a t i o n ⇋ C O_{3}^{2 -} + H_{2} O$
$H C O_{3}^{-} + H_{2} O i o n i z a t i o n ⇋ H_{2} C O_{3} + O H^{-}$
To calculate pH of the above solutions, the suitable approximation is

[C O_{3}^{2 -}] = [H_{2} C O_{3}]

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degree of ionization = degree of hydrolysis

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Both (A) and (B)

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Neither (A) nor (B)

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Solution

The correct option is B Both (A) and (B)
When the following two assumptions are taken,

(1) $[C O_{3}^{2 -}] = [H_{2} C O_{3}]$

(2) degree of ionization = degree of hydrolysis

The pH of the solution can be calculated using the expression:

$p H = \frac{p K_{a_{1}} + p K_{a_{2}}}{2}$

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Similar questions

N a H C O_{3}

H^{+}

O H^{-}

H C O_{3}^{-} + H_{2} O i o n i z a t i o n ⇋ C O_{3}^{2 -} + H_{3} O^{+}

H C O_{3}^{-} + H_{2} O h y d r o l y s i s ⇋ H_{2} C O_{3}^{-} + O H^{-}

An aqueous solution of carbonic acid ( $H_{2} C O_{3}$ ) contains

1

M

N a H C O_{3}

H_{2} C O_{3} + H_{2} O ⇌ H C O_{3}^{-} + H_{3} O^{+};

p K_{1} = 6.38

H C O_{3}^{-} + H_{2} O ⇌ C O_{3}^{2 -} + H_{3} O^{+};

p K_{2} = 10.26

C O_{2}

C O_{2}

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