A certain amount of H_2CO_3 and HCl are disso | vedclass

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(D) The overall dissociation reaction for $H_2CO_3$ is: $H_2CO_{3(aq)} \rightleftharpoons 2H^+{(aq)} + CO_3^{2-}{(aq)}$
The overall equilibrium const

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The overall dissociation reaction for $H_2CO_3$ is: $H_2CO_{3(aq)} ightleftharpoons 2H^+{(aq)} + CO_3^{2-}{(aq)}$
The overall equilibrium constant is $K_{overall} = K_{a_1} 	imes K_{a_2} = 10^{-5} 	imes 10^{-8} = 10^{-13}$.
Given $[CO_3^{2-}] = 0.01 \ M$ and $[H_2CO_3] = 0.1 \ M$.
The expression for $K_{overall}$ is $K_{overall} = \frac{[H^+]^2 [CO_3^{2-}]}{[H_2CO_3]}$.
Substituting the values: $10^{-13} = \frac{[H^+]^2 	imes 0.01}{0.1}$.
$[H^+]^2 = \frac{10^{-13} 	imes 0.1}{0.01} = 10^{-12}$.
$[H^+] = \sqrt{10^{-12}} = 10^{-6} \ M$.
Therefore,$pH = -\log[H^+] = -\log(10^{-6}) = 6$.

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