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(C) For the first ionization step of the weak dibasic acid: $H_{2}A \rightleftharpoons H^{+} + HA^{-}$.The equilibrium constant expression is $K_{a1}

Vedclass · Accepted Answer

$8.1 	imes 10^{-8} 	ext{ M}$

Vedclass · Answer

(C) For the first ionization step of the weak dibasic acid: $H_{2}A ightleftharpoons H^{+} + HA^{-}$.The equilibrium constant expression is $K_{a1} = \frac{[H^{+}][HA^{-}]}{[H_{2}A]}$.Given that $0.1 	ext{ mol}$ of $H_{2}A$ is dissolved in $1 	ext{ L}$ of solution,the initial concentration $[H_{2}A] = 0.1 	ext{ M}$.The solution also contains $0.1 	ext{ M}$ $HCl$,which is a strong acid and dissociates completely,providing $[H^{+}] = 0.1 	ext{ M}$.Since $K_{a1} = 8.1 	imes 10^{-8}$ is very small,the dissociation of $H_{2}A$ is negligible. Therefore,we can assume $[H_{2}A] \approx 0.1 	ext{ M}$ and $[H^{+}] \approx 0.1 	ext{ M}$ at equilibrium.Substituting these values into the expression: $8.1 	imes 10^{-8} = \frac{(0.1)[HA^{-}]}{0.1}$.Solving for $[HA^{-}]$,we get $[HA^{-}] = 8.1 	imes 10^{-8} 	ext{ M}$.

The first and second ionization constants of a weak dibasic acid $H_{2}A$ are $8.1 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. $0.1 \text{ mol}$ of $H_{2}A$ was dissolved in $1 \text{ L}$ of $0.1 \text{ M}$ $HCl$ solution. The concentration of $HA^{-}$ in the resultant solution is:

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