A weak acid of dissociation constant 10^{-5} | vedclass

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(B) For a weak acid $HA$ titrated with $NaOH$,the $pH$ is given by the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$.Give

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$5 - \log 2$

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(B) For a weak acid $HA$ titrated with $NaOH$,the $pH$ is given by the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$.Given $K_{a} = 10^{-5}$,so $pK_{a} = -\log(10^{-5}) = 5$.At one-third neutralization,the amount of salt formed is $1/3$ and the remaining acid is $1 - 1/3 = 2/3$.Substituting these values into the equation: $pH = 5 + \log \frac{1/3}{2/3} = 5 + \log(1/2) = 5 - \log 2$.

$A$ weak acid of dissociation constant $10^{-5}$ is being titrated with aqueous $NaOH$ solution. The $pH$ at the point of one-third neutralization of the acid will be:-

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