100 mL of 0.1 M HA (weak acid) and 100 mL | vedclass

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Question

(C) The mixture of a weak acid $(HA)$ and its salt with a strong base $(NaA)$ forms an acidic buffer solution.Using the Henderson-Hasselbalch equation

Vedclass · Accepted Answer

$5.3$

Vedclass · Answer

(C) The mixture of a weak acid $(HA)$ and its salt with a strong base $(NaA)$ forms an acidic buffer solution.Using the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$Given: $K_{a} = 10^{-5}$,so $pK_{a} = -\log(10^{-5}) = 5$.Since the volumes are equal,the ratio of concentrations is equal to the ratio of moles: $\frac{[Salt]}{[Acid]} = \frac{0.2 \ M}{0.1 \ M} = 2$.Substituting the values: $pH = 5 + \log(2)$.Given $\log 2 = 0.3$,so $pH = 5 + 0.3 = 5.3$.

$100 \ mL$ of $0.1 \ M$ $HA$ (weak acid) and $100 \ mL$ of $0.2 \ M$ $NaA$ are mixed. What is the $pH$ of the resultant solution? ($K_{a}$ of $HA$ is $10^{-5}$; $\log 2 = 0.3$)

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