The pH of the solution containing 50 mL each | vedclass

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(D) The mixture of a weak acid $(CH_{3}COOH)$ and its salt with a strong base $(CH_{3}COONa)$ forms an acidic buffer solution.The concentration of sal

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(D) The mixture of a weak acid $(CH_{3}COOH)$ and its salt with a strong base $(CH_{3}COONa)$ forms an acidic buffer solution.The concentration of salt $[Salt] = 0.10 \ M$.The concentration of acid $[Acid] = 0.01 \ M$.The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation:$pH = pK_{a} + \log \left( \frac{[Salt]}{[Acid]} \right)$Substituting the given values:$pH = 4.57 + \log \left( \frac{0.10}{0.01} \right)$$pH = 4.57 + \log(10)$Since $\log(10) = 1$,we get:$pH = 4.57 + 1 = 5.57$

The $pH$ of the solution containing $50 \ mL$ each of $0.10 \ M$ sodium acetate and $0.01 \ M$ acetic acid is $.......$
[Given $pK_{a}$ of $CH_{3}COOH = 4.57$] (in $.57$)

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The $pH$ of the solution containing $50 \ mL$ each of $0.10 \ M$ sodium acetate and $0.01 \ M$ acetic acid is $.......$[Given $pK_{a}$ of $CH_{3}COOH = 4.57$] (in $.57$)