The pH of an aqueous solution containing 1 M | vedclass

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(A) The buffer solution consists of benzoic acid (weak acid) and sodium benzoate (its conjugate base).Using the Henderson-Hasselbalch equation: $pH =

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$100$

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(A) The buffer solution consists of benzoic acid (weak acid) and sodium benzoate (its conjugate base).Using the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[	ext{salt}]}{[	ext{acid}]}$.Given $pH = 4.5$,$pK_{a} = 4.2$,$[	ext{salt}] = [	ext{sodium benzoate}]$,and $[	ext{acid}] = [	ext{benzoic acid}]$.$4.5 = 4.2 + \log \frac{[	ext{salt}]}{[	ext{acid}]}$$0.3 = \log \frac{[	ext{salt}]}{[	ext{acid}]}$Since $10^{0.3} \approx 2$,we have $\frac{[	ext{salt}]}{[	ext{acid}]} = 2$.Let $V_{a}$ be the volume of benzoic acid and $V_{s}$ be the volume of sodium benzoate.Since the molarities are equal $(1 \ M)$,the ratio of moles is equal to the ratio of volumes: $\frac{V_{s}}{V_{a}} = 2$,so $V_{s} = 2V_{a}$.Given the total volume $V_{s} + V_{a} = 300 \ mL$.Substituting $V_{s} = 2V_{a}$ into the equation: $2V_{a} + V_{a} = 300 \ mL$.$3V_{a} = 300 \ mL \implies V_{a} = 100 \ mL$.

The $pH$ of an aqueous solution containing $1 \ M$ benzoic acid $(pK_{a}=4.20)$ and $1 \ M$ sodium benzoate is $4.5$. The volume of benzoic acid solution in $300 \ mL$ of this buffer solution is . . . . . . $mL$.

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