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(A) For an acidic buffer,the Henderson-Hasselbalch equation is given by:$pH = pK_{a} + \log_{10} \frac{[	ext{Salt}]}{[	ext{Acid}]}$Since the solutio

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$4.75$

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(A) For an acidic buffer,the Henderson-Hasselbalch equation is given by:$pH = pK_{a} + \log_{10} \frac{[	ext{Salt}]}{[	ext{Acid}]}$Since the solution is prepared by mixing equimolar amounts,$[	ext{Salt}] = [	ext{Acid}]$,therefore $\frac{[	ext{Salt}]}{[	ext{Acid}]} = 1$.$pH = pK_{a} + \log_{10}(1) = pK_{a} + 0 = pK_{a}$Given $K_{a} = 1.78 	imes 10^{-5}$,we calculate $pK_{a}$:$pK_{a} = -\log_{10}(1.78 	imes 10^{-5})$$pK_{a} = -(\log_{10} 1.78 + \log_{10} 10^{-5})$$pK_{a} = -(0.25 - 5) = 4.75$Thus,the $pH$ of the buffer solution is $4.75$.

$A$ buffer solution is prepared by mixing equimolar acetic acid and sodium acetate. If $K_{a}$ of acetic acid is $1.78 \times 10^{-5}$,find the $pH$ of the buffer solution.

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