What is the pH of a buffer solution formed by | vedclass

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Question

(C) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log \left( \frac{[	ext{Salt}]}{[	ext{Acid}]}

Vedclass · Accepted Answer

$5.44$

Vedclass · Answer

(C) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log \left( \frac{[	ext{Salt}]}{[	ext{Acid}]} ight)$ Given: $pK_a = 4.7447$ $[	ext{Salt}] = 0.05 \ M$ $[	ext{Acid}] = 0.01 \ M$ Substituting the values: $pH = 4.7447 + \log \left( \frac{0.05}{0.01} ight)$ $pH = 4.7447 + \log(5)$ Since $\log(5) \approx 0.6990$: $pH = 4.7447 + 0.6990 = 5.4437$ Rounding to two decimal places,the $pH$ is $5.44$.

What is the $pH$ of a buffer solution formed by mixing $0.01 \ M$ acetic acid and $0.05 \ M$ sodium acetate $(pK_a = 4.7447)$?

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