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

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(A) For an acidic buffer solution,the Henderson-Hasselbalch equation is given by:$pH = pK_{a} + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$Given:$pK_{a} = 4.2$$[\text{Salt}] = 0.054 \ M$$[\text{Acid}] = 0.027 \ M$Substituting the values:$pH = 4.2 + \log_{10} \left( \frac{0.054}{0.027} \right)$$pH = 4.2 + \log_{10} (2)$Since $\log_{10} 2 \approx 0.3010$,$pH = 4.2 + 0.3010 = 4.5010$Thus,the $pH$ is approximately $4.5$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Buffer solution

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Calculate the $pH$ of a buffer solution containing $0.027 \ M$ weak acid and $0.054 \ M$ of its salt with a strong base,if the $pK_{a}$ is $4.2$.

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A
$4.5$
B
$3.2$
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$5.6$
D
$6.4$

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Calculate the $pH$ of a buffer solution containing $0.027 \ M$ weak acid and $0.054 \ M$ of its salt with a strong base,if the $pK_{a}$ is $4.2$.

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The $pH$ of a buffer solution made by mixing $25 \ mL$ of $0.02 \ M$ $NH_4OH$ and $25 \ mL$ of $0.2 \ M$ $NH_4Cl$ at $25^{\circ}C$ is ($pK_b$ of $NH_4OH = 4.8$).

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