The pH of a buffer solution made by mixing 25 | vedclass

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(B) The mixture of $NH_4OH$ (a weak base) and $NH_4Cl$ (its salt with a strong acid) forms a basic buffer solution.For a basic buffer, the $pOH$ is ca

Vedclass · Accepted Answer

$8.2$

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(B) The mixture of $NH_4OH$ (a weak base) and $NH_4Cl$ (its salt with a strong acid) forms a basic buffer solution.For a basic buffer, the $pOH$ is calculated using the Henderson-Hasselbalch equation:$pOH = pK_b + \log \frac{[	ext{salt}]}{[	ext{base}]}$Given: $[	ext{salt}] = 0.2 \ M$, $[	ext{base}] = 0.02 \ M$, and $pK_b = 4.8$.$pOH = 4.8 + \log \left( \frac{0.2}{0.02} ight) = 4.8 + \log(10) = 4.8 + 1 = 5.8$.Now, calculate the $pH$ using the relation $pH + pOH = 14$ at $25^{\circ}C$:$pH = 14 - 5.8 = 8.2$.

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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