A 0.15 text{ mol} of pyridinium chloride is a | vedclass

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(A) $1$. The solution is a basic buffer consisting of a weak base (pyridine) and its salt (pyridinium chloride).$2$. The concentration of the base $[B

Vedclass · Accepted Answer

$5.07$

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(A) $1$. The solution is a basic buffer consisting of a weak base (pyridine) and its salt (pyridinium chloride).$2$. The concentration of the base $[Base] = 0.2 	ext{ M}$.$3$. The concentration of the salt $[Salt] = \frac{0.15 	ext{ mol}}{0.5 	ext{ L}} = 0.3 	ext{ M}$.$4$. Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pK_b + \log\left(\frac{[Salt]}{[Base]}ight)$.$5$. $pOH = 8.75 + \log\left(\frac{0.3}{0.2}ight) = 8.75 + \log(1.5) = 8.75 + 0.176 = 8.926$.$6$. Since $pH + pOH = 14$, $pH = 14 - 8.926 = 5.074 \approx 5.07$.

$A$ $0.15 \text{ mol}$ of pyridinium chloride is added to $500 \text{ cm}^3$ of $0.2 \text{ M}$ pyridine solution. Assuming no change in volume upon mixing, what is the $pH$ of the resulting solution? (Given: $pK_b$ of pyridine $= 8.75$)

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