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Question

(C) The buffer is a basic buffer consisting of a weak base $(NH_3)$ and its salt $(NH_4Cl)$.Using the Henderson-Hasselbalch equation for basic buffers

Vedclass · Accepted Answer

$8.8117$

Vedclass · Answer

(C) The buffer is a basic buffer consisting of a weak base $(NH_3)$ and its salt $(NH_4Cl)$.Using the Henderson-Hasselbalch equation for basic buffers:$pOH = pK_b + \log \frac{[Salt]}{[Base]}$Given $K_b = 1.8 	imes 10^{-5}$,so $pK_b = -\log(1.8 	imes 10^{-5}) \approx 4.7447$.Calculate the moles of salt and base:$Moles \, of \, NH_4Cl = 500 \, mL 	imes 0.5 \, M = 250 \, mmol$.$Moles \, of \, NH_3 = 300 \, mL 	imes 0.3 \, M = 90 \, mmol$.$pOH = 4.7447 + \log \left( \frac{250}{90} ight) = 4.7447 + \log(2.777) \approx 4.7447 + 0.4437 = 5.1884$.Since $pH + pOH = 14$,$pH = 14 - 5.1884 = 8.8116$.

Find the $pH$ of a buffer solution prepared by mixing $300 \, mL$ of $0.3 \, M \, NH_3$ and $500 \, mL$ of $0.5 \, M \, NH_4Cl$. Given $K_b$ for $NH_3 = 1.8 \times 10^{-5}$.

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