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(C) For a basic buffer,the $pH$ is related to $pOH$ by the equation: $pH + pOH = 14$. Given $pH = 8.26$,so $pOH = 14 - 8.26 = 5.74$. The Henderson-Has

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$107$

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(C) For a basic buffer,the $pH$ is related to $pOH$ by the equation: $pH + pOH = 14$. Given $pH = 8.26$,so $pOH = 14 - 8.26 = 5.74$. The Henderson-Hasselbalch equation for a basic buffer is: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$. Substituting the given values: $5.74 = 4.74 + \log \frac{[NH_4^+]}{0.2}$. $1 = \log \frac{[NH_4^+]}{0.2}$,which implies $\frac{[NH_4^+]}{0.2} = 10^1 = 10$. Therefore,$[NH_4^+] = 10 	imes 0.2 = 2 \, M$. Since the volume is $1 \, L$,the number of moles of $NH_4Cl$ required is $2 \, mol$. Mass of $NH_4Cl = 	ext{moles} 	imes 	ext{molar mass} = 2 \, mol 	imes 53.5 \, g \, mol^{-1} = 107 \, g$.

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