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(C) buffer solution with $pH = 9$ is basic in nature. For a basic buffer,$pOH = 14 - pH = 14 - 9 = 5$. The $pKb$ of $NH_4OH$ is approximately $4.75$.

Vedclass · Accepted Answer

$NH_4Cl$ and $NH_4OH$

Vedclass · Answer

(C) buffer solution with $pH = 9$ is basic in nature. For a basic buffer,$pOH = 14 - pH = 14 - 9 = 5$. The $pKb$ of $NH_4OH$ is approximately $4.75$. Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pKb + \log \frac{[Salt]}{[Base]}$. Since $NH_4Cl$ (salt) and $NH_4OH$ (weak base) form a basic buffer with a $pOH$ close to $5$,this mixture is the correct choice. $CH_3COONa$ and $CH_3COOH$ form an acidic buffer $(pH $NaCl$ and $NaOH$ do not form a buffer. $KH_2PO_4$ and $K_2HPO_4$ form a buffer with $pH$ near $7.2$.

Which of the following mixtures will produce a buffer solution with $pH = 9$?

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