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(A) Initial $pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]} = 4.745 + \log \frac{0.10}{0.10} = 4.745$. Since $pH + pOH = 14$,initial $pH = 14 - 4.745 = 9.2

Vedclass · Accepted Answer

$48$

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(A) Initial $pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]} = 4.745 + \log \frac{0.10}{0.10} = 4.745$. Since $pH + pOH = 14$,initial $pH = 14 - 4.745 = 9.255$. On adding $0.05 \ mol$ of $HCl$,the reaction is: $NH_3 + H^+ ightarrow NH_4^+$. New moles: $NH_3 = 0.10 - 0.05 = 0.05 \ mol$,$NH_4^+ = 0.10 + 0.05 = 0.15 \ mol$. New $pOH' = 4.745 + \log \frac{0.15}{0.05} = 4.745 + \log 3 = 4.745 + 0.477 = 5.222$. New $pH' = 14 - 5.222 = 8.778$. Change in $pH = |pH' - pH| = |8.778 - 9.255| = 0.477$. $0.477 = 47.7 	imes 10^{-2} \approx 48 	imes 10^{-2}$.

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