A litre of buffer solution contains 0.1 mol | vedclass

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(B) The initial moles are $n_{NH_3} = 0.1 \ mol$ and $n_{NH_4^+} = 0.1 \ mol$.When $0.02 \ mol$ of $HCl$ is added,it reacts with $NH_3$ as follows:$NH

Vedclass · Accepted Answer

$9079$

Vedclass · Answer

(B) The initial moles are $n_{NH_3} = 0.1 \ mol$ and $n_{NH_4^+} = 0.1 \ mol$.When $0.02 \ mol$ of $HCl$ is added,it reacts with $NH_3$ as follows:$NH_3 + HCl ightarrow NH_4^+ + Cl^-$After the reaction:$n_{NH_3} = 0.1 - 0.02 = 0.08 \ mol$$n_{NH_4^+} = 0.1 + 0.02 = 0.12 \ mol$Using the Henderson-Hasselbalch equation for a basic buffer:$pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]}$$pOH = 4.745 + \log \frac{0.12}{0.08}$$pOH = 4.745 + \log(1.5)$$pOH = 4.745 + (\log 3 - \log 2)$$pOH = 4.745 + (0.477 - 0.301) = 4.745 + 0.176 = 4.921$Since $pH + pOH = 14$ at $298 \ K$:$pH = 14 - 4.921 = 9.079$Expressing as $9079 	imes 10^{-3}$,the value is $9079$.

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