What is the value of [H^{+}] when 0.1 mol of | vedclass

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(B) The reaction between the weak base $CH_3NH_2$ and the strong acid $HCl$ is: $CH_3NH_2 + HCl \rightarrow CH_3NH_3^{+} + Cl^{-}$. Initial moles: $CH

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$8 	imes 10^{-11}$

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(B) The reaction between the weak base $CH_3NH_2$ and the strong acid $HCl$ is: $CH_3NH_2 + HCl ightarrow CH_3NH_3^{+} + Cl^{-}$. Initial moles: $CH_3NH_2 = 0.1$,$HCl = 0.08$. After reaction: $CH_3NH_2 = 0.1 - 0.08 = 0.02 \ mol$,$CH_3NH_3^{+} = 0.08 \ mol$. This forms a basic buffer solution. The $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]} = -\log(5 	imes 10^{-4}) + \log \frac{0.08}{0.02} = (4 - \log 5) + \log 4 = 4 - 0.7 + 0.6 = 3.9$. $pH = 14 - pOH = 14 - 3.9 = 10.1$. Alternatively,using $[OH^{-}] = K_b 	imes \frac{[Base]}{[Salt]} = 5 	imes 10^{-4} 	imes \frac{0.02}{0.08} = 5 	imes 10^{-4} 	imes 0.25 = 1.25 	imes 10^{-4}$. $[H^{+}] = \frac{K_w}{[OH^{-}]} = \frac{10^{-14}}{1.25 	imes 10^{-4}} = 0.8 	imes 10^{-10} = 8 	imes 10^{-11}$.

What is the value of $[H^{+}]$ when $0.1 \ mol$ of $CH_3NH_2$ is mixed with $0.08 \ mol$ of $HCl$? Given $K_b = 5 \times 10^{-4}$.

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