0.1 mol of CH_3NH_2 (K_b = 5 times 10^{-4}) | vedclass

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(B) The reaction is: $CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^-$Initial moles: $CH_3NH_2 = 0.1 \ mol$,$HCl = 0.08 \ mol$.After reaction: $CH_3NH_2

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$8 	imes 10^{-11} \ M$

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(B) The reaction is: $CH_3NH_2 + HCl ightarrow CH_3NH_3^+ + Cl^-$Initial moles: $CH_3NH_2 = 0.1 \ mol$,$HCl = 0.08 \ mol$.After reaction: $CH_3NH_2 = 0.02 \ mol$,$CH_3NH_3^+ = 0.08 \ mol$.This forms a basic buffer solution.Using the Henderson-Hasselbalch equation for a basic buffer:$pOH = pK_b + \log \frac{[salt]}{[base]}$$pK_b = -\log(5 	imes 10^{-4}) = 4 - \log(5) = 4 - 0.699 = 3.301$.$pOH = 3.301 + \log \frac{0.08}{0.02} = 3.301 + \log(4) = 3.301 + 0.602 = 3.903$.$pH = 14 - pOH = 14 - 3.903 = 10.097$.$[H^{+}] = 10^{-pH} = 10^{-10.097} \approx 8 	imes 10^{-11} \ M$.

$0.1 \ mol$ of $CH_3NH_2$ $(K_b = 5 \times 10^{-4})$ is mixed with $0.08 \ mol$ of $HCl$ and diluted to $1 \ L$. What will be the $[H^{+}]$ concentration in the solution?

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