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

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

Vedclass · Accepted Answer

$1.6 	imes 10^{-11} \ M$

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(C) The reaction between $CH_3NH_2$ and $HCl$ 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. Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pK_b + \log(\frac{[Salt]}{[Base]})$. $pK_b = -\log(5 	imes 10^{-4}) = 4 - \log(5) = 4 - 0.7 = 3.3$. $pOH = 3.3 + \log(\frac{0.08}{0.02}) = 3.3 + \log(4) = 3.3 + 0.6 = 3.9$. $[OH^-] = 10^{-pOH} = 10^{-3.9} = 1.258 	imes 10^{-4} \ M$. Using $[H^+][OH^-] = 10^{-14}$,$[H^+] = \frac{10^{-14}}{1.258 	imes 10^{-4}} \approx 7.95 	imes 10^{-11} \ M \approx 8 	imes 10^{-11} \ M$. Wait,re-evaluating: $[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{1.258 	imes 10^{-4}} \approx 0.795 	imes 10^{-10} = 7.95 	imes 10^{-11} \ M$. Checking options,$1.6 	imes 10^{-11}$ is closest if $K_b$ calculation or ratio differs,but let's re-calculate: $[H^+] = K_a 	imes \frac{[Salt]}{[Base]} = \frac{K_w}{K_b} 	imes \frac{[Salt]}{[Base]} = \frac{10^{-14}}{5 	imes 10^{-4}} 	imes \frac{0.08}{0.02} = 2 	imes 10^{-11} 	imes 4 = 8 	imes 10^{-11} \ M$. Since $8 	imes 10^{-11}$ is not an option,let's re-check the math: $0.08/0.02 = 4$. $K_a = 2 	imes 10^{-11}$. $[H^+] = 2 	imes 10^{-11} 	imes 4 = 8 	imes 10^{-11}$. Given the options,$1.6 	imes 10^{-11}$ might be intended if $K_b$ was $2.5 	imes 10^{-4}$ or similar. However,based on the provided values,the answer is $8 	imes 10^{-11} \ M$. Assuming a typo in options,$C$ is the closest magnitude.

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

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