What is the pH of the solution obtained by mi | vedclass

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(C) The molar mass of methyl ammonium nitrate $(CH_3NH_3NO_3)$ is $14 + 3(1) + 14 + 3(1) + 14 + 3(16) = 108 \ g/mol$. (Correction: $12+3+14+3+14+48 =

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$10.3$

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(C) The molar mass of methyl ammonium nitrate $(CH_3NH_3NO_3)$ is $14 + 3(1) + 14 + 3(1) + 14 + 3(16) = 108 \ g/mol$. (Correction: $12+3+14+3+14+48 = 94 \ g/mol$ is correct as per input).Millimoles of salt $= \frac{5.076 \ g}{94 \ g/mol} 	imes 1000 = 54 \ mmol$.Millimoles of weak base $= 120 \ mL 	imes 0.225 \ M = 27 \ mmol$.Using the Henderson-Hasselbalch equation for a basic buffer:$pOH = pK_b + \log \frac{[Salt]}{[Base]}$$pK_b = -\log(4 	imes 10^{-4}) = 4 - \log 4 = 4 - 0.602 = 3.398 \approx 3.4$.$pOH = 3.4 + \log \left( \frac{54}{27} ight) = 3.4 + \log(2) = 3.4 + 0.3 = 3.7$.$pH = 14 - pOH = 14 - 3.7 = 10.3$.

What is the $pH$ of the solution obtained by mixing $5.076 \ g$ of methyl ammonium nitrate $(CH_3NH_3NO_3)$ into $120 \ mL$ of $0.225 \ M$ methylamine $(CH_3NH_2$; $K_b = 4 \times 10^{-4})$?

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