In the titration of \(50.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) \(\left(K_{\mathrm{b}}=4.4 \times 10^{-4}\right),\) with \(0.50 M\) HCl, calculate the pH under the following conditions. a. after \(50.0 \mathrm{mL}\) of \(0.50 \mathrm{M}\) HCl has been added b. at the stoichiometric point

First, let's calculate the moles of methylamine and HCl in the solution: Moles of methylamine = volume × concentration = 50.0 mL × 1.0 M = 50.0 mmol Moles of HCl = volume × concentration = 50.0 mL × 0.50 M = 25.0 mmol Now, we need to determine the amount of CH3NH2 that has reacted with HCl and the amount of unreacted CH3NH2 and CH3NH3+ formed as a result of partial neutralization: Moles CH3NH2 reacted = 25.0 mmol Moles unreacted CH3NH2 = 50.0 mmol - 25.0 mmol = 25.0 mmol Moles of CH3NH3+ formed = moles of HCl added = 25.0 mmol Next, we determine the concentration of CH3NH2 and CH3NH3+ in the reaction mixture after adding 50.0 mL of HCl: Total volume = volume of methylamine solution + volume of HCl solution = 50.0 mL + 50.0 mL = 100 mL \[ CH\textsubscript{3}NH\textsubscript{2\, } = \frac{25\ mmol}{100\ mL} =0.25\ M \] \[ CH\textsubscript{3}NH\textsubscript{3\, }^\textsubscript{+} = \frac{25\ mmol}{100\ mL} =0.25\ M \] Now, using the Kb expression to find the hydroxide ion concentration OH-: Kb = 4.4 x 10^-4 = \[\frac{[OH^{-}][CH_{3}NH_{3}^{+}]}{[CH_{3}NH_{2}]}\] \[ [OH^{-}] = \frac{Kb [CH_{3}NH_{2}]}{[CH_{3}NH_{3}^{+}]} = \frac{4.4 \times 10^{-4} (0.25)}{0.25} = 4.4 \times 10^{-4}\] Now, we can calculate the pOH: \[pOH = -log_{10} [OH^{-}] = -log_{10}(4.4 \times 10^{-4}) \] \[pOH \approx 3.36\] Now, the pH can be calculated by subtracting the pOH from 14: \[ pH = 14 - pOH = 14 - 3.36 \] \[ pH \approx 10.64 \]

At the stoichiometric point, equal moles of acid and base have reacted. Thus, Moles of HCl = Moles of CH3NH2 = 50.0 mmol To calculate the volume of HCl solution needed to react with all of the methylamine: Volume of HCl solution = \(\frac{moles\ of\ HCl}{concentration\ of\ HCl}\) = \(\frac{50.0\ mmol}{ 0.50\ M}\) = 100 mL At the stoichiometric point, the amount of CH3NH3+ formed is equal to the initial amount of CH3NH2: CH3NH3+ formed = moles of CH3NH2 = 50.0 mmol Now, we calculate the total volume at the stoichiometric point: Total volume = Volume of methylamine solution + Volume of HCl solution = 50.0 mL + 100 mL = 150 mL Thus, the concentration of CH3NH3+ is: \[CH\textsubscript{3}NH\textsubscript{3\, }^\textsubscript{+} = \frac{50.0\ mmol}{150\ mL} =0.333\ M \] We know that, Ka = Kw/Kb From that, we can calculate Ka for CH3NH3+: Ka = \(\frac{1.0 \times 10^{-14}}{4.4 \times 10^{-4}}\) ≈ 2.27 x 10^-11 Then, we can use the Ka expression to find the H+ concentration: \[ Ka = \frac{[H^{+}]}{[CH_{3}NH_{3}^{+}]} \] \[ [H^{+}] = Ka [CH_{3}NH_{3}^{+}] = 2.27 \times 10^{-11} \times 0.333 \] \[ [H^{+}] \approx 7.56 \times 10^{-11} \] Now, calculate the pH: \[ pH = -log_{10} [H^{+}] = -log_{10}(7.56 \times 10^{-11}) \] \[ pH \approx 10.12 \] So, the pH at the stoichiometric point is approximately 10.12.