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 \(0.50 \mathrm{M} \mathrm{HCl}\), calculate the \(\mathrm{pH}\) under the following conditions. a. after \(50.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) has been added b. at the stoichiometric point

First, we'll calculate the initial number of moles of methylamine and HCl at each point in the titration. Initial moles of methylamine: \( (50.0 \mathrm{~mL}) \times (1.0 \mathrm{M}) = 50.0 \mathrm{~mmol} \) a. Moles of HCl after 50.0 mL has been added: \( (50.0 \mathrm{~mL}) \times (0.50 \mathrm{M}) = 25.0 \mathrm{~mmol} \) b. At the stoichiometric point, all the methylamine has reacted with the HCl. The stoichiometric ratio between methylamine and HCl is 1:1, so the moles of HCl needed to reach the stoichiometric point is equal to the initial moles of methylamine. There are 50.0 mmol of HCl at the stoichiometric point.

Methylamine is a weak base, and it reacts with HCl, a strong acid, to form a salt and water: \[ CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^- \]

We'll set up an ICE table to find the concentration of the species involved in the reaction: ### CONDITION A ### | | CH3NH2 | HCl | CH3NH3+ | Cl- | |-------|--------|-----|---------|-----| | I | 50.0 |25.0 | 0 | 0 | | C | -25.0 | -25.0| +25.0 |+25.0| | E | 25.0 | 0 | 25.0 | 25.0| ### CONDITION B ### | | CH3NH2 | HCl | CH3NH3+ | Cl- | |-------|--------|-----|---------|-----| | I | 50.0 | 50.0| 0 | 0 | | C | -50.0 | -50.0| +50.0 |+50.0| | E | 0 | 0 | 50.0 | 50.0|

a. After 50.0 mL of 0.50 M HCl has been added: There are still some unreacted methylamine and an equal moles of its conjugate acid, CH3NH3+. We have to consider the buffer solution made by CH3NH2 and CH3NH3+ and use the Henderson-Hasselbalch equation to find the pH: \( pH = pK_b + \log\frac{[CH_3NH_2]}{[CH_3NH_3^+]} \) \( pK_b = -\log(4.4 \times 10^{-4}) = 3.36 \) \( pH = 3.36 + \log \frac{25.0}{25.0} = 3.36 \) b. At the stoichiometric point: There is no CH3NH2 remaining at this point. The solution contains the CH3NH3+ ion and its subsequent hydrolysis needs to be considered: \[ CH_3NH_3^+ + H_2O \rightleftharpoons CH_3NH_2 + H_3O^+ \] To find the pH, we can use the equation: \( K_a = \frac{[H_3O^+][CH_3NH_2]}{[CH_3NH_3^+]} \) \( K_a \) can be found using the relationship: \(K_w = K_a \times K_b \) \( K_a = \frac{1 \times 10^{-14}}{4.4 \times 10^{-4}} = 2.27 \times 10^{-11}\) Let the concentration of H3O+ be x. Then, we can rewrite the K_a equation as: \( 2.27 \times 10^{-11} = \frac{(x)(x)}{50.0 - x} \) The ionization is very small compared to the initial concentration of CH3NH3+, so x is very small. Hence, we can assume that 50.0 - x ≈ 50.0. Solving for x, we get: \( x = [H_3O^+] = 3.38 \times 10^{-6} M \) Finally, the pH can be calculated as: \( pH = - \log{[H_3O^+]} = - \log{3.38 \times 10^{-6}} = 5.47 \) So, the pH under condition a is 3.36 and under condition b is 5.47.