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

First, we need to write the balanced equation for the reaction between methylamine and hydrochloric acid: \[CH_{3}NH_{2}(aq) + HCl(aq) \rightarrow CH_{3}NH_{3}^{+}(aq) + Cl^{-}(aq)\]

We will calculate the initial moles of both methylamine and HCl: - Moles of methylamine: 1.0 M × 0.050 L = 0.050 mol - Moles of HCl (condition a): 0.050 L × 0.50 M = 0.025 mol

In this condition, there is still an excess of methylamine since only half the moles of HCl are added. First, we need to calculate the moles of methylamine after the reaction with HCl: - Moles of methylamine after the reaction: 0.050 mol - 0.025 mol = 0.025 mol Now, we can set up an equilibrium expression for the remaining methylamine: \[CH_{3}NH_{2}(aq) + H_{2}O(l) \rightleftharpoons CH_{3}NH_{3}^{+}(aq) + OH^{-}(aq)\] Let x represent the concentration of OH⁻ ions at equilibrium. Then the equilibrium equation becomes: \[\frac{x(0.025+x)}{0.025-x} = 4.4 \times 10^{-4}\] Since x is significantly smaller than 0.025, we can approximate the equation: \[\frac{x(0.025)}{0.025} = 4.4 \times 10^{-4}\] Solving for x: \[x = 4.4 \times 10^{-4}\] Now, find the pOH and pH: \[pOH = -\log{(4.4\times 10^{-4})} = 3.36\] \[pH = 14 - pOH = 14 - 3.36 = 10.64\] Thus, the pH after adding 50.0 mL of 0.50 M HCl is 10.64.

At the stoichiometric point, all of the methylamine has reacted with an equal amount of HCl. In this case, the pH is determined by the equilibrium of the methyl ammonium ion (CH₃NH₃⁺). Knowing that the total volume has doubled, we can find the concentration of methyl ammonium ion: \[C_{CH_{3}NH_{3}^{+}} = \frac{0.050 \, \text{mol}}{2 \cdot 0.050 \, \text{L}} = 0.50 \, \text{M}\] Now, we need to set up an equilibrium expression for the dissociation of the methyl ammonium ion: \[CH_{3}NH_{3}^{+}(aq) \rightleftharpoons CH_{3}NH_{2}(aq) + H^{+}(aq)\] Let x represent the concentration of H⁺ ions at equilibrium. Then the equilibrium equation becomes: \[\frac{x(0.50+x)}{0.50-x} = \frac{K_{w}}{K_{b}} = \frac{1.0 \times 10^{-14}}{4.4\times10^{-4}}\] Since x is significantly smaller than 0.50, we can approximate the equation: \[\frac{x(0.50)}{0.50} = \frac{1.0 \times 10^{-14}}{4.4\times10^{-4}}\] Solving for x: \[x = \frac{1.0 \times 10^{-14}}{4.4\times10^{-4}} = 2.27 \times 10^{-11}\] Now, find the pH: \[pH = -\log{(2.27\times 10^{-11})} = 10.64\] Thus, the pH at the stoichiometric point is 10.64.