What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M methylamine,CH₃NH₂,with 25.00 mL of 0.10 M methylammonium chloride,CH₃NH₃Cl? Assume that the volume of the solutions are additive and that K_b = 3.70 × 10^-4 for methylamine. A)10.27 B)10.57 C)10.87 D)11.78

Question

Quizplus · Accepted Answer

First, we need to find the concentration of the methylammonium ion in the solution:

methylammonium ion + water ⇌ methylamine + hydronium ion

Kb = Kw/Ka = 1.0 × 10^-14/3.70 × 10^-4 = 2.70 × 10^-11

CH3NH3Cl → CH3NH3+ + Cl-

Initial concentration of CH3NH3+ = 0.10 M × (25.00/50.00) = 0.05 M

Let x be the concentration of hydronium ion formed. Then, the concentration of methylamine is 0.10 M × (25.00/50.00) = 0.05 M, and the concentration of the methylammonium ion is 0.05 M + x.

Using the Kb expression:

Kb = [methylamine][hydronium ion]/[methylammonium ion]

2.70 × 10^-11 = (0.05)(x)/(0.05 + x)

Solving for x, we get:

x = 3.43 × 10^-6 M

Therefore, pH = -log(3.43 × 10^-6) = 10.57

The answer is choice B.

What Is the PH of a Solution Prepared by Mixing