What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine,CH₃NH₂,with 20.00 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)10.17 B)10.57 C)10.97 D)11.78

Question

Quizplus · Accepted Answer

Step 1: Calculate the moles of methylamine and methylammonium chloride
moles of methylamine = (50.00 mL) (0.10 mol/L) = 0.00500 mol
moles of methylammonium chloride = (20.00 mL) (0.10 mol/L) = 0.00200 mol

Step 2: Calculate the total volume of the solution
total volume = 50.00 mL + 20.00 mL = 70.00 mL = 0.07000 L

Step 3: Calculate the concentration of methylamine and methylammonium chloride in the final solution
[methylamine] = (0.00500 mol) / (0.07000 L) = 0.0714 M
[methylammonium chloride] = (0.00200 mol) / (0.07000 L) = 0.0286 M

Step 4: Calculate the dissociation constant for methylamine
Kb = Kw / Ka = 1.0 × 10^-14 / 3.70 × 10^-4 = 2.70 × 10^-11

Step 5: Calculate the concentration of hydroxide ions
Kb = [CH3NH2][OH-] / [CH3NH3+] = (x)(x) / (0.0286 - x) = 2.70 × 10^-11
Solving for x gives x = 3.17 × 10^-6 M

Step 6: Calculate the pH of the solution
pH = 14 - pOH = 14 - (-log[OH-]) = 10.97

What Is the PH of a Solution Prepared by Mixing