What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine,CH₃NH₂,with 15.00 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)10.04 B)10.57 C)11.09 D)11.78

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of methylammonium chloride that is hydrolyzed in solution:

Kb for CH3NH2 = 3.70×10^-4
pKb for CH3NH2 = 3.43
pKb = -log(Kb)

pKb for CH3NH3Cl can be calculated using the following equation:

pKb(CH3NH3Cl) = pKb(CH3NH2) + log([CH3NH2]/[CH3NH3Cl])

We know that [CH3NH2] = 0.10 M (from the amount of methylamine added to the solution) and [CH3NH3Cl] = 0.05 M (from the amount of methylammonium chloride added to the solution after it has completely dissociated). Therefore:

pKb(CH3NH3Cl) = 3.43 + log(0.10/0.05)
pKb(CH3NH3Cl) = 3.43 + 0.301
pKb(CH3NH3Cl) = 3.73

Finally, we can calculate the pOH of the solution:

pOH = pKb(CH3NH3Cl) + log([OH-]/[CH3NH3+])

We know that [OH-] = [CH3NH2] x Kb / [CH3NH3Cl] = (0.1 M x 3.70×10^-4) / 0.05 M = 7.4×10^-4 M

We also know that [CH3NH3+] = [CH3NH3Cl] = 0.05 M

Therefore:

pOH = 3.73 + log(7.4×10^-4 / 0.05)
pOH = 3.73 - 1.13
pOH = 2.6

Finally, we can calculate the pH:

pH = 14 - pOH
pH = 14 - 2.6
pH = 11.4

Thus, the best choice is C (11.09).

What Is the PH of a Solution Prepared by Mixing