What is the pH of a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 methylamine, CH₃NH₂, with 20.00 mL of 0.10 mol L^-1 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

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Quizplus · Accepted Answer

The solution is a buffer system consisting of a weak base (methylamine) and its conjugate acid (methylammonium chloride). Using the Henderson-Hasselbalch equation for a base, pH = pKa + log([base]/[acid]). First, calculate pKa from pKb: pKa = 14 - pKb = 14 - (-log(3.70 × 10^-4)) = 10.57. The concentrations after mixing are [base] = (0.10 mol/L * 50.00 mL) / 70.00 mL = 0.0714 mol/L and [acid] = (0.10 mol/L * 20.00 mL) / 70.00 mL = 0.0286 mol/L. Plug these into the equation: pH = 10.57 + log(0.0714/0.0286) = 10.97.

What Is the PH of a Solution Prepared by Mixing