What is the pH of a buffer made from 0.580 mol of CH₃NH₂ (Kb = 4.4 × 10⁻⁴) and 0.230 mol of CH₃NH₃I?

Answer 1

Henderson Hasselbalch equation for a basic buffer

pOH = pKb + log [CH₃NH₃⁺] /[CH₃NH₂)

pKb = - log Kb = -log 4.4x10^-4 = 3.36

pOH = 3.36 + log [0.230] / [0.580] = 3.36 + log 0.397

pOH = 3.36 -0.40

pOH = 2.96

pH = 14 - pOH

pH = 11.04