Calculate the pH at the equivalence point for the titration of 0.170 M methylamine (CH3NH2) with 0.170 M HCl. The 𝐾b of methylamine is 5.0×10−4.

Answer 1

CH₃NH₂ + HCl ==> CH₃NH₃⁺Cl^-

At the equivalence point, moles CH₃NH₂ = moles HCl and ALL of the CH₃NH₂ & HCl have been converted to CH₃NH₃⁺Cl^-

Moles CH₃NH₃⁺Cl^- formed = 0.170 moles

Since molarity of HCl and CH3NH2 are the same, the volume will double at the equivalence point

Final [CH₃NH₃⁺Cl^-] = 0.170 M / 2 = 0.085 M

Looking at the hydrolysis of CH₃NH₃⁺ (Cl^- is a spectator)

CH₃NH₃⁺ + H2O ==> H₃O⁺ + CH₃NH₂

Ka for CH₃NH₃⁺ = 1x10^-14 / 5.0x10^-4 = 2x10^-11

Ka = [H₃O⁺][CH₃NH₂] / [CH₃NH₃⁺] = (x)(x) / 0.085

x² = 1.7x10^-12

x = 1.30x10^-6 = [H₃O⁺]

pH = -log [H₃O⁺] = -log 1.30x10^-6

pH = 5.88