Titrations of Weak Acids

Question

Calculate the pH at the equivalence point for the titration of 0.210 M0.210 M methylamine (CH3NH2)(CH3NH2) with 0.210 M HCl.0.210 M HCl. The 𝐾b of methylamine is 5.0×10−4.5.0×10−4.

pH=?

Answer 1

Methylamine is a weak base. HCl is a strong acid. pH @ equivalence should have pH <7.

CH3NH2(aq) + HCl(aq) ==> CH3NH3⁺Cl^-(aq)

At equivalence, moles of acid = moles of base

moles CH3NH2 = 0.210 moles

moles HCl = 0.210 moles

moles CH3NH3+Cl- = 0.210 moles

Since the concentration of HCl and CH3NH2 are the same (0.210 M), the volume of HCl needed to reach equivalence will equal the volume of CH3NH2, and so the final volume at equivalence will be twice the original volume.

Final [CH3NH3⁺Cl^-] = 0.210 mol / 2 = 0.105 M

Looking at the hydrolysis of CH3NH3⁺], we have...

CH3NH3⁺(aq) + H2O(l) ==> H3O⁺(aq) + CH3NH2(aq)

Ka for CH3NH3⁺ = Kw/Kb = 1x10^-14 / 5x10^-4 = 2x10^-11

2x10^-11 = [H3O⁺][CH3NH2] / [CH3NH3⁺] = (x)(x) / 0.105 - x (assume x is small relative to 0.105)

x² = 2.1x10^-12

x = [H3O⁺] = 1.5x10^-6 (this is small compared to 0.105 so above assumption was vallid)

pH = -log [H3O⁺] = -log 1.5x10^-6

pH = 5.84 (acidic, as predicted)