A 15.4 mL aliquot of methylamine that has a concentration of 0.463 M will be titrated with 1.58 M HCl. Calculate the pH at the equivalence point.

Question

The Kb at the temperature of the experiment is equal to 4.27 x 10^-4. kw=1.00 x 10^-14

Answer 1

At the equivalence point, all of the methylamine will be converted to the conjugate acid.

Let methylamine be represented as B (a base)

B + HCl ==> BH+ + Cl^- (because we have a weak base + a strong acid, pH @ equivalence must be <7)

mols B = 15.4 ml x 1 L / 1000 ml x 0.463 mol/L = 0.00713 mols B

mols BH⁺ at equivalence = 0.00713 mols BH⁺

Volume of 1.58 M HCl needed = 0.00713 mols x 1 L / 1.58 mols = 0.00451 L = 4.51 mls

Total volume after addition of HCl = 15.4 ml + 4.51 ml = 19.9 mls = 0.0199 L

Fina [BH⁺] = 0.00713 mol / 0.0199 L = 0.358 M

Hydrolysis of BH⁺:

BH⁺ + H₂O ==> B + H₃O⁺

Since BH⁺ is acting as an acid, we need the Ka which can be obtained from the Kb and Kw

Ka = Kw/Kb = 1x10^-14 / 4.27x10^-4

Ka = 2.34x10^-11

Ka = [B][H3O⁺] / [BH⁺]

2.34x10^-11 = (x)(x) / 0.358

x² = 8.38x10^-12

x = 2.89x10^-6 = [H3O⁺]

pH = -log 2.89x10^-6

pH = 5.54 (note that as predicted, this pH is <7)