Consider the titration of 25 mL

Question

Consider the titration of 25 mL of 0.100M benzylamine with 0.0500 M HCl. What is the pH of the solution when 0.00 mL, 35.00 mL, 50.00 and 60.00 mL of acid has been added? Sketch the titration curve.(K_b=2.2x10^-5)

Answer 1

Benzylamine is a weak base which can be represented as B.

The reaction with HCl forms the conjugate acid

B + H⁺ ==> BH⁺ This creates a buffer of a weak base and the conjugate acid.

pOH = pKb + log [BH⁺]/[B]

pKb = -log Kb = -log 2.2x10^-5 = 4.66

pH before any HCl is added is determined from the Kb of the hydrolysis

B + H2O ==> BH⁺ + OH^-

Kb = [BH⁺][OH^-] / [B]

2.2x10^-5 = (x)(x) / 0.1

x = 1.48x10^-3 M = [OH-]

pOH = -log 1.48x10^-3 = 2.83

pH = 14 - pOH = 11.2

After addition of 35 ml 0.0500 M HCl:

0.035 L x 0.0500 mol/L = 0.00175 mols HCl added. This reacts with B to form BH⁺

mols B = 0.025 L x 0.100 mol/L = 0.0025 mol - 0.00175 = 0.00075 mol B

mols BH⁺ = 0.00175

pOH = pKb = log [BH⁺]/[B]

pOH = 4.66 + log (0.00175/0.00075) = 4.66 + 0.37

pOH = 5.03

pH = 8.97

After addition of 50 mls 0.0500 M HCl:

0.05 L x 0.0500 mol/L = 0.0025 mols HCl added. This reacts with B to form BH⁺

mols B = 0.0025 - 0.0025 = 0

mols BH⁺ = 0.0025

This is no longer a buffer as all of the B (benzylamine) is converted to the conjugate acid, BH⁺

BH⁺ + H₂O ==> B + H₃O⁺

Ka = [B][H⁺]/[B⁺] and Ka = 1x10^-14/2.2x10^-5 = 4.55x10^-10

4.55x10^-10 = (x)(x) / 0.033 (the 0.033 is from 0.0025 mol / 0.075 L)

x = 3.87x10^-6 M = H⁺

pH = -log H⁺

pH = 5.41

For 60 mls HCl, you can estimate pH as the excess HCl (10 ml) so that would be 0.001 L x 0.05 mol/L = 5x10^-5 moles H⁺ in a total volume of 85 ml

[H⁺] = 5x10^-5 mol / 0.085 L = 5.88x10^-4 M

pH = -log H⁺

pH = 3.23

Can't really draw a graph on this platform but you can look up a graph of a weak base titrated with a strong acid.