Question

Calculate the pH at the equivalence point when 12.35 mL of a 0.100 M Codeine solution (Kb = 1.6X10^-6) is titrated with 0.100 M hydrochloric acid.

Answer 1

Codeine is weak base.

pKb = - log Kb = - log (1.6 * 10^-6) = 5.796

one mole codeine react with one mole HCl.

thus

volume of HCl at equivalence point = 12.35 mL * 0.100 M /  0.100 M = 12.35 ml

total volume of solution = 12.35 + 12.35 = 24.70 ml

thus

concentration salt = 12.35 mL * 0.100 M /  24.70 = 0.05 M

salt hydrolysis occurs.

pH = 1/2 * [pKw - pkb - log C]

or

pH = 0.5 * [14 - 5.796 - log (0.05)]

or

pH = 4.75

pH at the equivalence point = 4.75