Calculate the pH of the solution at equivalence.

Question

A chemist titrates 90.0 mL of a 0.2315M pyridine (C₅H₅N) solution with 0.6080M HBr solution at 25 °C. Calculate the pH at equivalence. The pK_b of pyridine is 8.77.

Round your answer to 2 decimal places. You may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added

Answer 1

C₅H₅N is a weak base. HBr is a strong acid. This salt formed at equivalence C₅H₅NH⁺Br^- will be an acidic salt and the pH of the solution should be <7.

C₅H₅N + HBr ==> C₅H₅NH⁺ + Br^-

Initial moles C₅H₅N = 90.0 ml x 1 L / 1000 ml x 0.2315 mol / L = 0.02084 mols

Initial mole HBr = 0.02084 moles @ equivalence

Moles C₅H₅NH⁺ formed @ equivalence = 0.02084 mols

Volume of HBr needed for equivalence = 0.02084 mols x 1 L / 0.6080 mol = 0.03468 L

Total volume at equivalence = 0.090 L + 0.03468 L = 0.12 L

Final [C₅H₅NH⁺] @ equivalence = 0.02084 mols / 0.12 L = 0.174 M

For the pH, we must look at the hydrolysis of C₅H₅NH⁺

C₅H₅NH⁺ + H2O ==> C₅H₅N + H₃O⁺

Ka = [C₅H₅N][H₃O⁺] / [C₅H₅NH⁺ ] and Ka = 1x10^-14 / Kb = 1x10^-14 / 1x10^-8.77 = 5.89x10^-5

5.89x10^-5 = (x)(x) / 0.174 - x (ignore x in denominator since Ka is small)

5.89x10^-5 = x² / 0.174

x² = 1.02x20⁶

x = [H₃O⁺] = 1.01x10^-3 M

pH = -log [H₃O⁺]

pH = -log 1.01x10^-3

pH = 2.99