Given 100. mL of a 0.245 M solution of pyridine (C5H5N pKb = 8.77), what is the pH if 100. mL of 0.245 M HCl is added?

Question

Your help would be greatly appreciated.

Answer 1

Instead of writing out pyridine as C₅H₅N each time, since it is a base, we can just refer to it as B.

The reaction of this base with HCl would be...

B + HCl ==> BH⁺ + Cl^- or simply B + H⁺ ==> BH⁺

mols B = 100. ml x 1 L / 1000 ml x 0.245 mol/L = 0.0245 mols B

mols H⁺ added = 100.0 ml x 1 L / 1000 ml x 0.245 mol/L = 0.0245 mols H⁺

mols B left = 0

mols H⁺ left = 0

mols BH⁺ formed = 0.0245 mols BH+

Final volume = 200. mls = 0.200 L

Final [BH⁺] = 0.0245 mol / 0.200 L = 0.1225 M

Hydrolysis of BH⁺:

BH⁺ + H2O ==> B + H₃O+ here BH⁺ is acting as an acid and H₂O as the base

Since BH⁺ is an acid, we need to use the Ka for BH⁺

pKa + pKb = 14 and pKa = 14 - pKb

pKa = 14 - 8.77 = 5.23 and so Ka = 5.89x10^-6

Ka = [B][H₃O⁺] / [BH⁺] = (x)(x) / 0.1225

5.89x10^-6 = x² / 0.1225

x² = 7.22x10^-7

x = 8.5x10^-4 M = [H₃O⁺]

pH = -log [H3O+] = -log 8.5x10^-4

pH = 3.07

(note: be sure to check the math)