Question

38.50mL of a 0.23M weak base (Kb= 1.8 x 10^-5) are titrated with 0.37M sulfuric acid, and it took 15.67mL of the acid to reach the equivalence point. Calculate the pH of the solution after 22.00mL of the acid has been added.

Answer 1

mmol = Molarity (M) * Volume (in ml)

Initial mmol of weak base present = (0.23 M) * (50 mL) = 11.5

mmol of acid added to reach equivalence point = (0.37 M) * (15.67 mL) = 5.80

Since per mol of H2SO4, 2 mol of H+ are produced,

H2SO4(aq) + 2H+(aq) + SO (aq)

mmol of acid added when 22.00 mL of acid is added = (0.37 M) * (22.00 mL) = 8.14

So, excess mmol of acid = 8.14 - 5.80 = 2.34

and mmol of H+ present = 2 * 2.34 = 4.68

Total volume of solution = (50.0 + 22 ) mL = 72 mL

Therefore [H+] = 4.68 mmol / 72 mL = 0.065 M

therefore pH = -log [H+] = -log (0.065) =1.187         ..Answer

Note that mainly H+ will come from H2SO4 only at this point. we are neglecting hydrolysis of conjugate acid of the base, as the value will be small.