Question

20 mL of 0.25 M of NH3 is titrated with 0.40 M HCl. Calculate the pH of the solution after 20 mL HCl is added. Kb NH3 = 1.8 × 10−5

Answer 1

Number of mmols of NH₃ before addition of HCl = M x V = 0.25 x 20 = 5

Number of mmols pf HCl added = M x V = 0.40 x 20 = 8

Number of mmols of NH₄Cl formed = number of mmols of NH₃ before addition of HCl = 5

Number of mmoles of HCl remained = 8 – 5 = 3

Total volume of solution = 20 + 20 = 40 mL

Molarity of HCl = 3/40 = 0.075 M

Molarity of NH₄Cl = 5/40 = 0.125 M

Ka x Kb = 10^-14

Ka x 1.8 x 10^-5 = 10^-14

Ka (NH₄Cl) = 5.56 x 10^-10

[H⁺] from HCl = 0.075 M

[H⁺] from NH₄Cl = (Ka x C)^1/2 = (5.56 x 10^-10 x 0.125)^1/2 = 8.34 x 10^-6

So, [H⁺] from NH₄Cl is negligible.

pH = -log 0.075 = 1.12