Question

What is the pH of an aqueous solution made by combining 39.42 mL of a 0.3856 M ammonium chloride with 39.60 mL of a 0.3222 M solution of ammonia to which 4.802 mL of a 0.0876 M solution of HCl was added?

Answer 1

Kb of NH3 = 1.8*10^-5

pKb= -log Kb

     = -log (1.8*10^-5)

     = 4.745

before adding HCl:

moles of NH4Cl = 0.3856 M * 39.42 mL

                 = 15.2 mmol

moles of NH3 = 0.3222 M * 39.60 mL

                 = 12.759 mmol

moles of HCl added= 0.0876 M * 4.802 mL = 0.4207 mmol

after reaction,

moles of NH4Cl = 15.2 + 0.4207 = 15.6207 mmol

moles of NH3 = 12.759 – 0.4207 = 12.3383 mmol

use:

pOH = pKb + log (NH4Cl/NH3)

     = 4.745 + log (15.6207 / 12.3383)

     =4.847

pH = 14 – pOH

    = 14 – 4.847

   = 9.153

Answer: 9.153