Question

Calculate the pH of a solution made by mixing 100 mL of .300 M NH3 with 100 mL of .200 M HCl. (Kb for NH3 is 1.8 x 10 ^-5).

Answer 1

The reaction is:-

NH₃ + HCl ---------------> NH₄Cl + H₂O

The mole ratio between NH₃ and HCl is 1 : 1

mols of NH₃ = 0.100 L x 0.300 = 0.03 mol

mols of HCl = 0.100 L x 0.200= 0.02 mol

reaction will proceed until 0.02 mol of NH₃ gets reacted. That will leave behind excess of NH₃

0.03 - 0.02 = 0.01 mols excess NH₃

NH₃ reacts with water to produce NH₄⁺ and OH^- ions in a 1 : 1 ratio

Kb = [NH₄⁺][OH^-] / [NH₃]
Suppose concentration of [NH₄⁺] = [OH^-] = X moles
1.8 x 10^-5 = x² / 0.01
x = 4.24 x 10^-4
Therfore, [OH^-] =  4.24 x 10^-4
pOH = -log (OH^-) = -log(4.24 x 10^-4) = 3.37
pH = 14 - 3.37 = 10.63