Question

Calculate the pH of a solution obtained by mixing of 100 ml of 0.1 M HCI and 100 ml of 0.2 M NH₃ ? Kb, for NH₃ = 1.8 x 10^-5 ?

Answer 1

100 ml of 0∙1 M HCl + 100 ml of 0∙2 M NH₃

Kb for NH₃ = 1.8 x 10^-5

100 × 0∙1 = 10 m mol of HCl

100 × 0∙2 = 20 m mol of NH₃

NH₃ (1 mol) + HCl (1 mol) → NH₄Cl (1 mol)

i.e., 10 m mol 10 m mol 10 m mol

NH₃ left unreacted = 10 m mol

Volume of solution = 200 ml

Molarity of NH₃ = \(\frac{10}{200}\) M

= 0 ∙ 05 M

[NH₄⁺] = \(\frac{10}{200}\) M

= 0 ∙ 05 M

pOH = pK_b + log\(\frac{[Salt]}{[Acid]}\)

= -log (1.8 x 10^-5) + log \(\frac{0.05}{0.05}\)

= 4∙74

pH = 14 – pOH

= 14 − 4∙74

= 9∙26.