Question

3.Here you prepare a buffer solution using hypochlorous acid (HCIO₂) and potassium hypochlorite (KCIO₂).

a) If you mix 150ml of 0.15M HCIO and 2.00g of KCIO, what will be the pH of this buffer solution? Ka of hypochlorous acid is 2.5x10^-9.

b) If you add 10ml of 0.50 M HCl to the prepared buffer solution from a), what will be the pH of this solution?

c)If you add 1ml of 5M NaOH to the original buffer solution (from A)), what will be the pH of this solution?

Answer 1

pKa = -log(Ka )= -log(2.5*10^-9) = 8.60

a)

mmol of acid = MV = 0.15*150 = 22.5

mmol of KClO2 = mass/MW *1000 = 2/122.55 *1000 = 16.319

pH= pKa + log(ClO-/HClO)

pH = 8.60 + log(16.319/22.5) = 8.4605

b)

mmol of H+ = MV = 0.5*10 = 5

mmol of acid = MV = 22.5 + 5 = 27.5

mmol of KClO2 = 16.319 - 5 = 11.319

pH= pKa + log(ClO2-/HClO2

pH = 8.60+ log(11.319/27.5) = 8.214

c)

mmol of OH- = 5*1 = 5

mmol of acid = MV = 22.5 - 5 = 17.5

mmol of KClO2 = 16.319 + 5 = 21.319

pH= pKa + log(ClO2-/HClO2

pH = 8.60+ log(21.319/17.5) = 8.685