Question

The pKa of hypochlorous acid is 7.530. A 51.0 mL solution of 0.117 M sodium hypochlorite (NaOCl) is titrated with 0.336 M HCl. Calculate the pH of the solution a) after the addition of 6.45 mL of 0.336 M HCl. b)at the equivalence point with 0.336 M HCl.

Answer 1

    NaOCl(aq) + HCl(aq) ---> NaCl(aq) + HOCl(aq)

No of mole of NaOCl = 51*0.117 = 5.967 mmol

a) after addition of 6.45 ml of HCl

No of mole of HCl added = 6.45*0.336 = 2.167 mmol

pH = pka + log(salt/acid)

pKa of hypochlorous acid = 7.530

salt = NaOCl = 5.967-2.167 = 3.8 mmol

acid = HOCl = 2.167 mmol

pH = 7.53+log(3.8/2.167) = 7.77

b) at equivalence point ,

No of mole of HCl added = No of mole of NaOCl

volume of HCl added = 5.967/0.336 = 17.76 ml

concentration of HOCl formed = 5.967/(17.76+51) = 0.0868 M

pH of HOCl = 1/2(pka-logC)

            = 1/2(7.53-log0.0868)

            = 4.3