Question

The pKa of hypochlorous acid is 7.530. A 56.0 mL solution of 0.123 M sodium hypochlorite (NaOCl) is titrated with 0.255 M HCl. Calculate the pH of the solution

a) after the addition of 10.2 mL of 0.255 M HCl.

b) after the addition of 28.1 mL of 0.255 M HCl.

c) at the equivalence point with 0.255 M HCl.

Answer 1

pH of acidic bufefr = pka + log(salt/acid)

A) no of mol of sodium hypochlorite (NaOCl) = 56*0.123 = 6.888 mmol

   no of mol of HCl. = 10.2*0.255 = 2.601 mmol

pH = 7.53+log((6.888-2.601)/2.601)

    = 7.75

b) no of mol of sodium hypochlorite (NaOCl) = 56*0.123 = 6.888 mmol

   no of mol of HCl. = 28.1*0.255 = 7.1655 mmol

concentration of excess HCl = (7.1655-6.888)/(56+28.1) = 0.0033 M

pH = -log(0.0033) = 2.48

c) at the equivalence point with 0.255 M HCl.

no of mol of sodium hypochlorite (NaOCl) = 56*0.123 = 6.888 mmol

no of mol of HCl = 6.888 mmol

volume of HCl added = 6.888/0.255 = 27 ml

concentration of HOcl = 6.888/(56+27) = 0.083 M

pH =1/2(pka-logC)

    = 1/2(7.53-log0.083)

   = 4.3