Question

The pKa of hypochlorous acid is 7.530. A 54.0 mL solution of 0.147 M sodium hypochlorite (NaOCl) is titrated with 0.253 M HCl. Calculate the pH of the solution

a) after the addition of 12.5 mL of 0.253 M HCl.
pH= ?

b) after the addition of 33.5 mL of 0.253 M HCl.
pH=?

c) at the equivalence point with 0.253 M HCl.
pH=?

Answer 1

a)

mmol of OCl- = MV = 54*0.147 = 7.938

mmol of H+ = MV = 0.253*12.5 = 3.1625

then, HOCl forms:

mmol of OCl- left = 7.938-3.1625 = 4.7755

mmol of HCOl = 0+3.1625 = 3.1625

pH = pKa + log(A-/HA)

pH = 7.53 + log(4.7755/3.1625 )

ph = 7.71

b)

similarly

mmol of OCl- = MV = 54*0.147 = 7.938

mmol of H+ = MV = 0.253*33.5 = 8.4755

excess acid:

mmol of H+ = 8.4755-7.938 = 0.5375

Vtotal = 54.0 + 33.5 = 87.5

[H+] = 0.5375/87.5 = 0.00614

pH = -log(0.00614) = 2.211

c)

in equivalence

V = mmol/V = 7.938/0.253 = 31.3754 mL

Vtotal = 31.3754+54 = 85.3754

[HOCl] = 7.938/85.3754 = 0.09297

Ka = [H+][OCl-]/[HOCl]

10^-7.53 = x*x/(M-x)

10^-7.53 = x*x/(0.09297)

x = sqrt((10^-7.53) / 0.09297)

x = 0.000563

ph = -log(0.000563 = 3.25