Question

The pKa of hypochlorous acid is 7.530. A 58.0 mL solution of 0.101 M sodium hypochlorite (NaOCl) is titrated with 0.332 M HCl. Calculate the pH of the solution

a) after the addition of 6.90 mL of 0.332 M HCl.

b) after the addition of 18.8 mL of 0.332 M HCl.

c) at the equivalence point with 0.332 M HCl.

Answer 1

millimoles of NaOCl = 58 x 0.101 = 5.858

pKa = 7.530

a)

millimoles of HCl = 6.90 x 0.332 = 2.2908

OCl-   +     HCl    ---------------> HOCl

5.858     2.2908                            0

3.567         0                               2.2908

pH = pKa + log [salt / acid]

      = 7.530 + log [3.567 / 2.2908]

pH = 7.72

b)

millimoles of HCl = 18.8 x 0.332 = 6.2416

OCl-   +     HCl    ---------------> HOCl

5.858     6.2416                             0

   0           0.3836                      5.858

here strong acid reamins.

[H+] = 0.3836 / (18.8 + 58) = 4.99 x 10^-3 M

pH = -log [H+] = -log (4.99 x 10^-3)

pH = 2.30

c)

At equivalence point.

volume of HCl = 17.64

here weak acid remains.

concentration of HOCl = 5.858 / (58 + 17.64) = 0.0744 M

pH = 1/2 (pKa - log C)

    = 1/2 (7.530 - log 0.0744)

pH = 4.32