Question

The pKa of a hypochlorous acid is 7.530. A 60.0 mL solution of 0.138 M sodium hypochlorite (NaOCl) is titrate with 0.319 M HCl. Calculate the pH of the solution:

a) after the addition of 7.79 mL of 0.319 M HCl

b) after the addition of 27.2 mL of 0.319 M HCl

c) at the equivalence point with 0.319 M HCl

Answer 1

a)

mmoles of OCl- = 60 x 0.138 = 8.28

mmoles of HCl = 7.79 x 0.319 = 2.485

OCl-   +    HCl    -----------> HClO

8.28       2.485                     0

5.795          0                      2.485

pH = pKa + log [salt / acid]

    = 7.530 + log [2.485 / 5.795]

pH = 7.162

b)

mmoles of HCl = 27.2 x 0.319 = 8.6768

OCl-   +    HCl    -----------> HClO

8.28       8.68                    0

0          0.397                     8.28

here strong acid reamins.

[HCl] = 0.397 / (27.2 + 60) = 4.55 x 10^-3 M

pH = -log (4.55 x 10^-3)

pH = 2.34

c)

volume at equivalence point = 25.956

here weak acid remains.

concentration of acid = 8.28 / 60 + 25.956 = 0.0963 M

pH = 1/2 (pKa - log C)

     = 1/2 (7.530 - log 0.0963)

pH = 4.273