Question

The pKa of hypochlorous acid is 7.530. A 50.0 mL solution of 0.142 M sodium hypochlorite (NaOCl) is titrated with 0.338 M HCl. Calculate the pH of the solution:

a) after the addition of 7.12 mL of 0.338 M HCl.

b) after the addition of 21.8 mL of 0.338 M HCl.

c) at the equivalence point with 0.338 M HCl.

Answer 1

millimoles of NaOCl = 50 x 0.142 = 7.1

a) millimoles of HCl added = 7.12 x 0.338 = 2.41

7.1 - 2.41 = 4.69 millimoles NaOCl left

2.41 millimoles of HOCl formed

total volume = 50 + 7.12 = 57.12

[NaOCl] = 4.69 / 57.12 = 0.082 M

[HOCl] = 2.41 / 57.12 = 0.0422 M

pH = pKa + log [NaOCl] / [HOCl]

pH = 7.53 + log [0.082] / [0.0422]

pH = 7.82

b) millimoles of HCl added = 21.8 x 0.338 = 7.37

7.37 - 7.1 = 0.27 millimoles HCl left

total volume = 50 + 21.8 = 71.8 mL

[HCl] = 0.27 / 71.8 = 0.00376 M

pH = - log [H⁺]

pH = - log [0.00376]

pH = 2.42

c) 7.1 millimoles HCl must be added to reach equivalence point.

7.1 = V x 0.338

V = 21.0 mL HCl must be aded

at equivalence point all NaOCl convert to HOCl

[HOCl] = 7.1 / 71 = 0.1 M

HOCl is weak acid

pH = 1/2 [pKa - logC]

pH = 1/2 [7.53 - log 0.1]

pH = 4.26