Question

The pKa of hypochlorous acid is 7.530. A 59.0 mL solution of 0.137 M sodium hypochlorite (NaOCl) is titrated with 0.321 M HCl. Calculate the pH of the solution

a) after addition of 8.89 mL of 0.321 M HCl

b) after addition of 26.3 mL of 0.321 M HCl

c) at equivalence point with 0.321 M HCl

Answer 1

millimoles of NaOCl = 59 x 0.137 = 8.083

a) after 8.89 mL HCl added

millimoles of HCl = 8.89 x 0.321 = 2.854

8.083 - 2.854 = 5.229 millimoles NaOCl left

2.854 millimoles HOCl will form

[NaOCl] = 5.229 / 67.89 = 0.077 M

[HOCl] = 2.854 / 67.89 = 0.042 M

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

pH = 7.53 + log [0.077] / [0.042]

pH = 7.79

b) millimoles of HOCl = 26.3 x 0.321 = 8.44

8.44 - 8.083 = 0.357 millimoles HCl left

[HCl] = 0.357 / 85.3 = 0.0042 M

as HCl is strong acid [HCl] = [H⁺] = 0.0042 M

pH = - log [H⁺]

pH = - log [0.0042]

pH = 2.38

c) at equivalent point all NaOCl becomes HOCl

8.083 = V x 0.32

V = 25.26 mLtotal volume = 59 + 25.26 = 84.26 mL

[HCl] = 8.083 / 84.26 = 0.096 M

pH = 1/2 [pKa - logC]

pH = 1/2 [7.53 - log 0.096]

pH = 4.27