Question

The pKa of hypochlorous acid is 7.530. A 51.0 mL solution of 0.141 M sodium hypochlorite (NaOCl) is titrated with 0.319 M HCl. Calculate the pH of the solution

a) after the addition of 6.94 mL of 0.319 M HCl.

b) after the addition of 23.3 mL of 0.319 M HCl.

c) at the equivalence point with 0.319 M HCl.

Answer 1

millimoles of NaOCl = 51 x 0.141 = 7.191

a)

millimoles of HCl = 6.94 x 0.319 =2.214

NaOCl   + HCl --------------------------> HOCl + NaCl

7.191       2.214                                    0          0

4.977          0                                      2.214

pH = pKa + log [NaOCl / HOCl]

pH = 7.530 + log (4.977 / 2.214)

pH = 7.882

b)

millimoles of HCl = 0.319 x 23.3 = 7.433

[H+] = 7.433 - 7.191 / (51 + 23.3) = 0.2417 / (51 + 23.3)

        = 3.25 x 10^-3 M

pH = -log [H+]

pH = -log (3.25 x 10^-3)

pH = 2.49

c)

at equivalence point volume = 7.191 / 0.319 = 22.5 mL

[HOCl] = 7.191 / (22.5 + 51) = 0.0978 M

pH = 4.270