The pKa of hypochlorous acid is 7530 A 590 mL solution of 0

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The pKa of hypochlorous acid is 7.530. A 59.0 mL solution of 0.142 M sodium hypochlorite (NaOCI) is titrated with 0.272 M HCI. Calculate the pH of the solution a) after the addition of 11.3 mL of 0.272 M HC Number b) after the addition of 32.8 mL of 0.272 M HC Number pH-II c) at the equivalence point with 0.272 M HCI. Number

Solution

a)

mmoles of NaOCl = 59 x 0.142 = 8.378

mmoles of HCl = 11.3 x 0.272 = 3.074

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

8.378      3.074                               0

5.304          0                               3.074

pH = pKa + log [salt / acid]

     = 7.530 + log [5.304 / 3.074]

pH = 7.77

b)

mmoles of HCl = 32.8 x 0.272 = 8.9216

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

8.378      8.922                               0

   0          0.5436                            8.378

here strong acid reamins.

[H+] = 0.5436 / (59 + 32.8) = 5.92 x 10^-3 M

pH = -log (5.92 x 10^-3) = 2.23

pH = 2.23

c)

At equivalence point :

mmoles of OCl- = mmoles of HCl

8.378 = 0.272 x v

v = 30.8 mL

here acid remains.

concentration of HOCl = 8.378 / 59 + 30.8 = 0.0933 M

pH = 1/2 (pKa - log C)

    = 1/2 (7.530 - log 0.0933)

pH = 4.28