Question

The pKa of hypochlorous acid is 7.530. A 52.0 mL solution of 0.129 M sodium hypochlorite (NaOCl) is titrated with 0.255 M HCl. Calculate the pH of the solution

a) after the addition of 10.5 mL of 0.255 M HCl.

b) after the addition of 27.4 mL of 0.255 M HCl.

c) at the equivalence point with 0.255 M HCl.

Answer 1

NaOCl + 2HCl -------> H2O + Cl2 + NaCl

1 mole of NaOCl react with 2 mole of HCl

a) Concentration of HCl = 0.255M

No of mole of HCl for 10.5ml =( 0.255mole/1000ml )×10.5ml = 0.0026775

0.0026775 ml react with 0.00133875 mole of NaOCl

Initial mole of NaOCl = (0.129/1000)×52= 0.006708

Remaining mole of NaOCl = 0.006708 - 0.00133875=0.00536925

Total volume = 52ml + 10.5ml = 62.5ml

[ NaOCl ] =( 0.00536925/62.5)×1000= 0.0859M

pKa = 7.53

Ka = 2.95×10^-8

Kb = Kw / Ka

Kb = 1×10^-14/2.95×10^-8

= 3.39 × 10^-7

OCl- + H2O -----> HOCl + OH-

Kb = [ OH- ] [ HOCl ]/[ OCl- ]

3.39×10^-7 = X^2/(0.0859 - X)

X^2 + 3.39×10^-7X - 2.9×10^-8 = 0

X = 1.70×10^-4M

[ OH- ] = 1.70×10^-4M

pOH = -log[OH-]

= - log [ 1.70× 10^-4]

= 3.77

pH = 14 -pOH

= 14 - 3.77

= 10.23

b) No of mole of HCl =( 0.255/1000)× 27.4 = 0.006987

0.006987 mole of HCl react with 0.0034935mole of NaOCl

Remaining NaOCl = 0.006708 - 0.0034935 = 0.0032145

Total Volume = 52ml + 27.4ml = 79.4ml

[ NaOCl ] =( 0.0032145/79.4)×1000=0.0405M

3.39×10^-7 = X^2/ (0.0405 - X)

X^2 + 3.39×10^-7 X - 1.37×10^-8= 0

X = 1.17×10^-4

[ OH- ] = 1.17×10^-4 M

pOH = 3.93

pH = 10.07

c) at equivalence point

pH = 7

Because the salt NaCl formed will not hydrolysed by water