Question

The pKa of hypochlorous acid is 7.530. A 58.0 mL solution of 0.123 M sodium hypochlorite (NaOCl) is titrated with 0.255 M HCl.

Calculate the pH of the solution after the addition of 8.48 mL of 0.255 M HCl.

pH=

Calculate the pH of the solution after the addition of 29.7 mL of 0.255 M HCl.

pH=

Calculate the pH of the solution at the equivalence point with 0.255 M HCl.

pH=

Answer 1

The reaction of titration is

NaOCl + H+ -------------------------------------> HOCl + Na+

Q1)

After 8.48mL of 0.255M HCl added

NaOCl + H+ -------------------------------------> HOCl + Na+

58x0.123=7.134 8.48x0.255=2.1624 0 0 initial mmoles

4.9716 0 2.1624 - aftr rxn

So the solution is a buffer and its pH is given by Hendersen equation as

pH = pKa + log [conjugate base]/[acid]

=7.530 + log [4.9716/2.1624]

=7.8916

Q2)

After 29.7mL of 0.255M HCl added

NaOCl + H+ -------------------------------------> HOCl + Na+

58x0.123=7.134 29.7x0.255= 7.573 0 0 initial mmoles

0 0.4395 7.134 - aftr rxn

Now the solution has excess acid , thus pH is decided by [H+] =[Hcl] in solution

[Hcl] = mmoles/ total volume= 0.4395/(58+29.7) =0.0050 M

thus pH = -log 0.0050 =2.300

Q3) At equivalence

VHcl = 58.0x0.123 /0.255 =27.97 mL

NaOCl + H+ -------------------------------------> HOCl + Na+

58x0.123=7.134 27.97x0.255= 7.134 0 0 initial mmoles

0 0 7.134

nOw the solution has only weak acid HOCl

[HOCl] = mmoles/V = 7.134/(58+27.97) =0.0810M

pH of weak acid is given by  

pH = 1/2[ pKa-log C]

= 1/2[ 7.530-log 0.081]

= 4.31