Question

The pKa of hypochlorous acid is 7.530. A 60.0 mL solution of 0.116 M sodium hypochlorite (NaOCl) is titrated with 0.291 M HCl. Calculate the pH of the solution:

a) after the addition of 7.18 mL of 0.291 M HCl.

I figured this one out by using henderson hasselbalch and got pH=7.90

b) after the addition of 25.2 mL of 0.291 M HCl

c) at the equivalence point with 0.291 M HCl.

I would appreciate it worked out! Thanks!

Answer 1

Hi,

Sodium hypochlorite is a basic salt (formed by a strong base NaOH and a weak acid HOCl) and it produces OH- ion when dissolved in water because of the hydrolysis of OCl- ion;

Dissociation:
NaOCl(aq) -------> Na+(aq) + OCl-(aq)

Hydrolysis:
OCl-(aq) + H2O(l) <----> HOCl(aq) + OH-(aq)

On the other hand, HCl is a strong acid and ionizes completely;
HCl(aq) ------> H+(aq) + Cl-(aq)

Neutralization reaction:
NaOCl(aq) + HCl(aq) -----> NaCl(aq) + HOCl(aq)

Net ionic equation of neutralization;
OCl-(aq) + H+(aq) ------> HOCl(aq)

A) moles ClO- = 0.0600 L x 0.116 M=0.00696

moles H+ = 7.18 x 10^-3 L x 0.291 M=0.00209

moles ClO- = 0.00696 - 0.00209 = 0.00487

moles HClO = 0.00209

total volume = 0.06718 L

[ClO-]= 0.00487/ 0.06718=0.0725 M

[HClO]= 0.00209/ 0.06718=0.0311 M

pKa = 14 - 7.530 =6.47

pH = 6.47 + log 0.0725/ 0.0311= 7.90

B) We will do calculation similar as A

moles ClO- = 0.0600 L x 0.116 M=0.00696

moles H+ = 25.2 x 10^-3 L x 0.291 M=0.00733

Since all ClO- is neutralized, excess moles of H+ = 0.00733-0.00696 =0.00037 moles

therefore, pH = -log [0.00037] = 3.43

C) According to the equations, since the mole of NaOCl (and consequently the mole of OH-) is equal to the mole of HCl (and consequently the mole of H+), The mole of NaOCl and OCl- ion is also 0.00696 mole. [This is the requirement to reach the equivalence point of titration]

Net ionic equation of neutralization;
OCl-(aq) + H+(aq) ------> HOCl(aq)
0.00696 ... 0.00696 .......0.00696 mol

Since the moles of HCl and NaOCl are the same, their molarities and volumes should also be same.
Total volume at the equivalence point;
60.00 cm3 + 60.00 cm3 = 120.00 cm3 = 0.120 L

The molarity of HOCl formed = n / V
[HOCl] = 0.00696 mol / 0.120 L = 0.058 M

Since HOCl is a weak acid it dissociates in small extent;
HOCl(aq) <-------> H+(aq) + OCl-(aq)
0.058 - x M ............. x M ....... x M

Ka = [H+] [OCl-] / [HOCl]

pka=7.530

therefre, ka= 2.95