Question

The pKa of hypochlorous acid is 7.530. A 50.0 mL solution of 0.108 M sodium hypochlorite (NaOCl) is titrated with 0.336 M HCl. Calculate the pH of the solution

a) after the addition of 5.40 mL of 0.336 M HCl.

b) after the addition of 16.8 mL of 0.336 M HCl.

c) at the equivalence point with 0.336 M HCl.

Answer 1

use:

pKa = -log Ka

7.53 = -log Ka

Ka = 2.951*10^-8

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/2.951*10^-8

Kb = 3.388*10^-7

a)when 5.4 mL of HCl is added

Given:

M(HCl) = 0.336 M

V(HCl) = 5.4 mL

M(OCl-) = 0.108 M

V(OCl-) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.336 M * 5.4 mL = 1.8144 mmol

mol(OCl-) = M(OCl-) * V(OCl-)

mol(OCl-) = 0.108 M * 50 mL = 5.4 mmol

We have:

mol(HCl) = 1.8144 mmol

mol(OCl-) = 5.4 mmol

1.8144 mmol of both will react

excess OCl- remaining = 3.5856 mmol

Volume of Solution = 5.4 + 50 = 55.4 mL

[OCl-] = 3.5856 mmol/55.4 mL = 0.0647 M

[HOCl] = 1.8144 mmol/55.4 mL = 0.0328 M

They form basic buffer

base is OCl-

conjugate acid is HOCl

Kb = 3.388*10^-7

pKb = - log (Kb)

= - log(3.388*10^-7)

= 6.47

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 6.47+ log {3.275*10^-2/6.472*10^-2}

= 6.174

use:

PH = 14 - pOH

= 14 - 6.1742

= 7.8258

Answer: 7.83

b) when 16.8 mL of HCl is added

Given:

M(HCl) = 0.336 M

V(HCl) = 16.8 mL

M(OCl-) = 0.108 M

V(OCl-) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.336 M * 16.8 mL = 5.6448 mmol

mol(OCl-) = M(OCl-) * V(OCl-)

mol(OCl-) = 0.108 M * 50 mL = 5.4 mmol

We have:

mol(HCl) = 5.6448 mmol

mol(OCl-) = 5.4 mmol

5.4 mmol of both will react

excess HCl remaining = 0.2448 mmol

Volume of Solution = 16.8 + 50 = 66.8 mL

[H+] = 0.2448 mmol/66.8 mL = 0.0037 M

use:

pH = -log [H+]

= -log (3.665*10^-3)

= 2.436

Answer: 2.44

c)

find the volume of HCl used to reach equivalence point

M(OCl-)*V(OCl-) =M(HCl)*V(HCl)

0.108 M *50.0 mL = 0.336M *V(HCl)

V(HCl) = 16.0714 mL

Given:

M(HCl) = 0.336 M

V(HCl) = 16.0714 mL

M(OCl-) = 0.108 M

V(OCl-) = 50 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.336 M * 16.0714 mL = 5.4 mmol

mol(OCl-) = M(OCl-) * V(OCl-)

mol(OCl-) = 0.108 M * 50 mL = 5.4 mmol

We have:

mol(HCl) = 5.4 mmol

mol(OCl-) = 5.4 mmol

5.4 mmol of both will react to form HOCl and H2O

HOCl here is strong acid

HOCl formed = 5.4 mmol

Volume of Solution = 16.0714 + 50 = 66.0714 mL

Ka of HOCl = Kw/Kb = 1.0E-14/3.388E-7 = 2.952*10^-8

concentration ofHOCl,c = 5.4 mmol/66.0714 mL = 0.0817 M

HOCl + H2O -----> OCl- + H+

8.173*10^-2 0 0

8.173*10^-2-x x x

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

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.952*10^-8)*8.173*10^-2) = 4.912*10^-5

since c is much greater than x, our assumption is correct

so, x = 4.912*10^-5 M

[H+] = x = 4.912*10^-5 M

use:

pH = -log [H+]

= -log (4.912*10^-5)

= 4.3088

Answer: 4.31