Question

The pK, of hypochlorous acid is 7.530. A 56.0 mL solution of 0.142 M sodium hypochlorite (NaOCI) is titrated with 0.292 MHCI. Calculate the pH of the solution after the addition of 8.44 mL of 0.292 M HCI. pH = Calculate the pH of the solution after the addition of 28.6 ml of 0.292 M HCI. pH = Calculate the pH of the solution at the equivalence point with 0.292 M HCI. pH =

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

1)when 8.44 mL of HCl is added

Given:

M(HCl) = 0.292 M

V(HCl) = 8.44 mL

M(ClO-) = 0.142 M

V(ClO-) = 56 mL

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

mol(HCl) = 0.292 M * 8.44 mL = 2.4645 mmol

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

mol(ClO-) = 0.142 M * 56 mL = 7.952 mmol

We have:

mol(HCl) = 2.4645 mmol

mol(ClO-) = 7.952 mmol

2.4645 mmol of both will react

excess ClO- remaining = 5.4875 mmol

Volume of Solution = 8.44 + 56 = 64.44 mL

[ClO-] = 5.4875 mmol/64.44 mL = 0.0852 M

[HClO] = 2.4645 mmol/64.44 mL = 0.0382 M

They form basic buffer

base is ClO-

conjugate acid is HClO

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 {3.824*10^-2/8.516*10^-2}

= 6.122

use:

PH = 14 - pOH

= 14 - 6.1224

= 7.8776

Answer: 7.88

2)when 28.6 mL of HCl is added

Given:

M(HCl) = 0.292 M

V(HCl) = 28.6 mL

M(ClO-) = 0.142 M

V(ClO-) = 56 mL

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

mol(HCl) = 0.292 M * 28.6 mL = 8.3512 mmol

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

mol(ClO-) = 0.142 M * 56 mL = 7.952 mmol

We have:

mol(HCl) = 8.3512 mmol

mol(ClO-) = 7.952 mmol

7.952 mmol of both will react

excess HCl remaining = 0.3992 mmol

Volume of Solution = 28.6 + 56 = 84.6 mL

[H+] = 0.3992 mmol/84.6 mL = 0.0047 M

use:

pH = -log [H+]

= -log (4.719*10^-3)

= 2.3262

Answer: 2.33

3)

find the volume of HCl used to reach equivalence point

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

0.142 M *56.0 mL = 0.292M *V(HCl)

V(HCl) = 27.2329 mL

Given:

M(HCl) = 0.292 M

V(HCl) = 27.2329 mL

M(ClO-) = 0.142 M

V(ClO-) = 56 mL

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

mol(HCl) = 0.292 M * 27.2329 mL = 7.952 mmol

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

mol(ClO-) = 0.142 M * 56 mL = 7.952 mmol

We have:

mol(HCl) = 7.952 mmol

mol(ClO-) = 7.952 mmol

7.952 mmol of both will react to form HClO and H2O

HClO here is strong acid

HClO formed = 7.952 mmol

Volume of Solution = 27.2329 + 56 = 83.2329 mL

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

concentration ofHClO,c = 7.952 mmol/83.2329 mL = 0.0955 M

HClO + H2O -----> ClO- + H+

9.554*10^-2 0 0

9.554*10^-2-x x x

Ka = [H+][ClO-]/[HClO]

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)*9.554*10^-2) = 5.31*10^-5

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

so, x = 5.31*10^-5 M

[H+] = x = 5.31*10^-5 M

use:

pH = -log [H+]

= -log (5.31*10^-5)

= 4.2749

Answer: 4.27