Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.160 M HClO(aq) with 0.160 M KOH(aq). The ionization constant for HClO is 4.0×10^-8.
(a) before addition of any KOH
pH=
(b) after addition of 25.0 mL of KOH
pH=
(c) after addition of 40.0 mL of KOH
pH=
(d) after addition of 50.0 mL of KOH
pH=
(e) after addition of 60.0 mL of KOH
pH=

Answer 1

a)when 0.0 mL of KOH is added

HClO dissociates as:

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

0.16 0 0

0.16-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 ((4*10^-8)*0.16) = 8*10^-5

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

so, x = 8*10^-5 M

use:

pH = -log [H+]

= -log (8*10^-5)

= 4.0969

Answer: 4.10

b)when 15.0 mL of KOH is added

Given:

M(HClO) = 0.16 M

V(HClO) = 50 mL

M(KOH) = 0.16 M

V(KOH) = 15 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.16 M * 50 mL = 8 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.16 M * 15 mL = 2.4 mmol

We have:

mol(HClO) = 8 mmol

mol(KOH) = 2.4 mmol

2.4 mmol of both will react

excess HClO remaining = 5.6 mmol

Volume of Solution = 50 + 15 = 65 mL

[HClO] = 5.6 mmol/65 mL = 0.0862M

[ClO-] = 2.4/65 = 0.0369M

They form acidic buffer

acid is HClO

conjugate base is ClO-

Ka = 4*10^-8

pKa = - log (Ka)

= - log(4*10^-8)

= 7.398

use:

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

= 7.398+ log {3.692*10^-2/8.615*10^-2}

= 7.03

Answer: 7.03

c)when 40.0 mL of KOH is added

Given:

M(HClO) = 0.16 M

V(HClO) = 50 mL

M(KOH) = 0.16 M

V(KOH) = 40 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.16 M * 50 mL = 8 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.16 M * 40 mL = 6.4 mmol

We have:

mol(HClO) = 8 mmol

mol(KOH) = 6.4 mmol

6.4 mmol of both will react

excess HClO remaining = 1.6 mmol

Volume of Solution = 50 + 40 = 90 mL

[HClO] = 1.6 mmol/90 mL = 0.0178M

[ClO-] = 6.4/90 = 0.0711M

They form acidic buffer

acid is HClO

conjugate base is ClO-

Ka = 4*10^-8

pKa = - log (Ka)

= - log(4*10^-8)

= 7.398

use:

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

= 7.398+ log {7.111*10^-2/1.778*10^-2}

= 8.00

Answer: 8.00

d)when 50.0 mL of KOH is added

Given:

M(HClO) = 0.16 M

V(HClO) = 50 mL

M(KOH) = 0.16 M

V(KOH) = 50 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.16 M * 50 mL = 8 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.16 M * 50 mL = 8 mmol

We have:

mol(HClO) = 8 mmol

mol(KOH) = 8 mmol

8 mmol of both will react to form ClO- and H2O

ClO- here is strong base

ClO- formed = 8 mmol

Volume of Solution = 50 + 50 = 100 mL

Kb of ClO- = Kw/Ka = 1*10^-14/4*10^-8 = 2.5*10^-7

concentration ofClO-,c = 8 mmol/100 mL = 0.08M

ClO- dissociates as

ClO- + H2O -----> HClO + OH-

0.08 0 0

0.08-x x x

Kb = [HClO][OH-]/[ClO-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.5*10^-7)*8*10^-2) = 1.414*10^-4

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

so, x = 1.414*10^-4 M

[OH-] = x = 1.414*10^-4 M

use:

pOH = -log [OH-]

= -log (1.414*10^-4)

= 3.8495

use:

PH = 14 - pOH

= 14 - 3.8495

= 10.1505

Answer: 10.15

e)when 60.0 mL of KOH is added

Given:

M(HClO) = 0.16 M

V(HClO) = 50 mL

M(KOH) = 0.16 M

V(KOH) = 60 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.16 M * 50 mL = 8 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.16 M * 60 mL = 9.6 mmol

We have:

mol(HClO) = 8 mmol

mol(KOH) = 9.6 mmol

8 mmol of both will react

excess KOH remaining = 1.6 mmol

Volume of Solution = 50 + 60 = 110 mL

[OH-] = 1.6 mmol/110 mL = 0.0145 M

use:

pOH = -log [OH-]

= -log (1.455*10^-2)

= 1.8373

use:

PH = 14 - pOH

= 14 - 1.8373

= 12.1627

Answer: 12.16