Titrations of acids and bases. HA + H 2 O H 3 O + + A -

Titrations of acids and bases


HA + H2O H3O+ + A-


HA + H2O H3O+ + A-

B + H2O OH- + HB+


HA + H2O H3O+ + A-

B + H2O OH- + HB+

H3O+ + OH- 2 H2O

H3O+ + OH- 2 H2O

At equivalence point [H3O+] = [OH-]

For strong acid – strong base

H3O+ + OH- 2 H2O


Kw = [ H3O +] = [OH-] = 10-14

H3O+ + OH- 2 H2O


Kw = [ H3O +] = [OH-] = 10-14

pH = 7


HA + H2O H3O+ + A-

B + H2O OH- + HB+

H3O+ + OH- 2 H2O

Volume of base added

Start with acid solution

pH

7

pH

4

10

100 ml 0.1 M HCl

100 ml 0.1 M HCl

Titrate with 0.1 M NaOH

100 ml 0.1 M HCl


H3O+ + OH- 2 H2O

100 ml 0.1 M HCl


H3O+ + OH- 2 H2O

Every part of a mole of NaOH added

reduces the moles of H3O+ by an equal amount.

100 ml 0.1 M HCl


H3O+ + OH- 2 H2O

As NaOH solution is added, the overall volume

increases. This further decreases [H3O+].

100 ml 0.1 M HCl


Neutralization and dilution; both increase

the pH of the solution.

100 ml 0.1 M HCl


Add 50 ml of NaOH solution.

100 ml 0.1 M HCl



Start with 0.1 L x 0.1 M H3O+

100 ml 0.1 M HCl



Start with 0.1 L x 0.1 M H3O+

0.01 moles H3O+

100 ml 0.1 M HCl



0.01 moles H3O+

Add 0.05 L x 0.1 M OH-

100 ml 0.1 M HCl



0.01 moles H3O+

Add 0.05 L x 0.1 M OH-

0.005 moles OH-

100 ml 0.1 M HCl



0.01 moles H3O+

0.005 moles OH-

H3O+ is neutralized by OH- 1:1 ratio

100 ml 0.1 M HCl



0.01 moles H3O+

0.005 moles OH-

Remaining H3O+ = 0.01 moles – 0.005 moles

100 ml 0.1 M HCl



0.01 moles H3O+

0.005 moles OH-

Remaining H3O+ = 0.005 moles

Total volume = 150 ml = 0.15 L

100 ml 0.1 M HCl



0.01 moles H3O+

0.005 moles OH-

[H3O+]= 0.005 moles0.15 L

100 ml 0.1 M HCl



0.01 moles H3O+

0.005 moles OH-

[H3O+]= 0.005 moles0.15 L

= 0.033 M

100 ml 0.1 M HCl



0.01 moles H3O+

0.005 moles OH-

[H3O+]= 0.005 moles0.15 L

= 0.033 M

pH = 1.48

100 ml 0.1 M HCl


Add 99.9 ml of NaOH solution.

100 ml 0.1 M HCl



0.01 moles H3O+

100 ml 0.1 M HCl



0.01 moles H3O+

0.0999 L x 0.1 M = 0.00999 moles OH-

100 ml 0.1 M HCl



0.01 moles H3O+

0.0999 L x 0.1 M = 0.00999 moles OH-

Moles H3O+ = 0.01 – 0.00999 = 1 x 10-5

100 ml 0.1 M HCl



0.01 moles H3O+

0.0999 L x 0.1 M = 0.00999 moles OH-

Moles H3O+ = 0.01 – 0.00999 = 1 x 10-5

Total volume = 0.1 L + 0.0999 L = 0.1999 L

100 ml 0.1 M HCl



0.01 moles H3O+

0.0999 L x 0.1 M = 0.00999 moles OH-

[H3O+]= 1 x 10-5 moles

0.1999 L

100 ml 0.1 M HCl



0.01 moles H3O+

0.0999 L x 0.1 M = 0.00999 moles OH-

[H3O+]= 1 x 10-5 moles

0.1999 L = 5 x 10-5

100 ml 0.1 M HCl



0.01 moles H3O+

0.0999 L x 0.1 M = 0.00999 moles OH-

[H3O+]= 1 x 10-5 moles

0.1999 L = 5 x 10-5

pH = 4.3

4.3

+ 0.1 ml will give

pH = 7

If more 0.1 M NaOH solution is added

after the equivalence point, there is no

H3O+ to neutralize it.

At pH = 7 the volume is 0.2 L


Add 0.1 mL 0.1 M NaOH



Add 0.0001 L x 0.1 M = 10-5 moles OH-

[OH-] = 1 x 10-5 moles

0.201 L= 5 x 10-5 M



Add 0.0001 L x 0.1 M = 10-5 moles OH-

[OH-] = 1 x 10-5 moles

0.2001 L= 5 x 10-5 M

pOH = 4.3 pH = 9.7

4.3

pH = 7

V = 199.9 ml

V = 200 ml

pH = 9.7 V = 200.1 ml

Titrating a weak acid with a strong base


HA + H2O H3O+ + A-

Ka < 1


HA + H2O H3O+ + A-

Ka < 1

Much less than 100% dissociation.


HA + H2O H3O+ + A-

Ka < 1

Much less than 100% dissociation.

Every OH- added neutralizes an H3O+

and shifts the equilibrium to the right.

Titrate 100 ml 0.10 M CH3COOH

With 0.10 M NaOH


With 0.10 M NaOH

Ka = 1.8 x 10-5


With 0.10 M NaOH

Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

0 ml NaOH


With 0.10 M NaOH

Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

0 ml NaOH

pH = 2.87


With 0.10 M NaOH

Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

50 ml NaOH

.1 L x 0.1 M = 0.01 moles CH3COOH

0.05L x 0.1 M = 0.005 moles NaOH

CH3COOH + H2O H3O+ + CH3COO-

+OH-

2 H2O

CH3COOH + H2O H3O+ + CH3COO-

+OH-

2 H2O

Net result: for every OH- added, there is

one less CH3COOH and one more CH3COO-


With 0.10 M NaOH

Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

50 ml NaOH


0.05L x 0.1 M = 0.005 moles NaOH

0.01 - 0.005 moles CH3COOH = 0.005 moles


With 0.10 M NaOH

Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

50 ml NaOH


0.05L x 0.1 M = 0.005 moles NaOH


0.005 moles CH3COO-


With 0.10 M NaOH

Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

50 ml NaOH


0.005 moles CH3COO-

0.15 L solution

[CH3COOH] = 0.005 moles

0.005 moles CH3COO-

0.15 L= 0.033 M

= 0.033 M

[CH3COOH] = 0.005 moles

0.005 moles CH3COO-

0.15 L= 0.033 M

= 0.033 M

Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

Buffer solution: pKa = pH = 4.74


With 0.10 M NaOH

Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

100 ml NaOH

0.01 moles CH3COO-

0.20 L solution


With 0.10 M NaOH

Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

100 ml NaOH

0.01 moles CH3COO-

0.20 L solution

[CH3COO-] = 0.05 M


Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

[CH3COO-] = 0.05 M

Kb = Kw

Ka

= 10-14

1.8 x 10-5

= 5.6 x 10-10

Kb =[CH3COOH][OH-]

[CH3COO-]


Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

[CH3COO-] = 0.05 M

Kb = Kw

Ka

= 10-14

1.8 x 10-5

= 5.6 x 10-10

Kb =[CH3COOH][OH-]

[CH3COO-]=

y2

0.05


Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

Kb = Kw

Ka

= 10-14

1.8 x 10-5

= 5.6 x 10-10

Kb =[CH3COOH][OH-]

[CH3COO-]=

y2

0.05

y2 = (5.6 x 10-10)(0.05) = 2.8 x 10-11


Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

Kb = Kw

Ka

= 10-14

1.8 x 10-5

= 5.6 x 10-10

Kb =[CH3COOH][OH-]

[CH3COO-]=

y2

0.05

y2 = (5.6 x 10-10)(0.05) = 2.8 x 10-11

y = 5.3 x 10-6


Ka = 1.8 x 10-5 = [H3O+][CH3COO-]

[CH3COOH]

Kb = Kw

Ka

= 10-14

1.8 x 10-5

= 5.6 x 10-10

Kb =[CH3COOH][OH-]

[CH3COO-]=

y2

0.05

y2 = (5.6 x 10-10)(0.05) = 2.8 x 10-11

y = 5.3 x 10-6 pOH =5.3; pH = 8.7

After the equivalence point,

OH- is being added to a saturated

buffer system.

After the equivalence point,

OH- is being added to a saturated

buffer system.

pH increases rapidly

Titrations of acids and bases. HA + H 2 O H 3 O + + A -

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