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Titration Curves I. Strong Acid + Strong Base 0.1 M HCl 0.1 M NaOH 25.0 mL 25.0 mL 2.5 x 10 -3 mol 2.5 x 10 -3 mol Initial pH HCl H + + Cl- 0.1 M0.1 M [H + ] =0.1 M pH =- log H + = 1.00 pH volume NaOH added (mL) 5.0 10.0 0 10 20 30 40 50 .

Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

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Page 1: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

Titration Curves

I. Strong Acid + Strong Base

0.1 M HCl 0.1 M NaOH

25.0 mL 25.0 mL

2.5 x 10-3mol 2.5 x 10-3mol

1. Initial pH HCl H++ Cl-0.1 M 0.1 M[H+] = 0.1 M

pH = - log H+ = 1.00

pH

volume NaOH added (mL)

5.0

10.0

010 20 30 40 50

.

Page 2: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

pH

volume NaOH added (mL)

5.0

10.0

010 20 30 40 50

.

Strong Acid + Strong Base

0.1 M HCl 0.1 M NaOH

25.0 mL 10.0 mL

2.5 x 10-3mol 1.0x 10-3 mol- = 1.5 x 10-3 mol

V = 25 + 10 mL

[H+] = 1.5 x 10-3 mol35 x 10-3 L

[H+] = 4.28 x 10-2 M

pH =1.37 .

Page 3: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

pH

volume NaOH added (mL)

5.0

10.0

010 20 30 40 50

.

Strong Acid + Strong Base

0.1 M HCl 0.1 M NaOH

25.0 mL

2.5 x 10-3mol

.

20.0 mL

2.0 x 10-3 mol- = 0.5 x 10-3 mol

V = 25 + 20 mL

[H+] = 0.5 x 10-3 mol45 x 10-3 L

[H+] = 1.11 x 10-2 M

pH =1.95 .

Page 4: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

pH

volume NaOH added (mL)

5.0

10.0

010 20 30 40 50

.

Strong Acid + Strong Base

0.1 M HCl 0.1 M NaOH

25.0 mL

2.5 x 10-3mol

.

25.0 mL

2.5 x 10-3 mol- = 0.0 mol

[H+] =

pH =7.00

.

1.00 x 10-7 M

.

Page 5: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

pH

volume NaOH added (mL)

5.0

10.0

010 20 30 40 50

.

Strong Acid + Strong Base

0.1 M HCl 0.1 M NaOH

25.0 mL

2.5 x 10-3mol

.

35.0 mL

V = 25 + 35 mL

[OH-] = 1.0 x 10-3 mol60 x 10-3 L

[OH-] = 1.67 x 10-2 M

pOH =1.78 .

3.5 x 10-3 mol OH-

.

pH = 12.22

.

pH

volume NaOH added (mL)

5.0

10.0

010 20 30 40 50

Page 6: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

Titration CurvesWeak Acid + Strong Base

0.1 M CH3COOH 0.1 M NaOH

25.0 mL 25.0 mL

pH

volume NaOH added (mL)

5.0

10.0

010 20 30 40 50

Initial weak acid

half-way point pH = pKa

Ka = 1.8 x 10-5

equivalence point

= [H+] [CH3COO-]

[CH3COOH]

= 4.74

CH3COO- + H2OCH3COOH + OH-

Kb = 5.6 x 10-10 = [OH-] [CH3COOH]

[CH3COO-]

strong base

pH = pKa + log [CH3COO-] [CH3COOH]

Page 7: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

Titration Curves

Weak Base + Strong Acid

0.1 M NH3 0.1 M HCl

25.0 mL 25.0 mL

1. Initial pH

NH3 NH4+ + OH-

[OH-] = 1.34 x 10-3 M

pOH = 2.87

Kb= 1.8 x 10-5 = [OH-][NH4+]

[NH3]

+ H2O

pH = 11.12

pH

volume HCl added (mL)

5.0

10.0

010 20 30 40 50

Page 8: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

Titration Curves

Weak Base + Strong Acid0.1 M NH3 0.1 M HCl

25.0 mL2.5 x 10-3mol

NH3 NH4+ + OH-

Kb= 1.8 x 10-5 = [NH4+] [OH-]

[NH3]

10.0 mL1.0 x 10-3 mol- = 1.5 x 10-3 mol

V = 25 + 10 mL

[NH3] [NH4+] [OH-]

0.043 0.029 0.00.043 -x 0.029 + x x

1.8 x 10-5 = [x] [0.029 + x]

[0.043 - x]

x = 2.67 x 10-5 pOH = 4.57+ H2O

pH = 9.43

pH

volume HCl added (mL)

5.0

10.0

010 20 30 40 50

.

Page 9: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

Titration Curves

Weak Base + Strong Acid0.1 M NH3 0.1 M HCl

25.0 mL2.5 x 10-3mol

NH3 NH4+ + OH-

Kb= 1.8 x 10-5 = [NH4+] [OH-]

[NH3]

20.0 mL2.0 x 10-3 mol- = 5.0 x 10-4 mol

V = 25 + 20 mL

[NH3] [NH4+] [OH-]

0.011 0.044 0.00.011 -x 0.044 + x x

1.8 x 10-5 = [x] [0.044 + x]

[0.011 - x]

x = 4.5 x 10-6 pOH = 5.35+ H2O pH = 8.65

pH

volume HCl added (mL)

5.0

10.0

010 20 30 40 50

. .

Page 10: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

Titration Curves

Weak Base + Strong Acid0.1 M NH3 0.1 M HCl

25.0 mL2.5 x 10-3mol

NH3NH4+ +H+

Ka= 5.6 x 10-10 = [NH3] [H+]

[NH4+]

25.0 mL2.5 x 10-3 mol- = 0.00

V = 25 + 25 mL

[NH4] [NH3] [H+]

0.05 0.00 0.00.05 -x x x

5.6 x 10-10 = [x2]

[0.05 - x]

x = 5.9 x 10-6 pH = 5.27

pH

volume HCl added (mL)

5.0

10.0

010 20 30 40 50

. .

.pH

volume HCl added (mL)

5.0

10.0

010 20 30 40 50

Page 11: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

Titration Curves

Weak Base + Strong Acid0.1 M NH3 0.1 M HCl

25.0 mL

2.5 x 10-3mol

20.0 mL

pOH = pKb + log [NH4+]

[NH3]

5.0 x 10-4 mol NH3

2.0 x 10-3 mol NH4+

V = 45 x 10-3 L

pOH =

Ka = 1.8 x 10-5

4.74 + log (0.44) (0.11)

= 5.34

pH

volume HCl added (mL)

5.0

10.0

010 20 30 40 50

pH = 8.65

Page 12: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

Polyprotic AcidH2SO3 HSO3

- + H+ Ka1 = 1.4 x 10-2

HSO3- SO3

2- + H+ Ka2 = 6.5 x 10-8

2 equivalents of base

0.10 M H2SO3 0.10 M NaOH

40 mL 80 mL

Initial pH

1.4 x 10-2 = [HSO3-] [H+]

[H2SO3]= x2

0.1 - x

x = 0.03 pH = 1.51

equivalents of base

1 2

pH

5.0

10.0

Page 13: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

Polyprotic AcidH2SO3 HSO3

- + H+ Ka1 = 1.4 x 10-2

HSO3- SO3

2- + H+ Ka2 = 6.5 x 10-8

2 equivalents of base

0.10 M H2SO3 0.1 M NaOH

equivalents of base

1 2

pH

5.0

10.0

half-way point pH = pKa

- log 1.4 x 10-2 = 1.85

.

- log 6.5 x 10-8 = 7.19

.1st equivalence point

1.84 + 7.192

= 4.52 .

2nd equivalence point

conjugate base, SO3-

.

buffering regions

equivalents of base

1 2

pH

5.0

10.0

Page 14: Titration Curves I. Strong Acid+ Strong Base 0.1 M HCl0.1 M NaOH 25.0 mL 2.5 x 10 -3 mol 1. Initial pH HCl  H+H+ + Cl- 0.1 M [H + ] = 0.1 M pH =- log

Polyprotic Acid

H2SO3 HSO3- + H+ Ka1 = 1.4 x 10-2

HSO3- SO3

2- + H+ Ka2 = 6.5 x 10-8

2 equivalents of base

0.10 M H2SO3 0.1 M NaOH40 mL 80 mL

Initial pH

1.4 x 10-2 = [HSO3-][H+]

[H2SO3]

= x2

0.1 - x

x = 0.03

pH = 1.51