IB Chemistry on Titration Curves between Acids and Bases

http://lawrencekok.blogspot.com

Prepared by

Lawrence Kok

Tutorial on Titration Curves, Acid Base Titration and

Buffer Regions.

http://lawrencekok.blogspot.com/

NEUTRALIZATION

Neutral salt

Strong acid and Strong base Strong acid and Weak base Weak acid and Strong base

Acidic salt Basic salt

NH4+ + H2O ↔ NH3 + H3O

+ CH3COO- + H2O ↔ CH3COOH + OH-

lose H+ to produce H+ gain H+ to produce OH-

NH4+ + H2O → NH3 + H3O

+

NH4CI → NH4+ + CI-

H3O+ (Acidic)

Cation hydrolysis Anion hydrolysis

CH3COONa → CH3COO- + Na+

CH3COO- + H2O→ CH3 COOH + OH-

OH- (Alkaline)

NaCI → Na+ + CI-

No H2O hydrolysis

H2O (Neutral)

HCI + NaOH → NaCI + H2O

Neutralization Reaction Salt Salt hydrolysis Type salt pH salt

Strong acid+

Strong base

HCI+

NaOHNaCI

No hydrolysis Neutral salt 7

Strong acid+

Weak base

HCI+

NH3

NH4CICation

hydrolysisAcidic salt < 7

Weak acid+

Strong base

CH3COOH+

NaOHCH3COONa

Anionhydrolysis

Basic salt > 7

Weak acid+

Weak base

CH3COOH+

NH3

CH3COONH4

Anion/Cationhydrolysis

Depends ?

Click here on acidic buffer simulation

Click here buffer simulation

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/acidbasepH/pHbuffer20.html



http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf



CH3COO- + H2O → CH3 COOH + OH-

Salt Hydrolysis

Neutralization Reaction Salt Salt hydrolysis Type salt pH salt

Strong acid+

Strong base

HCI+

NaOHNaCI

No hydrolysis Neutral salt 7

Strong acid+

Weak base

HCI+

NH3

NH4CICation

hydrolysisAcidic salt < 7

Weak acid+

Strong base

CH3COOH+

NaOHCH3COONa

Anionhydrolysis

Basic salt > 7

Weak acid+

Weak base

CH3COOH+

NH3

CH3COONH4

Anion/Cationhydrolysis

Depends ?

Weak acid and Weak base

CH3COOH + NH3 → CH3COONH4

Acidicity depend on Ka and Kb

Ka > Kb – Acidic – H+ ions producedKb < Ka – Basic – OH- ions producedKa = Kb – Neutral – hydrolyzed same extent.

CH3COONH4 → CH3COO- + NH4+

NH4+ + H2O → NH3 + H3O

+

salt

anion cation

OH- - Basic H3O+ - AcidicKb Ka

Ka = Kb

NEUTRAL

NH3 + HF → NH4F

salt

NH4F → NH4+ + F-

NH4+ + H2O → NH3 + H3O

+ F- + H2O → HF + OH-

cation anion

KaH3O

+ - Acidic KbOH- - Basic

Acidicity depend on Ka and Kb

Ka > Kb – Acidic – H+ ions producedKb < Ka – Basic – OH- ions producedKa = Kb – Neutral – hydrolyzed same extent.

Kb > Ka

BASIC

Weak acid

+

Weak base

Titration bet strong acid with strong baseHCI + NaOH → NaCI + H2O

Titration curves Strong Acid with Strong Base

Click here titration simulation

NaOHM = 0.1MV = 0 ml

HCIM = 0.1MV = 25ml

7

HCI + NaOH → NaCI + H2OM = 0.1M M = 0.1MV = 25ml V = 25ml

2.7

11.3

• Rapid jump in pH (2.7 – 11.3)• Rapid change at equivalence pt• Equivalence pt → amt acid = amt base• pH at equivalence pt = 7• Neutral salt, NaCI - neutral

1

HCIM = 0.1MV = 25ml

NaOHM = 0.1MV = 25ml

HCI left → 25 ml, 0.1M Conc H+ = 0.1M

HCIM = 0.1MV = 1ml left

NaOHM = 0.1MV = 24 ml add

HCI left → 1ml, 0.1MMole H+ = (0.1 x 1)/1000

= 0.0001molConc H+ = Mole/Vol

= 0.0001/0.049= 0.002M

NaOHM = 0.1MV = 25ml add


Neutral Salt, NaCIConc H+ = 1 x 10-7M (Dissociation of water)


NaOH left → 1ml left, 0.1MMoles OH-= (0.1 x 1)/1000

= 0.0001molConc OH- = Moles/Vol

= 0.0001/0.051= 0.002M

NaOHV = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml

11.3

2.7

Neutralization

Mole ratio

1: 1

Equivalent point pH 7

Dilution factor!

1

]1.0lg[

]lg[

pH

pH

HpH

7.2

]002.0lg[

]lg[

pH

pH

HpH7

]101lg[

]lg[

7

pH

pH

HpH

3.117.214

7.2

)002.0lg(

)lg(

pH

pOH

pOH

OHpOH

7

http://chem-ilp.net/labTechniques/AcidBaseIdicatorSimulation.htm



Titration bet strong acid with weak baseHCI + NH4OH → NH4CI + H2O

Titration curves Strong Acid with Weak Base

NH4OHM = 0.1MV = 0 ml

HCIM = 0.1MV = 25ml

5.3

HCI + NH4OH → NH4CI + H2OM = 0.1M M = 0.1MV = 25ml V = 25ml

2.7

7.8

• Rapid jump in pH (2.7 – 7.8)• Rapid change at equivalence pt• Equivalence pt → amt acid = amt base• pH at equivalence pt = 5.3• Acidic salt, NH4CI – pH = 5.3

1

HCIM = 0.1MV = 25ml

NH4OHM = 0.1MV = 25ml

HCI left → 25 ml, 0.1M Conc H+ = 0.1M


NH4OHM = 0.1MV = 24 ml add

HCI left → 1ml, 0.1MMole H+ = (0.1 x 1)/1000


= 0.0001/0.049= 0.002M

NH4OHM = 0.1MV = 25ml add

HCIM = 0.1M

V = 0ml left

Acidic Salt, NH4CINH4

+ hydrolysis to produce H+

pH = 5.3


NH4OH left → 1ml left, 0.1MMoles OH-= (0.1 x 1)/1000


= 0.0001/0.051= 0.002M

Conc NH4CI = Moles/Vol= 2.5 x 10-3/0.051= 0.05M

NH4OHV = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml

7.8

2.7

Neutralization


pOH = pKb -lg (base)(salt)

pOH = 4.74 – lg (0.002)(0.05)

pOH = 6.13pH + pOH = 14pH = 7.8

pH buffer – salt and weak base

pH buffer region

salt and weak base

1

]1.0lg[

]lg[

pH

pH

HpH

7.2

]002.0lg[

]lg[

pH

pH

HpH

5.3

Mole ratio

1: 1




Titration between weak acid with strong baseCH3COOH + NaOH → CH3COONa + H2O

Titration curves Weak Acid with Strong Base



9

CH3COOH + NaOH → CH3COONa + H2OM = 0.1M M = 0.1MV = 25ml V = 25ml

11.3

6.11

• Rapid jump in pH (6.11 – 11.3 )• Rapid change at equivalence pt• Equivalence pt → amt acid = amt base• pH at equivalence pt = 9• Basic salt, CH3COONa = pH 9

2.87

CH3COOHM = 0.1MV = 25ml


CH3COOH left → 25 ml, 0.1M CH3COOH ↔ (CH3COO- )(H+)Ka = (CH3COO-) (H+)

CH3COOH(H+) = √Ka x CH3COOH(H+) = 1.34 x 10-3

CH3COOHM = 0.1MV = 1ml left




Basic Salt (CH3COONa)

CH3COO- hydrolysis produce OH-

pH = 9


NaOHV = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml


CH3COOH left → 1ml, 0.1M Mole CH3COOH = (MV)/1000

= (1 x 0.1)/1000= 0.0001mol

Conc CH3COOH = Mole/Vol= 0.0001/0.049= 2.04 x 10-3

Conc CH3COONa = Mole/Vol= 2.4 x 10-3/0.049= 0.048M

Ka = 1.8 x 10-5

NaOH left → 1ml left, 0.1MMoles OH-= (0.1 x 1)/1000


= 0.0001/0.051= 0.002M

11.3

6.11

Neutralization

pH buffer region

weak acid and salt

pH buffer – salt and weak acid

pH = pKa -lg [acid][salt]

pH = 4.74 – lg [2.04 x 10-3][0.048]

pH = 4.74 + 1,37pH = 6.11

Weak Acid

87.2

]1034.1lg[

]lg[

3

pH

pH

HpH

9

3.117.214

7.2

)002.0lg(

)lg(

pH

pOH

pOH

OHpOH

Mole ratio

1: 1


Titration curves Weak Acid with Weak Base



7

CH3COOH + NH4OH → CH3COONH4 + H2OM = 0.1M M = 0.1MV = 25ml V = 25ml

7.8

• No sharp rise in pH • pH changes gradually over a range• No inflection point

2.87





Neutral Salt CH3COONH4

pH = 7


NH4OHV = 1ml left

Total = 25 + 26

Vol = 51ml

7.8

6.11

Neutralization

Titration between weak base with weak acidCH3COOH + NH4OH → CH3COONH4 + H2O

6.11




CH3COOH left → 25 ml, 0.1M CH3COOH ↔ (CH3COO- )(H+)Ka = (CH3COO-) (H+)

CH3COOH(H+) = √Ka x CH3COOH


= (1 x 0.1)/1000= 0.0001mol


Conc CH3COONH4 = Moles/Vol= 2.4 x 10-3/0.049= 0.048M


pH = 4.74 – lg [2.04 x 10-3][0.048]

pH = 4.74 + 1,37pH = 6.11

pH buffer – weak acid and salt

pH buffer region

weak acid and salt

NH4OH left → 1ml, 0.1MpH = 7.8

87.2

]1034.1lg[

]lg[

3

pH

pH

HpH

Click here for notes

Mole ratio

1: 1




http://www.chem.uci.edu/~lawm/4-8-13.pdf



Titration bet strong acid with strong baseHCI + NaOH → NaCI + H2O

Titration bet weak acid with strong baseCH3COOH + NaOH → CH3COONa + H2O

Titration curves Acid with Base


11.3

2.7

Titration bet weak acid with weak baseCH3COOH + NH4OH → CH3COONH4 + H2O


6.11

11.3NaOHM = 0.1MV = 25ml

2.87

1

Vs

•Start at pH = 1 → End at 11.3• Rapid change at equivalence pt• Rapid jump in pH (2.7 – 11.3)• Equivalence pt → amt acid = amt base• pH at equivalence pt = 7• Neutral salt, NaCI - neutral

•Start at pH = 2.87 → End at 11.3• Rapid change at equivalence pt• Rapid jump in pH (6.11 – 11.3)• Equivalence pt → amt acid = amt base• pH at equivalence pt = 9• Basic salt, CH3COONa - basic

9

7

Vs

•Start at pH = 1 → End at 7.8• Rapid change at equivalence pt• Rapid jump in pH (2.7 – 7.8)• Equivalence pt → amt acid = amt base• pH at equivalence pt = 5.3• Acidic salt, NH4CI - acidic

1

2.7

5.3

7.8

2.87



•Start at pH = 2.87 → End at 7.8• pH changes gradually over a range• No sharp rise in pH• Equivalence pt → amt acid = amt base• pH at equivalence pt = 7• Neutral salt, CH3COONH4 - neutral

6.11

7

7.8

pH buffer region

salt and weak base

pH buffer region

salt and weak acid

Buffer region form• Slow gradual increase pH

due to buffering effect

Buffer region form• Slow gradual increase pH


HCIM = 0.1MV = 25ml

HCIM = 0.1MV = 25ml



Strong acid vs Strong base

Titration Acid Base

Strong acid vs Weak base Weak acid vs Strong base Weak acid vs Weak base Acid

BaseIndicator

2.7

11.3

7.8

2.7

11.3

6.11

7.8

6.11

HCIM = 0.1MV = 25ml

Dilution Factor

Water

Adding 20 ml water

What is conc H+ and pH?

3105.2..

025.01.0..

..

HMole

HMole

VMHMole

1

]1.0lg[

]lg[

pH

pH

HpH

055.0......

045.0

105.2..

3

HConc

Volume

MoleHConc

Before adding Water After adding WaterVol/Conc change

25.1

]055.0lg[

]lg[

pH

pH

HpH

pH drop due dilution Factor

Adding 20 ml base NaOH

What is conc H+ and pH?

NaOH

HCIM = 0.1MV = 25ml

Total vol = 25 + 20 = 45ml

Dilution/Neutralization Factor

Before adding NaOH

3105.2..

025.01.0..

..

HMole

HMole

VMHMole

1

]1.0lg[

]lg[

pH

pH

HpH

Mole change

After adding NaOH

3105.0... leftHMole

011.0.

045.0

105.0..

3

HConc

Volume

MoleHConc

95.1

]011.0lg[

]lg[

pH

pH

HpH

Total vol = 25 + 20 = 45ml

pH drop due dilution/Neutralization Factor

Dilution Factor during Titration

Why adding water and base causes pH to increase?

Titration curves Strong Acid with Strong Base

HCIM = 0.1MV = 0 ml

HCIM = 0.1MV = 25ml

7

HCI + NaOH → NaCI + H2OM = 0.1M M = 0.1MV = 25ml V = 25ml

2.7

11.3

• Rapid drop in pH (11.3– 2.7 )• Rapid change at equivalence pt• Equivalence pt → amt acid = amt base• pH at equivalence pt = 7• Neutral salt, NaCI - neutral



NaOH left → 25 ml, 0.1M Conc OH- = 0.1M

NaOHM = 0.1MV = 1ml left

HCIM = 0.1MV = 24 ml add

NaOH left → 1ml, 0.1MMole OH- = (0.1 x 1)/1000

= 0.0001molConc OH- = Mole/Vol

= 0.0001/0.049= 0.002M

HCIM = 0.1MV = 25ml add

NaOHM = 0.1MV = 0 ml left

Neutral Salt, NaCIConc H+ = 1 x 10-7M (dissociation of water)


HCI left → 1ml left, 0.1MMoles H+ = (0.1 x 1)/1000

= 0.0001molConc H+ = Moles/Vol

= 0.0001/0.051= 0.002M

HCIV = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml


13

11.3

2.7

Neutralization

Titration bet strong base with strong acidHCI + NaOH → NaCI + H2O

13114

1

)1.0lg(

)lg(

pH

pOH

pOH

OHpOH

3.117.214

7.2

)002.0lg(

)lg(

pH

pOH

pOH

OHpOH7

]101lg[

]lg[

7

pH

pH

HpH

7.2

]002.0lg[

]lg[

pH

pH

HpH

Mole ratio

1: 1




Titration curves Weak Acid with Strong Base

CH3COOHM = 0.1MV = 0 ml


9


11.3

6.13

• Rapid drop in pH ( 11.3 - 6.13)• Rapid change at equivalence pt• Equivalence pt → amt acid = amt base• pH at equivalence pt = 9• Basic salt, CH3COONa = pH 9

13

NaOHM = 0.1MV = 25.0ml



CH3COOHM = 0.1MV = 24 ml add

CH3COOHM = 0.1MV = 25ml add


Basic Salt (CH3COONa)

CH3COO- hydrolysis to produce OH-

pH = 9


CH3COOHV = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml


NaOH left → 25 ml, 0.1M Conc OH- = 0.1M

NaOH left → 1ml, 0.1MMoles OH- = (0.1 x 1)/1000

= 0.0001molConc OH- = Mole/Vol

= 0.0001/0.049= 0.002M


= (1 x 0.1)/1000= 0.0001mol


Conc CH3COONa = Mole/Vol= 2.5 x 10-3/0.051= 0.049M

11.3

6.13

Neutralization

pH buffer region

weak acid + salt

Titration bet strong base with weak acidCH3COOH + NaOH → CH3COONa + H2O


pH = 4.74 – lg [1.96 x 10-3][0.049]

pH = 4.74 + 1,39pH = 6.13

pH buffer – salt and weak acid

13114

1

)1.0lg(

)lg(

pH

pOH

pOH

OHpOH

3.117.214

7.2

)002.0lg(

)lg(

pH

pOH

pOH

OHpOH

Mole ratio

1: 1




Titration curves Strong Acid with Weak Base

HCI M = 0.1MV = 0 ml

HCIM = 0.1MV = 25ml

5.3

HCI + NH4OH → NH4CI + H2OM = 0.1M M = 0.1MV = 25ml V = 25ml

7.8

2.7

• Rapid drop in pH (7.8 – 2.7)• Rapid change at equivalence pt• Equivalence pt → amt acid = amt base• pH at equivalence pt = 5.3• Acidic salt, NH4CI – pH = 5.3

11.1


NH4OHM = 0.1MV = 1ml left



Acidic Salt, NH4CINH4

+ hydrolysis to produce H+

pH = 5.3


NH4OH left → 1ml left, 0.1MMole NH4OH = (0.1 x 1)/1000

= 0.0001 molConc NH4OH = Mole/Vol

= 0.0001/0.049= 0.002M

Conc NH4CI = Mole/Vol= 2.4 x 10-3/0.049 = 0.048

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml

7.8

2.7

Neutralization


pOH = 4.74 – lg (0.002) = 6.12(0.048)

pH + pOH = 14pH = 7.8

pH buffer – Weak base + salt


NH4OH left → 25 ml, 0.1M NH4OH ↔ NH4

+ + OH-

Kb = (NH4+) (OH-)

(NH4OH) 1.8 x 10-5 = (OH-)2

0.1OH- = √0.1 x 1.8 x 10-5

OH- = 1.34 x 10-3

pH buffer region

weak base + salt

HCI left → 1ml left, 0.1MMole H+ = (0.1 x 1)/1000


= 0.0001/0.051= 0.002M

Titration bet weak base with strong acidHCI + NH4OH → NH4CI + H2O


1.1187.214

87.2

)1034.1lg(

)lg(

3

pH

pOH

pOH

OHpOH


HCIV = 1ml left

7.2

]002.0lg[

]lg[

pH

pH

HpH

Mole ratio

1: 1




Titration curves Weak Acid with Weak Base


CH3COOHM = 0.1MV = 25ml 7

CH3COOH + NH4OH → CH3COONH4 + H2OM = 0.1M M = 0.1MV = 25ml V = 25ml

6.11

• No sharp drop in pH • pH changes gradually over a range• no inflection point

11.1



Neutral Salt CH3COONH4

pH = 7


Total = 25 + 26

Vol = 51ml

7.8

6.11

Neutralization


NH4OH left → 25 ml, 0.1M Conc NH4OH = 0.1MNH4OH ↔ NH4

+ + OH-

Kb = (NH4+) (OH-)

(NH4OH) 1.8 x 10-5 = (OH-)2

0.1OH- = √0.1 x 1.8 x 10-5

OH- = 1.34 x 10-3

Titration bet weak base with weak acidCH3COOH + NH4OH → CH3COONH4 + H2O

CH3COOH left → 1ml, 0.1M pH = 6.11

7.8


NH4OH left → 1ml left, 0.1MMole NH4OH = (0.1 x 1)/1000

= 0.0001molConc NH4OH = Mole/Vol

= 0.0001/0.049= 0.002M

Conc NH4CI = Mole/Vol= 2.4 x 10-3/0.049= 0.048M


pOH = 4.74 – lg (0.002)(0.048)

pOH = 6.12pH + pOH = 14pH = 7.8

pH buffer – weak base + salt

pH buffer region

weak base + salt




1.1187.214

87.2

)1034.1lg(

)lg(

3

pH

pOH

pOH

OHpOH

CH3COOHV = 1ml left

Mole ratio

1: 1




Titration bet strong base with strong acidHCI + NaOH→ NaCI + H2O

Titration bet strong base with weak acidCH3COOH + NaOH → CH3COONa + H2O

Titration curves Acid with Base

Titration bet weak base with strong acidHCI + NH4OH → NH4CI + H2O

11.3

2.7

Titration bet weak base with weak acidCH3COOH + NH4OH → CH3COONH4 + H2O


HCIM = 0.1MV = 25ml

6.13

11.3



13

Vs

•Start at pH = 1 3 → End at 2.7• Rapid change at equivalence pt• Rapid drop in pH (11.3 – 2.7)• Equivalence pt → amt acid = amt base• pH at equivalence pt = 7• Neutral salt, NaCI - neutral

•Start at pH = 13 → End at 6.13• Rapid change at equivalence pt• Rapid drop in pH (11.3 – 6.13)• Equivalence pt → amt acid = amt base• pH at equivalence pt = 9• Basic salt, CH3COONa - basic

9

7

Vs

•Start at pH = 11.1 → End at 2.7• Rapid change at equivalence pt• Rapid drop in pH (7.8 – 2.7)• Equivalence pt → amt acid = amt base• pH at equivalence pt = 5.3• Acidic salt, NH4CI - acidic

11.1

2.7

5.3

7.8 11.1


HCIM = 0.1MV = 25ml


•Start at pH = 11.1 → End at 6.11• pH changes gradually over a range• No sharp drop in pH• Equivalence pt → amt acid = amt base• pH at equivalence pt = 7• Neutral salt, CH3COONH4 - neutral

6.11

7

7.8

13

Buffer region form• Slow gradual drop pH


Buffer region form• Slow gradual drop pH


pH buffer region

weak acid + salt

pH buffer region

weak base + salt


Acidic Buffer Region CH3COOH (acid)/CH3COO-(salt)

Titration bet weak acid + strong baseCH3COOH + NaOH → CH3COONa + H2O


CH3COOH + NaOH → CH3COONa + H2OInitial 0.0025mol 0.00125mol added 0

Change (0.0025-0.00125)mol 0 mol 0.00125 mol form

Final 0.00125mol left 0 mol 0.00125 mol form

At half equivalence point :• Amt acid = Amt salt : ( 0.00125 = 0.00125)• pH = pKa -lg [acid] → pH = pKa → 4.74 = 4.74

[salt]Buffer at pH = 4.74 form when half amt of acid neutralize by baseor at half equivalence pt when amt acid = amt salt

Prepare Acidic Buffer pH = 4.74• Choose pKa acid closest pH 4.74• pKa = 4.74 (ethanoic acid) chosen• pH = pKa -lg [acid]

[salt]• 4.74 = 4.74 – lg [acid]

[salt]• [acid] = 1.00 (amt acid = amt salt)

[salt]

Buffer region at half equivalence pt

Amt acid = Amt salt

Weak acid 25ml, 0.1M

(0.0025 mol)

Titration bet weak acid with strong base CH3COOH + NaOH → CH3COONa + H2O

Buffer region at half equivalence point

Amt acid = Amt salt

At equivalence point(Neutralization)

Amt acid = Amt base


At half equivalence point :• Vol base = 12.5ml• pH = pKa

• Most effective buffering capacity

At equivalence point:• vol base = 25ml• Amt acid = amt base • Neutralization• Salt and water


NaOHM = 0.1MV = 12.5 ml add

CH3COOHM = 0.1MV = 12.5ml left


Strong base

12.5ml, 0.1M add

(0.00125 mol)

pH buffer region

weak acid + salt

Titration curve use to find pKa or Ka for weak acid

pH = pKa

Completely

neutralize

Half

neutralize

Mole ratio

1: 1




NH4OH + HCI → NH4CI + H2OInitial 0.0025 mol 0.00125 mol add 0.

Change (0.0025-0.00125)mol 0 0.00125 mol form

Final 0.00125 mol left 0 0.00125 mol form

At half equivalence point :• Amt base = Amt salt : (0.00125 = 0.00125)• pOH = pKb - lg [base] → pOH = pKb → 4.74 = 4.74

[salt]Buffer at pOH = 4.74 form when half amt of base neutralise by acidor at half equivalence point when amt base = amt salt

Prepare Buffer pH = 9.26 /pOH = 4.74• Choose pKb base closest to pOH = 4.74• pKb = 4.74 (NH3) chosen• pOH = pKb -lg [base]

[salt]• 4.74 = 4.74 – lg [base]

[salt]• [base] = 1.00 (amt base = amt salt)

[salt]

NH4OH + HCI → NH4CI + H2OM = 0.1M M = 0.1MV = 25ml V = 25ml

Titration bet weak base + strong acidNH4OH + HCI → NH4CI + H2O


Buffer region at half equivalence pt

Amt acid = Amt salt

Weak base 25ml, 0.1M

(0.0025 mol)

Titration bet weak base with strong acidNH4OH + HCI → NH4CI + H2O


Amt acid = Amt salt

At equivalence point(Neutralization)

Amt acid = Amt base

At half equivalence point :• Vol acid = 12.5ml• pH = pKb

• Most effective buffering capacity

At equivalence point:• vol acid = 25ml• Amt acid = amt base • Neutralization• Salt and water


HCIM = 0.1MV = 12.5 ml add

NH4OHM = 0.1MV = 12.5ml left


Strong acid

12.5ml, 0.1M add

(0.00125 mol)

pH buffer region

weak base + salt

Titration curve use to find pKb or Kb for weak base

pH = pKb

Basic Buffer Region NH3(base)/NH4CI(salt)

Completely

neutralize

Half

neutralize

Mole ratio

1: 1




Click here view buffering

Concept Map Buffer

pH

Proton availability Stable

Buffer solution

Weak acid ↔ Conjugate base

][

][lg

salt

acidpKpH a

pH = -lg[H+]

made up of

HA ↔ H+ + A-

Weak base ↔ Conjugate acid

or

Buffering capacity highest

Buffer formula

pH = pKa

1][

][

Salt

Acid

B + H2O ↔ BH+ + OH-

or

Ratio of acidsalt

DilutionAdd water

pH buffer

pH will not change

Temperature affect pH

pH change

Strong acidStrong base

Titration Acid Base

Strong acidWeak base

Weak acidStrong base

Weak acidWeak base

Neutralization Titration curve

Base

AcidIndicator

2.7

11.3

7.8

2.7

11.3

6.11

7.8

6.11

11.3

2.7

7.8

2.7

11.3

6.11

7.8

6.11

Indicator

Acid

Base

Adding base to acid

Adding acid to base

https://www.youtube.com/watch?v=rIvEvwViJGk



Concept Map

Strong acidStrong base

Titration Acid Base

Strong acidWeak base

Weak acidStrong base

Weak acidWeak base

Neutralization Titration curve

Base

Acid Indicator2.7

11.3

7.8

2.7

11.3

6.11

7.8

6.11

Adding base to acid

End point

pH range at equivalent pt

Equivalent ptStoichiometric pt

Point of inflection

Buffer region

Titration bet weak acid with strong baseCH3COOH + NaOH → CH3COONa + H2O

pH buffer region

weak acid and salt

6.11

11.3

9

• Amt acid = Amt base• Vol at equivalence pt = 25ml

25ml12.5ml

Half Equivalent pt

• Amt acid = Amt salt• Vol is = 12.5ml• Buffer region form

Indicator change

colour


25ml

Equivalent ptStoichiometric pt

• Amt acid = Amt base• Vol at equivalence pt = 25ml

pH = pKa6.3

2.7

7.8

pH buffer region

weak base and salt

Buffer region

50ml

pOH = pKb

• Amt base = Amt salt• Vol is = 50 ml• Buffer region form

CH3COOH + NaOH → NaCI + H2OM = 0.1M M = 0.1MV = 25 ml V = ? ml

Titration between Strong Acid with Strong Base

HCI + NaOH → NaCI + H2OM = 0.1M M = 0.1MV = 25 ml V = 25 ml

HCIM = 0.1MV = 25 ml

NaOHM = 0.1MV = ? ml

25 ml, 0.1M strong acid need to neutralize, 25ml, 0.1M strong base?

STRONG BASE

Mole ratio – 1: 1

Mole ratio – 1: 1


2.5 x 10-3 mol NaOH

2.5 x 10-3 H+ ion

2.5 x 10-3 mol HCI

Strong acid/base

dissociate completely

Titration between Weak Acid with Strong Base

What vol of 0.1M strong base need to neutralize, 25ml, 0.1M weak acid?


Will 2.5 x 10-3 mol weak acid

dissociate completely?

Answer – Yes 25ml base needed

Regardless whether strong or weak acid/base • Stoichiometric mole ratio is follow for neutralization• 2.5 x 10-3 mol weak acid neutralize 2.5 x 10-3 mol strong base

CH3COOH + H2O ↔ H3O+ + CH3COO-

H3O+ + OH- ↔ 2H2O

CH3COOH + OH- → CH3COO- + H2O

Le Chatelier principle

↓

Addition OH- remove H+

↓

Conc H+ reduced

↓

Shift equilibrium right

↓

Weak acid, CH3COOH dissociate completely

↓

Produce 2.5 x 10-3 mol H+

↓

Volume Strong base = 25 ml

WEAK ACID

CH3COOH

STRONG BASE

NaOH

CH3COOH ↔ CH3COO- + H+ OH- CH3COOH + OH- → CH3COO- + H2O

5108.1 aK14100.1 wK

9

145

108.1

100.1108.1

rxn

rxn

wbrxn

K

K

KKK

ANSWER

Effect on Kc

Inverse rxn

Add 2 rxn

wK

1

wb KK

+

Krxn HIGH

•Shift to right

•Weak acid dissociate completely

STRONG ACID

HCI + NH4OH → NH4CI + H2OM = 0.1M M = 0.1MV = ? ml V = 25 ml

Titration between Strong Acid with Strong Base

HCI + NaOH → NaCI + H2OM = 0.1M M = 0.1MV = 25 ml V = 25 ml

HCIM = 0.1MV = 25 ml


25 ml, 0.1M strong acid need to neutralize, 25ml, 0.1M strong base?

STRONG ACID

STRONG BASE

Mole ratio – 1: 1

Mole ratio – 1: 1


2.5 x 10-3 mol NaOH

2.5 x 10-3 OH- ion

2.5 x 10-3 mol HCI

Strong acid/base

dissociate completely

Titration between Strong Acid with Weak Base

What vol of 0.1M strong acid need to neutralize, 25ml, 0.1M weak base?

HCIM = 0.1MV = ? ml

Will 2.5 x 10-3 mol weak base

dissociate completely?

Answer – Yes 25ml acid needed

Regardless whether strong or weak acid/base • Stoichiometric mole ratio is follow for neutralization• 2.5 x 10-3 mol strong acid neutralize 2.5 x 10-3 mol weak base

NH3 + H2O ↔ NH4+ + OH-

H+ + OH- ↔H2ONH3 + H+→ NH4

+

Le Chatelier principle

↓

Addition H+ remove OH-

↓

Conc OH- reduced

↓

Shift equilibrium right

↓

Weak base, NH4OH dissociate completely

↓

Produce 2.5 x 10-3 mol OH-

↓

Volume Strong acid = 25 ml

STRONG ACID

HCI

WEAK BASE

NH4OH → NH4+ + OH-

NH4OH

H+ NH4OH + HCI → NH4CI + H2O

5108.1 bK14100.1 wK

9

145

108.1

100.1108.1

rxn

rxn

wbrxn

K

K

KKK

ANSWER

Effect on Kc

Inverse rxn

Add 2 rxn

wK

1

wb KK

+

Krxn HIGH

•Shift to right

•Weak base dissociate completely

NH3 molecules dissociates completely

Neutralization acid and base

Strong base with Strong acid

HCI + NaOH → NaCI + H2OM = 0.1M M = 0.1MV = 25 ml V = ?

CH3COOH + NH4OH → CH3COONH4 + H2OM = 0.1M M = 0.1MV = 25ml V = ?

CH3COOH + NaOH → CH3COONa + H2OM = 0.1M M = 0.1MV = 25ml V = ?

Weak base with Strong acid Weak base with Weak acid

HCIM = 0.1MV = 25ml

HCI + NH4OH → NH4CI + H2OM = 0.1M M = 0.1MV = 25 ml V = ?


HCIM = 0.1MV = 25ml


Will volume used the same?

Will volume used the same?

Yes

25ml

Yes

25ml

Stoichiometric mole ratio is followed for neutralization• 1 mole weak/strong acid will neutralize 1 mole weak/strong base

STRONG ACID WEAK ACID

WEAK BASE

STRONG ACIDWEAK ACID

Mole ratio – 1: 1Mole ratio – 1: 1

Mole ratio – 1: 1 Mole ratio – 1: 1

NaOHM = 0.1MV = ? STRONG BASE STRONG BASE

NaOHM = 0.1MV = ?

Strong base with Weak acid

NH4OHM = 0.1MV = ? WEAK BASE

NH4OHM = 0.1MV = ?

IB Buffer Calculation

Find pH buffer , titration by adding 18ml, 0.10M HCI to 32ml, 0.10M NH3 pKb = 4.751 Titration bet strong acid with weak baseNH3 + HCI → NH4CI + H2O

Buffer regionWeak base + salt

NH3 + HCI → NH4CI + H2OInitial 3.2 x 10-3 mol 1.8 x 10-3 mol add 0 0

Change (3.2 – 1.8) x 10-3 0 1.8 x 10-3 mol form

Final 1.4 x 10-3 0 1.8 x 10-3 mol form

Change mole to Conc → Mole ÷ Total vol

Conc (1.4 x 10-3)/ 0.05 (1.8 x 10-3)/0.05

Conc 2.8 x 10-2 M 3.6 x 10-2 M

(base) (salt)

• pOH = pKb - lg [base][salt]

• pOH = 4.75 – lg [2.8 x 10-2]/[3.6 x 10-2]• pOH = 4.86pH + pOH = 14pH = 9.14


Strong acid 18ml, 0.1M HCI add

Weak base32ml, 0.1M NH3

Total vol = 50 ml or 0.05 dm3

NH3 + H2O ↔ NH4+ + OH-

Kb = (NH4+) (OH-)

(NH3)1.77 x 10-5 = 3.6 x 10-2 x OH-

2.8 x 10-2

OH- = 2.8 x 10-2 x 1.77 x 10-5

3.6 x 10-2

OH- = 1.37 x 10-5

pOH = -lg[OH-]pOH = -lg 1.37 x 10-5

pOH = 4.86pH + pOH = 14pH = 9.14

1st method (formula) 2nd method (Kb)

pH calculation

molVMmole 3108.11000

1.00.18

molVMmole 3102.31000

1.032




2 Find pH when 50ml 0.1M NaOH add to 100ml, 0.1M CH3COOH Ka CH3COOH = 1.8 x 10-5M, pKa = 4.74

Titration bet strong base with weak acidNaOH + CH3COOH → CH3COONa + H2O



Amt base = Amt salt

Buffer Calculation

1st method (formula)2nd method (Ka)

NaOH + CH3COOH → CH3COONa H2OInitial 5 x 10-3 mol add 10 x 10-3 mol 0 0

Change 0 (10-5) x 10-3 5 x 10-3 mol form

Final 0 5 x 10-3 5 x 10-3 mol form

Change mole to Conc→Mole ÷ Total Vol

Conc (5 x 10-3)/0.15 (5 x 10-3)/0.15

Conc 3.3 x 10-2 M 3.3 x 10-2 M

(acid) (salt)

• pH = pKa - lg [acid][salt]

• pH = 4.74 – lg [3.3 x 10-2]/[3.3 x 10-2]• pH = 4.74


CH3COOH ↔ CH3COO- + H+

Ka = (CH3COO-)(H+)(CH3COOH)

1.8 x 10-5 = 3.3 x 10-2 x (H+)3.3 x 10-2

H+ = 1.8 x 10-5

pH = -lg [H+]pH = -lg(1.8 x 10-5 )pH = 4.74

pH calculation

molVMmole 31051000

1.050

molVMmole 310101000

1.0100

Strong base 50ml, 0.1M NaOH add

Weak acid 100ml, 0.1M CH3COOH




pH Calculation

3 Find pH when 50ml 0.1M NaOH add to 100ml, 0.1M HCI

Strong base 50ml, 0.1M NaOH add

Strong acid100ml, 0.1M HCI


Titration bet strong base with strong acidNaOH + HCI → NaCI + H2O

pH calculation

NaOH + HCI → NaCI + H2OInitial 5 x 10-3 mol added 10 x 10-3 mol 0 0

Change 0 (1 0 - 5) x 10-3

Final 0 5 x 10-3

Change mole to Conc→Mole ÷ Total Vol

Conc (5 x 10-3)/ 0.15

Conc 3.3 x 10-2 M

• pH = -lg[H+]• pH = -lg 3.3 x 10-2

pH = 1.48


NH4+ + H2O ↔ NH3 + H3O

+

Ka = (NH3)(H3O+)

(NH4+)

(H3O+)2 = Ka x NH4

+

H+ = √5.56 x 10-10 x 0.10H+ = 7.45 x 10-6

pH = -lg 7.45 x 10-6

pH = 5.13

Find pH 0.10M NH4CI in water. Kb NH3 = 1.8 x 10-5 M4

Acid dissociation constant

0.10M NH4CI

Ka (NH4) x Kb(NH3) = Kw

Ka = Kw /Kb

Ka = 10-14/ 1.8 x 10-5

Ka = 5.56 x 10-10

Using Ka

Find pH 0.50M NH3 in water. Kb NH3 = 1.8 x 10-5 M5

NH3 + H2O ↔ NH4+ + OH-

Kb = (NH4+) (OH-)

(NH3)1.8 x 10-5 = (OH-)2

0.50OH- = √0.50 x 1.8 x 10-5

OH- = 3.0 x 10-3

pOH = -lg 3.0 x 10-3

pOH = 2.52pH = 14 – 2.52pH = 11.48

0.50M NH3

Base Dissociation constantUsing Kb

molVMmole 31051000

1.050

molVMmole 310101000

1.0100




IB Questions

6 Table shows Ka values for acids at 298K. Write expression for Ka. Arrange acid in order of increasing strength

7

Acid CH3COOH HCN HSO4-

Ka 1.8 x 10-5 4.9 x 10-10 1.2 x 10-2

Table shows Kb values for base at 298K. Write expression for Kb. Arrange base in order of increasing strength

Base C2H5NH2 N2H4 NH3

Kb 4.7 x 10-4 4.9 x 10-10 1.2 x 10-2

CH3COOH ↔ CH3COO- + H+

COOHCH

HCOOCHKa

3

3

HCN + H2O ↔ CN- + H3O+

HCN

OHCNK a

3

HSO4- + H2O ↔ SO4

2- + H3O+

4

3

2

4

HSO

OHSOKa

Ka – Highest – Strongest acidKa – Lowest – Weakest acid

C2H5NH2 + H2O ↔ C2H5NH3+ + OH-

252

352

NHHC

OHNHHCKb

NH3 + H2O ↔ NH4+ + OH-

3

4

NH

OHNHKb

N2H4 + H2O ↔ N2H5+ + OH-

42

52

HN

OHHNKb

Kb – Highest – Strongest base Kb – Lowest – Weakest base

Table shows Ka/Kb values for diff substances at 298K. Arrange in order of increasing strength of acidity 8

Substance A B C D E

pKa/ pKb pKa = 3.4

pKa =6.7

pKa= 2.1

PKb = 6

pKb= 5

pKa = - lg10Ka

pKb = - lg10Kb

pKa + pKb = pKw

pKa + pKb = 14

Low pKa – High Ka – Strong acid

pKa = 14-6 = 8 pKa = 14-5 = 9C > A > B > D > E

A. 0.3 mol NH3 and 0.3 mol HCIB. 0.3 mol NH3 and 0.15 mol HCIC. 0.3 mol NH3 and 0.6 mol HCID. 0.3 mol NH3 and 0.15 mol H2SO4

I. 50ml, 0.1M CH3COONaII. 25ml, 0.1M NaOHIII. 50ml, 0.1M NaOH

IB Questions

9

Adding weak acid and its conjugate base/salt Titrating

Buffer preparation

CH3COOH / CH3COONa

Acidic buffer Basic buffer

NH3 / NH4CI

How buffer solution are prepared?

Weak acid with strong base Weak base with strong acid

Weak acid

Strong base Strong acid

Weak base

Buffer can be prepared by adding which of the following to 50ml, 0.1M CH3COOH 10

Which solution will produce a buffer in 1dm3 of water?11

12 Which substance could be added to ethanoic acid to prepare acidic buffer?

I. Hydrochloric acidII. Sodium ethanoateIII. Sodium hydroxide

25ml weak acid, HA titrated with 0.155M NaOH and graph is shown belowa) Determine pH at equivalence pointb) Explain using eqn, why equivalence point is not at pH = 7c) Calc conc of weak acid bef addition of any NaOHd) Estimate, using data from graph, Ka of weak acid

Sample IB Question on Acid Base Titration

a) pH is = 9At equivalence pointAmt acid = Amt base

HAM = ? MV = 25.0ml

NaOHM = 0.155MV = 22 ml

c) HA + NaOH → NaA + H2O Moles of Base = MV

= (0.155 x 0.022)= 3.41 x 10-3

Mole ratio ( 1 : 1)• 1 mole base neutralize 1 mole acid• 3.41 x 10-3 base neutralize 3.41 x 10-3 acidMoles Acid = MV = M x 0.025M x 0.025 = 3.41 x 10-3

M = 0.136M

d) pH = 5.3At half equivalence pt:• Vol base 11ml• Amt acid = amt saltpH = pKa - lg [acid]

[salt]pH = pKa = 5.3pKa = -lg Ka

5.3 = -lg Ka

Ka = 5 x 10-6

b) Neutralization - strong base with weak acidHA + NaOH→ A- + H2OA- is a strong conjugate baseA- + H2O →HA + OH- (basic salt)

3.41 x 10-3 base added

Graph below shows variation pH for titration bet 25.0ml H3PO4 with 0.1M NaOH. Write balanced eqn for rxn at pH=4.7, pH = 9.6, pH = 12.5. What vol of NaOH needed to neutralize 25ml H3PO4.

13

11

Click here notes

Polyprotic Acid – dissociate 1 proton stepwise1st H+ dissociation - H3PO4 ↔ H+ + H2PO4

-

2nd H+ dissociation – H2PO4- ↔ H+ + HPO4

2-

3rd H+ dissociation – HPO42- ↔ H+ + PO4

3-

13

3 108.4 aK

Ka get smaller – Weaker acid – Difficult remove H+ from an increasingly negatively charged.Stronger ESF bet H+ and anion

pH = 4.7

pH = 9.6

pH = 12.5

3

1 105.7 aK

8

2 102.6 aK

http://www.titrations.info/acid-base-titration-phosphoric-acid

Click here on titration animationClick here on titration simulation

Simulation and Animation on Buffer and Titrations

Click here for videos from Khan Academy

Click here acidic buffer animationClick here titration animationClick here titration simulation

Click here titration animation Click here titration animation

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/simDownload/index4.html



http://www.virtlab.com/index.aspx



http://www.youtube.com/watch?v=BBIGR0RAMtY&list=PL39E45DB611F4C7FE

http://www.media.pearson.com.au/schools/cw/au_sch_derry_ibchl_1/int/titrationCurve/1502.html



http://auth.mhhe.com/physsci/chemistry/animations/chang_7e_esp/crm3s5_5.swf






http://www.youtube.com/watch?v=yirkozUyG74

http://www.youtube.com/watch?v=JiSI9MWFbvk



Acknowledgements

Thanks to source of pictures and video used in this presentation

Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/

Prepared by Lawrence Kok

Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com

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http://lawrencekok.blogspot.com/

IB Chemistry on Titration Curves between Acids and Bases

Education

weak baseweak acid

strong basestrong acid

acid base titration

weak base nh4ohm

h gain h

molconc oh

molconc h

amt base ph