IB Chemistry on Titration Curves between Acids and Bases

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 +

NaOH

NaCI

No hydrolysis Neutral salt 7

Strong acid +

Weak base

HCI +

NH3

NH4CI

Cation hydrolysis

Acidic salt < 7

Weak acid +

Strong base

CH3COOH +

NaOH

CH3COONa

Anion hydrolysis

Basic salt > 7

Weak acid +

Weak base

CH3COOH +

NH3

CH3COONH4

Anion/Cation hydrolysis

Depends ?

Click here on acidic buffer simulation

Click here buffer simulation

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

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

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

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 +

NaOH

NaCI

No hydrolysis Neutral salt 7

Strong acid +

Weak base

HCI +

NH3

NH4CI

Cation hydrolysis

Acidic salt < 7

Weak acid +

Strong base

CH3COOH +

NaOH

CH3COONa

Anion hydrolysis

Basic salt > 7

Weak acid +

Weak base

CH3COOH +

NH3

CH3COONH4

Anion/Cation hydrolysis

Depends ?

Weak acid and Weak base

CH3COOH + NH3 → CH3COONH4

Acidicity depend on Ka and Kb

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

CH3COONH4 → CH3COO- + NH4+

NH4+ + H2O → NH3 + H3O

+

salt

anion cation

OH- - Basic H3O+ - Acidic Kb Ka

Ka = Kb

NEUTRAL

NH3 + HF → NH4F

salt

NH4F → NH4+ + F-

NH4+ + H2O → NH3 + H3O

+ F- + H2O → HF + OH-

cation anion

Ka H3O

+ - Acidic Kb OH- - Basic

Acidicity depend on Ka and Kb

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

Kb > Ka

BASIC

Weak acid

+

Weak base

Titration bet strong acid with strong base HCI + NaOH → NaCI + H2O

Titration curves Strong Acid with Strong Base

Click here titration simulation

NaOH M = 0.1M V = 0 ml

HCI M = 0.1M V = 25ml

7

HCI + NaOH → NaCI + H2O M = 0.1M M = 0.1M V = 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

HCI

M = 0.1M V = 25ml

NaOH M = 0.1M V = 25ml

HCI left → 25 ml, 0.1M Conc H+ = 0.1M

HCI

M = 0.1M V = 1ml left

NaOH M = 0.1M V = 24 ml add

HCI left → 1ml, 0.1M Mole H+ = (0.1 x 1)/1000 = 0.0001mol Conc H+ = Mole/Vol = 0.0001/0.049 = 0.002M

NaOH M = 0.1M V = 25ml add

HCI


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


NaOH left → 1ml left, 0.1M Moles OH-= (0.1 x 1)/1000 = 0.0001mol Conc OH- = Moles/Vol = 0.0001/0.051 = 0.002M

NaOH V = 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 base HCI + NH4OH → NH4CI + H2O

Titration curves Strong Acid with Weak Base

NH4OH M = 0.1M V = 0 ml

HCI M = 0.1M V = 25ml

5.3

HCI + NH4OH → NH4CI + H2O M = 0.1M M = 0.1M V = 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

HCI

M = 0.1M V = 25ml

NH4OH M = 0.1M V = 25ml

HCI left → 25 ml, 0.1M Conc H+ = 0.1M

HCI


NH4OH M = 0.1M V = 24 ml add

HCI left → 1ml, 0.1M Mole H+ = (0.1 x 1)/1000 = 0.0001mol Conc H+ = Mole/Vol = 0.0001/0.049 = 0.002M

NH4OH M = 0.1M V = 25ml add

HCI


Acidic Salt, NH4CI NH4

+ hydrolysis to produce H+

pH = 5.3


NH4OH left → 1ml left, 0.1M Moles OH-= (0.1 x 1)/1000 = 0.0001mol Conc OH- = Moles/Vol = 0.0001/0.051 = 0.002M Conc NH4CI = Moles/Vol = 2.5 x 10-3/0.051 = 0.05M

NH4OH V = 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 = 6.13 pH + pOH = 14 pH = 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 base CH3COOH + NaOH → CH3COONa + H2O

Titration curves Weak Acid with Strong Base

NaOH M = 0.1M V = 0 ml


9

CH3COOH + NaOH → CH3COONa + H2O M = 0.1M M = 0.1M V = 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

CH3COOH M = 0.1M V = 25ml


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

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

CH3COOH M = 0.1M V = 1ml left




Basic Salt (CH3COONa) CH3COO- hydrolysis produce OH-

pH = 9


NaOH V = 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, 0.1M Moles OH-= (0.1 x 1)/1000 = 0.0001mol Conc OH- = Moles/Vol = 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 + 1,37 pH = 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

NH4OH M = 0.1M V = 0 ml


7

CH3COOH + NH4OH → CH3COONH4 + H2O M = 0.1M M = 0.1M V = 25ml V = 25ml

7.8

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

2.87


NH4OH M = 0.1M V = 25ml



Neutral Salt CH3COONH4 pH = 7


NH4OH V = 1ml left

Total = 25 + 26

Vol = 51ml

7.8

6.11

Neutralization

Titration between weak base with weak acid CH3COOH + NH4OH → CH3COONH4 + H2O

6.11




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

(H+) = √Ka x CH3COOH


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


pH buffer – weak acid and salt

pH buffer region

weak acid and salt

NH4OH left → 1ml, 0.1M pH = 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

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

Titration bet strong acid with strong base HCI + NaOH → NaCI + H2O

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

Titration curves Acid with Base


11.3

2.7

Titration bet weak acid with weak base CH3COOH + NH4OH → CH3COONH4 + H2O


6.11

11.3 NaOH M = 0.1M V = 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

NH4OH M = 0.1M V = 25 ml

NH4OH M = 0.1M V = 25 ml

•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 due to buffering effect

HCI M = 0.1M V = 25ml

HCI M = 0.1M V = 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

Base Indicator

2.7

11.3

7.8

2.7

11.3

6.11

7.8

6.11

HCI M = 0.1M V = 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 Water Vol/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

HCI M = 0.1M V = 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

HCI M = 0.1M V = 0 ml

HCI M = 0.1M V = 25ml

7

HCI + NaOH → NaCI + H2O M = 0.1M M = 0.1M V = 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

NaOH M = 0.1M V = 1ml left

HCI M = 0.1M V = 24 ml add

NaOH left → 1ml, 0.1M Mole OH- = (0.1 x 1)/1000 = 0.0001mol Conc OH- = Mole/Vol = 0.0001/0.049 = 0.002M

HCI M = 0.1M V = 25ml add

NaOH M = 0.1M V = 0 ml left

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


HCI left → 1ml left, 0.1M Moles H+ = (0.1 x 1)/1000 = 0.0001mol Conc H+ = Moles/Vol = 0.0001/0.051 = 0.002M

HCI V = 1ml left

Total = 24 + 25

Vol = 49ml

Total = 25 + 26

Vol = 51ml


13

11.3

2.7

Neutralization

Titration bet strong base with strong acid HCI + 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

CH3COOH M = 0.1M V = 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

NaOH M = 0.1M V = 25.0ml



CH3COOH M = 0.1M V = 24 ml add

CH3COOH M = 0.1M V = 25ml add


Basic Salt (CH3COONa) CH3COO- hydrolysis to produce OH-

pH = 9


CH3COOH V = 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, 0.1M Moles OH- = (0.1 x 1)/1000 = 0.0001mol Conc OH- = Mole/Vol = 0.0001/0.049 = 0.002M


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 acid CH3COOH + NaOH → CH3COONa + H2O


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 = 0.1M V = 0 ml

HCI M = 0.1M V = 25ml

5.3

HCI + NH4OH → NH4CI + H2O M = 0.1M M = 0.1M V = 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

NH4OH M = 0.1M V = 25ml

NH4OH M = 0.1M V = 1ml left



Acidic Salt, NH4CI NH4

+ hydrolysis to produce H+

pH = 5.3


NH4OH left → 1ml left, 0.1M Mole NH4OH = (0.1 x 1)/1000 = 0.0001 mol Conc 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 = pKb -lg (base) (salt) pOH = 4.74 – lg (0.002) = 6.12 (0.048) pH + pOH = 14 pH = 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

0.1 OH- = √0.1 x 1.8 x 10-5 OH- = 1.34 x 10-3

pH buffer region

weak base + salt

HCI left → 1ml left, 0.1M Mole H+ = (0.1 x 1)/1000 = 0.0001mol Conc H+ = Mole/Vol = 0.0001/0.051 = 0.002M

Titration bet weak base with strong acid HCI + NH4OH → NH4CI + H2O


1.1187.214

87.2

)1034.1lg(

)lg(

3

pH

pOH

pOH

OHpOH

NH4OH M = 0.1M V = 25ml

HCI V = 1ml left

7.2

]002.0lg[

]lg[

pH

pH

HpH

Mole ratio

1: 1



Titration curves Weak Acid with Weak Base


CH3COOH M = 0.1M V = 25ml 7

CH3COOH + NH4OH → CH3COONH4 + H2O M = 0.1M M = 0.1M V = 25ml V = 25ml

6.11

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

11.1

NH4OH M = 0.1M V = 25ml


Neutral Salt CH3COONH4 pH = 7


Total = 25 + 26

Vol = 51ml

7.8

6.11

Neutralization


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

+ + OH-

Kb = (NH4+) (OH-)

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

0.1 OH- = √0.1 x 1.8 x 10-5 OH- = 1.34 x 10-3

Titration bet weak base with weak acid CH3COOH + NH4OH → CH3COONH4 + H2O

CH3COOH left → 1ml, 0.1M pH = 6.11

7.8


NH4OH left → 1ml left, 0.1M Mole NH4OH = (0.1 x 1)/1000 = 0.0001mol Conc NH4OH = Mole/Vol = 0.0001/0.049 = 0.002M Conc NH4CI = Mole/Vol = 2.4 x 10-3/0.049 = 0.048M

pOH = pKb -lg (base) (salt) pOH = 4.74 – lg (0.002) (0.048) pOH = 6.12 pH + pOH = 14 pH = 7.8

pH buffer – weak base + salt

pH buffer region

weak base + salt

NH4OH

M = 0.1M V = 25ml



1.1187.214

87.2

)1034.1lg(

)lg(

3

pH

pOH

pOH

OHpOH

CH3COOH V = 1ml left

Mole ratio

1: 1



Titration bet strong base with strong acid HCI + NaOH → NaCI + H2O

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

Titration curves Acid with Base

Titration bet weak base with strong acid HCI + NH4OH → NH4CI + H2O

11.3

2.7

Titration bet weak base with weak acid CH3COOH + NH4OH → CH3COONH4 + H2O


HCI M = 0.1M V = 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

NH4OH M = 0.1M V = 25 ml

HCI M = 0.1M V = 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 due to buffering effect

Buffer region form

• Slow gradual drop pH due to buffering effect

pH buffer region

weak acid + salt

pH buffer region

weak base + salt

NH4OH M = 0.1M V = 25 ml

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

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


CH3COOH + NaOH → CH3COONa + H2O Initial 0.0025 mol 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 base or 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)


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


NaOH M = 0.1M V = 12.5 ml add

CH3COOH M = 0.1M V = 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 + H2O Initial 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 acid or 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 + H2O M = 0.1M M = 0.1M V = 25ml V = 25ml

Titration bet weak base + strong acid NH4OH + 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 acid NH4OH + 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


HCI M = 0.1M V = 12.5 ml add

NH4OH M = 0.1M V = 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 acid salt

Dilution Add water

pH buffer

pH will not change

Temperature affect pH

pH change

Strong acid Strong base

Titration Acid Base

Strong acid Weak base

Weak acid Strong base

Weak acid Weak base

Neutralization Titration curve

Base

Acid Indicator

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

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

Concept Map

Strong acid Strong base

Titration Acid Base

Strong acid Weak base

Weak acid Strong base

Weak acid Weak base

Neutralization Titration curve

Base

Acid Indicator 2.7

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 pt Stoichiometric pt

Point of inflection


pH buffer region

weak acid and salt

6.11

11.3

9

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

25ml 12.5ml

Half Equivalent pt

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

Indicator change

colour


25ml

Equivalent pt Stoichiometric pt

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

pH = pKa 6.3

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 + H2O M = 0.1M M = 0.1M V = 25 ml V = ? ml

Titration between Strong Acid with Strong Base

HCI + NaOH → NaCI + H2O M = 0.1M M = 0.1M V = 25 ml V = 25 ml

HCI M = 0.1M V = 25 ml

NaOH M = 0.1M V = ? 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

NaOH M = 0.1M V = 25 ml

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 + H2O M = 0.1M M = 0.1M V = ? ml V = 25 ml

Titration between Strong Acid with Strong Base

HCI + NaOH → NaCI + H2O M = 0.1M M = 0.1M V = 25 ml V = 25 ml

HCI M = 0.1M V = 25 ml

NH4OH M = 0.1M V = 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

NaOH M = 0.1M V = 25 ml

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?

HCI M = 0.1M V = ? 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- ↔ H2O NH3 + 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 + H2O M = 0.1M M = 0.1M V = 25 ml V = ?

CH3COOH + NH4OH → CH3COONH4 + H2O M = 0.1M M = 0.1M V = 25ml V = ?

CH3COOH + NaOH → CH3COONa + H2O M = 0.1M M = 0.1M V = 25ml V = ?

Weak base with Strong acid Weak base with Weak acid

HCI M = 0.1M V = 25ml

HCI + NH4OH → NH4CI + H2O M = 0.1M M = 0.1M V = 25 ml V = ?


HCI M = 0.1M V = 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 ACID WEAK ACID

Mole ratio – 1: 1 Mole ratio – 1: 1

Mole ratio – 1: 1 Mole ratio – 1: 1

NaOH M = 0.1M V = ? STRONG BASE STRONG BASE

NaOH M = 0.1M V = ?

Strong base with Weak acid

NH4OH M = 0.1M V = ? WEAK BASE

NH4OH M = 0.1M V = ?

IB Buffer Calculation

Find pH buffer , titration by adding 18ml, 0.10M HCI to 32ml, 0.10M NH3 pKb = 4.75 1 Titration bet strong acid with weak base NH3 + HCI → NH4CI + H2O

Buffer region Weak base + salt

NH3 + HCI → NH4CI + H2O Initial 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 = 4.86 pH + pOH = 14 pH = 9.14


Strong acid 18ml, 0.1M HCI add

Weak base

32ml, 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 = 4.86 pH + pOH = 14 pH = 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 acid NaOH + CH3COOH → CH3COONa + H2O



Amt base = Amt salt

Buffer Calculation

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

NaOH + CH3COOH → CH3COONa H2O Initial 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 acid

100ml, 0.1M HCI


Titration bet strong base with strong acid NaOH + HCI → NaCI + H2O

pH calculation

NaOH + HCI → NaCI + H2O Initial 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 0.10 H+ = 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 M 4

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 M 5

NH3 + H2O ↔ NH4+ + OH-

Kb = (NH4+) (OH-)

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

0.50 OH- = √0.50 x 1.8 x 10-5 OH- = 3.0 x 10-3

pOH = -lg 3.0 x 10-3 pOH = 2.52 pH = 14 – 2.52 pH = 11.48

0.50M NH3

Base Dissociation constant

Using 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

OHCNKa

3

HSO4- + H2O ↔ SO4

2- + H3O+

4

3

2

4

HSO

OHSOKa

Ka – Highest – Strongest acid Ka – 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 = 9 C > A > B > D > E

A. 0.3 mol NH3 and 0.3 mol HCI B. 0.3 mol NH3 and 0.15 mol HCI C. 0.3 mol NH3 and 0.6 mol HCI D. 0.3 mol NH3 and 0.15 mol H2SO4

I. 50ml, 0.1M CH3COONa II. 25ml, 0.1M NaOH III. 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 acid II. Sodium ethanoate III. Sodium hydroxide

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

Sample IB Question on Acid Base Titration

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

HA M = ? M V = 25.0ml

NaOH M = 0.155M V = 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 acid Moles Acid = MV = M x 0.025 M x 0.025 = 3.41 x 10-3

M = 0.136M

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

5.3 = -lg Ka

Ka = 5 x 10-6

b) Neutralization - strong base with weak acid HA + NaOH → A- + H2O A- is a strong conjugate base A- + 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 stepwise 1st H+ dissociation - H3PO4 ↔ H+ + H2PO4

-

2nd H+ dissociation – H2PO4

- ↔ H+ + HPO42-

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

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Simulation and Animation on Buffer and Titrations

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

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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

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

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IB Chemistry on Titration Curves between Acids and Bases

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Transcript of IB Chemistry on Titration Curves between Acids and Bases