IB Chemistry on Titration Curves between Acids and Bases

1. 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

2. 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+ - AcidicKb 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

3. Titration bet strong acid with strong base
HCI + NaOH → NaCI + H2O
TitrationcurvesStrong 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
M = 0.1M
V = 0ml left
Neutral Salt, NaCI
Conc H+ = 1 x 10-7M
(Dissociation of water)
NaOH
M = 0.1M
V = 26ml add
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
HpH
7
]101lg[
]lg[
7





pH
pH
HpH
3.117.214
7.2
)002.0lg(
)lg(



 
pH
pOH
pOH
OHpOH
7

4. Titration bet strong acid with weak base
HCI + NH4OH → NH4CI + H2O
TitrationcurvesStrong 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
M = 0.1M
V = 1ml left
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
M = 0.1M
V = 0ml left
Acidic Salt, NH4CI
NH4
+ hydrolysis to
produce H+
pH = 5.3
NH4OH
M = 0.1M
V = 26ml add
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
Click here titration simulation
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

5. Titration between weak acid with strong base
CH3COOH + NaOH → CH3COONa + H2O
TitrationcurvesWeak Acid with Strong Base
NaOH
M = 0.1M
V = 0 ml
NaOH
M = 0.1M
V = 25ml
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
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
NaOH
M = 0.1M
V = 24 ml add
NaOH
M = 0.1M
V = 25ml add
CH3COOH
M = 0.1M
V = 0ml left
Basic Salt (CH3COONa)
CH3COO- hydrolysis produce OH-
pH = 9
NaOH
M = 0.1M
V = 26ml add
NaOH
V = 1ml left
Total = 24 + 25
Vol = 49ml
Total = 25 + 26
Vol = 51ml
Click here titration simulation
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

6. TitrationcurvesWeak Acid with Weak Base
NH4OH
M = 0.1M
V = 0 ml
CH3COOH
M = 0.1M
V = 25ml
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
CH3COOH
M = 0.1M
V = 25ml
NH4OH
M = 0.1M
V = 25ml
NH4OH
M = 0.1M
V = 25ml add
CH3COOH
M = 0.1M
V = 0ml left
Neutral Salt
CH3COONH4
pH = 7
NH4OH
M = 0.1M
V = 26ml add
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
NH4OH
M = 0.1M
V = 24ml add
CH3COOH
M = 0.1M
V = 1ml left
Click here titration simulation
CH3COOH left → 25 ml, 0.1M
CH3COOH ↔ (CH3COO- )(H+)
Ka = (CH3COO-) (H+)
CH3COOH
(H+) = √Ka x CH3COOH
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 CH3COONH4 = Moles/Vol
= 2.4 x 10-3/0.049
= 0.048M
pH = pKa -lg [acid]
[salt]
pH = 4.74 – lg [2.04 x 10-3]
[0.048]
pH = 4.74 + 1,37
pH = 6.11
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

7. Titration bet strong acid with strong base
HCI + NaOH → NaCI + H2O
Titration bet weak acid with strong base
CH3COOH + NaOH → CH3COONa + H2O
Titrationcurves Acid with Base
Titration bet strong acid with weak base
HCI + NH4OH → NH4CI + H2O
11.3
2.7
Titration bet weak acid with weak base
CH3COOH + NH4OH → CH3COONH4 + H2O
NaOH
M = 0.1M
V = 25ml
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
CH3COOH
M = 0.1M
V = 25ml
CH3COOH
M = 0.1M
V = 25ml

8. 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?
3
105.2..
025.01.0..
..






HMole
HMole
VMHMole
1
]1.0lg[
]lg[


 
pH
pH
HpH
055.0......
045.0
105.2
..
3






HConc
Volume
Mole
HConc
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
3
105.2..
025.01.0..
..






HMole
HMole
VMHMole
1
]1.0lg[
]lg[


 
pH
pH
HpH
Mole change
After adding NaOH
3
105.0... 
leftHMole
011.0.
045.0
105.0
..
3






HConc
Volume
Mole
HConc
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?

9. TitrationcurvesStrong 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
M = 0.1M
V = 25ml
NaOH
M = 0.1M
V = 25ml
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
M = 0.1M
V = 26ml add
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
Click here titration simulation
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
OHpOH
7
]101lg[
]lg[
7





pH
pH
HpH
7.2
]002.0lg[
]lg[


 
pH
pH
HpH
Mole ratio
1: 1

10. TitrationcurvesWeak Acid with Strong Base
CH3COOH
M = 0.1M
V = 0 ml
NaOH
M = 0.1M
V = 25ml
9
CH3COOH + NaOH → CH3COONa + H2O
M = 0.1M M = 0.1M
V = 25ml V = 25ml
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 = 25ml
NaOH
M = 0.1M
V = 1ml left
CH3COOH
M = 0.1M
V = 24 ml add
CH3COOH
M = 0.1M
V = 25ml add
NaOH
M = 0.1M
V = 0ml left
Basic Salt (CH3COONa)
CH3COO- hydrolysis to
produce OH-
pH = 9
CH3COOH
M = 0.1M
V = 26ml add
CH3COOH
V = 1ml left
Total = 24 + 25
Vol = 49ml
Total = 25 + 26
Vol = 51ml
Click here titration simulation
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
CH3COOH left → 1ml, 0.1M
Mole CH3COOH = (MV)/1000
= (1 x 0.1)/1000
= 0.0001mol
Conc CH3COOH = Mole/Vol
= 0.0001/0.051
= 1.96 x 10-3
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 = pKa -lg [acid]
[salt]
pH = 4.74 – lg [1.96 x 10-3]
[0.049]
pH = 4.74 + 1,39
pH = 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

11. TitrationcurvesStrong 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
HCI
M = 0.1M
V = 24 ml add
HCI
M = 0.1M
V = 25ml add
Acidic Salt, NH4CI
NH4
+ hydrolysis to
produce H+
pH = 5.3
HCI
M = 0.1M
V = 26ml add
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
Click here titration simulation
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
NH4OH
M = 0.1M
V = 0ml left
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

12. TitrationcurvesWeak Acid with Weak Base
CH3COOH
M = 0.1M
V = 0 ml
CH3COOH
M = 0.1M
V = 25ml7
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
CH3COOH
M = 0.1M
V = 25ml add
Neutral Salt
CH3COONH4
pH = 7
CH3COOH
M = 0.1M
V = 26ml add
Total = 25 + 26
Vol = 51ml
7.8
6.11
Neutralization
Click here titration simulation
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
CH3COOH
M = 0.1M
V = 24ml add
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
NH4OH
M = 0.1M
V = 1ml left
NH4OH
M = 0.1M
V = 0ml left
1.1187.214
87.2
)1034.1lg(
)lg(
3






pH
pOH
pOH
OHpOH
CH3COOH
V = 1ml left
Mole ratio
1: 1

13. Titration bet strong base with strong acid
HCI + NaOH → NaCI + H2O
Titration bet strong base with weak acid
CH3COOH + NaOH → CH3COONa + H2O
Titrationcurves 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
NaOH
M = 0.1M
V = 25ml
HCI
M = 0.1M
V = 25ml
6.13
11.3
NaOH
M = 0.1M
V = 25ml
CH3COOH
M = 0.1M
V = 25ml
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
CH3COOH
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

14. Acidic Buffer Region CH3COOH (acid)/CH3COO-
(salt)
Titration bet weak acid + strong base
CH3COOH + NaOH → CH3COONa + H2O
Click here buffer simulation
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 equivalencepoint :
• 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 equivalencept 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
CH3COOH + NaOH → CH3COONa + H2O
M = 0.1M M = 0.1M
V = 25ml V = 25ml
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 = 25 ml add
NaOH
M = 0.1M
V = 12.5 ml add
CH3COOH
M = 0.1M
V = 12.5ml left
CH3COOH
M = 0.1M
V = 0ml 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

15. 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 equivalencepoint :
• 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
Click here buffer simulation
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
Buffer region at half
equivalence point
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 = 25 ml add
HCI
M = 0.1M
V = 12.5 ml add
NH4OH
M = 0.1M
V = 12.5ml left
NH4OH
M = 0.1M
V = 0ml 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

16. Click here view buffering
Concept Map Buffer
pH
Proton availability Stable
Buffer solution
Weak acid ↔ Conjugate base
][
][
lg
salt
acid
pKpH 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

17. 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
Stoichiometricpt
Point of inflection
Titration bet weak acid with strong base
CH3COOH + NaOH → CH3COONa + H2O
pH buffer region
weak acid and salt
6.11
11.3
9
• Amt acid = Amt base
• Vol at equivalencept = 25ml
25ml12.5ml
Half Equivalent pt
• Amt acid = Amt salt
• Vol is = 12.5ml
• Buffer region form
Indicator change
colour
Titration bet strong acid with weak base
HCI + NH4OH → NH4CI + H2O
25ml
Equivalent pt
Stoichiometric pt
• Amt acid = Amt base
• Vol at equivalencept = 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

18. CH3COOH + NaOH → NaCI + H2O
M = 0.1M M = 0.1M
V = 25 ml V = ? ml
Titration betweenStrong 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?
CH3COOH
M = 0.1M
V = 25 ml
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
5
108.1 
aK
14
100.1 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

19. HCI + NH4OH → NH4CI + H2O
M = 0.1M M = 0.1M
V = ? ml V = 25 ml
Titration betweenStrong 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
5
108.1 
bK
14
100.1 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

20. 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 = ?
CH3COOH
M = 0.1M
V = 25ml
HCI
M = 0.1M
V = 25ml
CH3COOH
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 = ?

21. 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 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
Click here buffer simulation
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 3
108.1
1000
1.00.18 



molVMmole 3
102.3
1000
1.032 




22. 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
Click here buffer simulation
Buffer region at half
equivalence point
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
Total vol = 150 ml or 0.15 dm3
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 3
105
1000
1.050 



molVMmole 3
1010
1000
1.0100 



Strong base
50ml, 0.1M NaOH add
Weak acid
100ml, 0.1M CH3COOH

23. 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
Click here buffer simulation
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
Total vol = 150 ml or 0.15 dm3
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 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
0.50
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 3
105
1000
1.050 



molVMmole 3
1010
1000
1.0100 




24. 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
HCOOCH
Ka
3
3


HCN + H2O ↔ CN- + H3O+
  
 HCN
OHCN
Ka

 3
HSO4
- + H2O ↔ SO4
2- + H3O+
  
 


4
3
2
4
HSO
OHSO
Ka
Ka – Highest – Strongest acidKa – Lowest – Weakest acid
C2H5NH2 + H2O ↔ C2H5NH3
+ + OH-
  
 252
352
NHHC
OHNHHC
Kb


NH3 + H2O ↔ NH4
+ + OH-
  
 3
4
NH
OHNH
Kb


N2H4 + H2O ↔ N2H5
+ + OH-
  
 42
52
HN
OHHN
Kb


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 acidity8
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

25. 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 followingto 50ml, 0.1M CH3COOH10
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

26. 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 Questionon Acid Base Titration
a) pH is = 9
At equivalencepoint
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
PolyproticAcid – dissociate1 protonstepwise
1st H+ dissociation - H3PO4 ↔ H+ + H2PO4
-
2nd H+ dissociation – H2PO4
- ↔ H+ + HPO4
2-
3rd H+ dissociation – HPO4
2- ↔ 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

27. Click here on titration animationClick here on titration simulation
Simulation and Animation on Buffer and Titrations
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