1. Strong acids such as HI, HBr, HCl, and HNO3 ionize completely in water, yielding high concentrations of hydrogen ions and high electrical conductivity. Weak acids such as acetic acid and carbonic acid ionize only partially. 2. Strong bases such as LiOH, NaOH, and Ca(OH)2 contain hydroxide ions that fully dissociate in water. Weak bases such as ammonia and amines contain electron-rich nitrogen and ionize only partially. 3. The pH of a solution indicates its acidity, with pH less than 7 being acidic and pH greater than 7 being alkaline. A change of one pH unit corresponds to a ten

1. Strong/Weak Acid and Base
Strong Acid/Weak Acid
Strong acid - HI, HBr, HCI, HNO3, H2SO4, HCIO3, HCIO4
Weak Acid - CH3COOH, HF, HCN, H2CO3, H3BO3, H3PO4
Strong Base/ Weak Base
Strong base - LiOH, KOH, NaOH, CsOH, Ca(OH)2
Weak Base - NH3, C2H5NH2, (CH3)2NH, C3H5O2NH2
Distinguishbet strong and weak acid
ElectricalconductivityRate of rxn pH
Strongacid
Strong acid → High ionization → High conc H+ → High conductivity→ High rate rxn → Lower pH
Strong acid
Oxoacid
O atom > number ionizable proton
HNO3, H2SO4, HCIO3,HCIO4
Hydrohalicacid
HI, HBr, HCI
Weak acid
Hydrohalicacid
HF
Oxoacid
O atom ≥ number ionizable protonby 1
HCIO, HNO2, H3PO4
Carboxylicacid
COOH
Strong base – containOH- or O2-
LiOH, NaOH, CaO, K2O Ca(OH)2, Ba(OH)2
Weak base – contain electronrich nitrogen, N
NH3, C2H5NH2, (CH3)2NH, C3H5O2NH2
Strong base Weak base
1 2 3
Weak acid
0.1 M HCI 0.1 M CH3COOH
H+ 0.1 mole 0.0013 mole
pH 1 (Low) 2.87 (High)
Electrical conductivity High (Ionize completely) Low (Ionize partially)
Rate with magnesium Fast Slow
Rate with calcium
carbonate
Fast Slow
Weaker acid → Low ionization → Low conc H+ → Low conductivity→ Low rate rxn → High pH
Strong acid
HA A-H+
H+ H+
H+
H+ H+
H+
H+A-
A-
A-
A- A-
A-
Ionizes completely
Weak acid
HA
HA
H+ A-
H+
H+
A-
A-
HA
HA
HA
HA
HA
HA
Ionizes partially

2. Easier using pH scale than Conc [H+]
• Conc H+ increase 10x from 0.0001(10-4) to 0.001(10-3)
- pH change by 1 unit from pH 4 to 3
• pH 3 is (10x) more acidic than pH 4
• 1 unit change in pH is 10 fold change in Conc [H+]
Conc OH-
increase ↑ by 10x
pH increase ↑ by 1 unit
pOH with Conc OH-
pOH = -log [OH-
]
[OH-
] = 0.0000001M
pOH = -log [0.0000001]
pOH = -log1010-7
pOH = 7
pH + pOH = 14
pH + 7 = 14
pH = 7 (Neutral)
pH with Conc H+
pH = -log [H+
]
[H+
] = 0.0000001M
pH = -log [0.0000001]
pH = -log1010-7
pH = 7 (Neutral)
Conc H+
increase ↑ by 10x
pH decrease ↓ by 1 unit
pH measurement of Acidity of solution
• pH is the measureof acidity of solutionin logarithmicscale
• pH = powerof hydrogenor minuslogarithmto base ten of hydrogenion concentration
← Acidic – pH < 7 Alkaline – pH > 7 →
pOH with Conc OH-
pOH = -log [OH-
]
[OH-
] = 0.1M
pOH = -log[0.1]
pOH = 1
pH + pOH = 14
pH + 1 = 14
pH = 13 (Alkaline)
pH with Conc H+
pH = -log [H+
]
[H+
] = 0.01M
pH = -log [0.01]
pH = -log1010-2
pH = 2 (Acidic)
Easier pH scaleConc H+

3. Formula for acid/basecalculation
[OH-][H+]
Kw = [H+] x [OH-] = 1 x 10-14
[OH-] = 10-pOHpOH = -lg [OH-]
pOHpH
pH = -lg [H+] [H+] = 10-pH
pH + pOH = 14
Formula for acid/basecalculation
DissociationConstant for Weak Acid
pH = -log10[H+]
pOH = -log10[OH-]
pH + pOH = 14
pH + pOH = pKw
Kw = [H+][OH-]
Ka x Kb = Kw
Ka x Kb = 1 x 10-14
pKa = - lg10Ka
pKb = - lg10Kb
pKa + pKb = pKw
pKa + pKb = 14

 AHHA
  
 HA
AH
Ka



 HCOOCHCOOHCH 33
  
 
 
 COOHCH
H
COOHCH
HCOOCH
Ka
3
2
3
3


DissociationConstant for Weak Base

 OHBHOHB 2
  
 B
OHBH
Kb



 OHNHOHNH 423
  
 
 
 3
2
3
4
NH
OH
NH
OHNH
Kb



 OHCOOCHOHCOOHCH 3323

 OHCOOCHOHCOOHCH 3323
  
 COOHCH
OHCOOCH
Ka
3
33



 OHCOOHCHOHCOOCH 323
  
 


COOCH
OHCOOHCH
Kb
3
3
Derive Ka x Kb = Kw
Relationship bet Weak acid and its conjugate base
Weak acid Conjugate Base
  
 
  
 


COOCH
OHCOOHCH
COOHCH
OHCOOCH
3
3
3
33
  
 
  
    


 OHOH
COOCH
OHCOOHCH
COOHCH
OHCOOCH
3
3
3
3
33
wba KKK 

4. Formula for acid/basecalculation
Ka /Kb measureequilibriumposition
Ka/Kb large ↑ – ↑ dissociation– shift to right – favour product
Ka/Kb large ↑ – pKa /pKb small ↓ – Strongeracid/base
Strongacid
Large ↑ Ka
Weak acid
Small ↓ Ka
Strongbase
Large ↑ Kb
Weak base
Small ↓Kb
↑ Ka → ↓ pKa
Ka /Kb measureequilibriumposition
Ka /Kb small ↓ – ↓ dissociation– shift to left – reactant favour
Ka /Kb small ↓ – pKa /pKb high ↑– Weak acid/base
↑ Kb → ↓ pKb
↓ Ka → ↑ pKa
↓ Kb →↑ pKb
For weak acid/ base

 CIHHCI 
 OHNHOHNH 423
Shift right Shift left
CH3COOH + H2O ↔ CH3COO- + H3O+
CH3COOH CH3COO-CH3COOH ↔ CH3COO-
Strong Acid Weak conjugate BaseConjugate acid base pair
Small dissociation
constant
Strong Acid Weak base
ba KK /
Strongacid
Strongbase

5. Formula for acid/basecalculation
[OH-][H+]
Kw = [H+] x [OH-] = 1 x 10-14
[OH-] = 10-pOHpOH = -lg [OH-]
pOHpH
pH = -lg [H+] [H+] = 10-pH
pH + pOH = 14
Formula for acid/basecalculation
DissociationConstant for Weak Acid
pH = -log10[H+]
pOH = -log10[OH-]
pH + pOH = 14
pH + pOH = pKw
Kw = [H+][OH-]
Ka x Kb = Kw
Ka x Kb = 1 x 10-14
pKa = - lg10Ka
pKb = - lg10Kb
pKa + pKb = pKw
pKa + pKb = 14

 AHHA
  
 HA
AH
Ka



 HCOOCHCOOHCH 33
  
 
 
 COOHCH
H
COOHCH
HCOOCH
Ka
3
2
3
3


DissociationConstant for Weak Base

 OHBHOHB 2
  
 B
OHBH
Kb



 OHNHOHNH 423
  
 
 
 3
2
3
4
NH
OH
NH
OHNH
Kb


Dissociatepartially ↔ used
Weak acid/base
Ka /Kb value pKa /pKb value easier!
Click here weak acid dissociation Click here weak acid dissociation Click here CH3COOH dissociation Click here strong acid ionization
Weak acid/base Animation

6. NH3 ↔ NH4
+
Buffer Solution
Acid part
Neutralize
each other
Salt part
Base part
- NH3(weakbase) + NH4CI (salt)
- NH3 + H2O ↔ NH4
+ + OH− → NH3 moleculeneutralise added H+
- NH4CI → NH4
+ + CI− → NH4
+ neutralise added OH−
- Effectivebuffer equal amt weak base NH3 and conjugate acid NH4
+
Acidic Buffer Basic Buffer
Resist a change in pH when small amt acid/base is added.
CH3COOH + H2O ↔ CH3COO- + H3O+
Acidic Buffer - weak acid and its salt/conjugatebase
CH3COOH ↔ CH3COO-
Conjugate acid base pair
CH3COOH CH3COO-
Weak Acid Conjugate Base
BUFFER
Dissociate fully

 HCOOCHCOOHCH 33
COOHCH3 COONaCH3

 NaCOOCHCOONaCH 33
Dissociate partially
- CH3COOH (weakacid) + CH3COONa (salt)
- CH3COOH ↔ CH3COO- + H+ → CH3COOH neutraliseadded OH−
- CH3COONa → CH3COO- + Na+ → CH3COO- neutraliseadded H+
- Effectivebuffer equal amt weak acid CH3COOH and base CH3COO-
COOHCH3

COOCH3
BUFFER
Add acid H+Add alkaline OH-
Neutralize
each other
Basic buffer - weak base and its salt/conjugateacid

 OHNHOHNH 423
NH3 + H2O ↔ NH4
+ + OH-
NH3
Weak Base
NH4
+
Conjugate acid
CINH43NH
BUFFER
Conjugate acid base pair
Add acid H+ Add alkaline OH-
Neutralize
each other
Neutralize
each other
Dissociate partially

 CINHCINH 44
3NH

4NH
Base part
Salt part
Acid part
Dissociate fully
BUFFER

7. How to prepareacidic/ basic buffer
Acid Dissociationconstant
CH3COOH + H2O ↔ CH3COO-
+ H3O+
Ka = (CH3COO-
) (H3O+
)
(CH3COOH)
-lgKa = -lgH+ -lg (CH3COO-)
(CH3COOH)
-lgH+ = -lg Ka + lg (CH3COO-)
(CH3COOH)
pH = pKa + lg (CH3COO-)
(CH3COOH)
Acidic BufferFormula
• Mixture Weak acid + Salt/Conjugatebase
• CH3COOH ↔ CH3COO-
+ H+
(dissociate partially)
• CH3COONa → CH3COO-
+ Na+
(dissociatefully)
Basic BufferFormula
• Mixture Weak base + Salt/Conjugateacid
• NH3 + H2O ↔ NH4
+
+ OH_
(dissociate partially)
• NH4CI → NH4
+
+ CI_
(dissociate fully)
pH = pKa - lg (acid)
(salt)
pH = pKa + lg (salt)
(acid)
Base Dissociationconstant
NH3 + H2O ↔ NH4
+
+ OH-
Kb = (NH4
+
) (OH-
)
(NH3)
-lgKb = -lgOH- -lg (NH4
+)
(NH3)
-lgOH- = -lgKb + lg (NH4
+)
(NH3)
pOH = pKb + lg (NH4
+)
(NH3)
pOH = pKb + lg (salt)
(base)
pOH = pKb - lg (base)
(salt)
Basic BufferAcidic Buffer
salt salt
acid base
Henderson Hasselbalch Equation
multiply -lg
both sides
Henderson Hasselbalch Equation

8. Basic Buffer PreparationAcidic Buffer Preparation
Prepare Acidic Buffer pH = 5.2
• Choose pKa acid closest to pH 5.2
• pKa = 4.74 (ethanoic acid) chosen
• pH = pKa -lg [acid]
[salt]
• 5.2 = 4.74 – lg [acid]
[salt]
• [acid] = 0.35
[salt]
Ratio of [acid] = 0.35
[salt]
Use same conc acid/salt but different vol ratio
• 1M, 35ml (acid) = 0.35 or 0.1M, 35ml (acid) = 0.35
1M, 100ml (salt) 0.1M, 100ml (salt)
Use same vol acid/salt but different conc ratio
• 3.5M, 10ml (acid) = 0.35 or 0.35M, 10ml (acid) = 0.35
10M, 10ml (salt) 1M, 10ml (salt)
Buffer capacity
• Adding water will not change the pH of acidic buffer
• Ratio of acid/salt still the same
• Ka acid remain same
Prepare Basic Buffer pH = 9.5 or pOH = 4.5
• Choose pKb base closest to pOH = 4.5
• pKb = 4.74 (NH3) chosen
• pOH = pKb -lg [base]
[salt]
• 4.5 = 4.74 – lg [base]
[salt]
• [base] = 1.74
[salt]
Ratio of [base] = 1.74
[salt]
Use same conc base/salt but different vol ratio
• 1M, 174ml (base) = 1.74 or 0.1M, 174ml (base) = 1.74
1M, 100ml (salt) 0.1M, 100ml (salt)
Use same vol base/salt but different conc ratio
• 1.74M, 10ml (base) = 1.74 or 0.174M, 10ml (base) = 1.74
1M, 10ml (salt) 0.1M, 10ml (salt)
Buffer capacity
• Adding water will not change the pH of basic buffer
• Ratio of base/salt still the same
• Kb base remainsame
Buffer solution
Buffer Preparation
1 1
2 2
3 Use fix vol, 1dm3 and use differentmole ratio (Acid/salt)
• 0.35 mole acid + 1 mole salt to 1 dm3 solvent= 0.35
Use fix vol, 1dm3 and use differentmole ratio (base/salt)
• 1.74 mole base + 1 mole salt to 1 dm3 solvent= 1.74
3
3 ways to prepare buffer 3 ways to prepare buffer

9. Basic Buffer PreparationAcidic Buffer Preparation
Prepare Acidic Buffer pH = 5.2
• Choose pKa acid closest to pH 5.2
• pKa = 4.74 (ethanoic acid) chosen
• pH = pKa -lg [acid]
[salt]
• 5.2 = 4.74 – lg [acid]
[salt]
• [acid] = 0.35
[salt]
Ratio of [acid] = 0.35
[salt]
Use same conc acid/salt but different vol ratio
BufferA BufferB
• 1M, 35ml (acid) = 0.35 or 0.1M, 35ml (acid) = 0.35
1M, 100ml (salt) 0.1M, 100ml (salt)
Prepare Basic Buffer pH = 9.5 or pOH = 4.5
• Choose pKb base closest to pOH = 4.5
• pKb = 4.74 (NH3) chosen
• pOH = pKb -lg [base]
[salt]
• 4.5 = 4.74 – lg [base]
[salt]
• [base] = 1.74
[salt]
Ratio of [base] = 1.74
[salt]
Use same conc base/salt but different vol ratio
Buffer A Buffer B
• 1M, 174ml (base) = 1.74 or 0.1M, 174ml (base) = 1.74
1M, 100ml (salt) 0.1M, 100ml (salt)
Buffer solution
Buffering Capacity
1 1
1M, 35ml
(acid)
1M, 100ml
(salt)
0.1M, 35ml
(acid)
0.1M, 100ml
(salt)
BA
1M, 174ml
(base)
1M, 100ml
(salt)
0.1M, 174ml
(base)
0.1M, 100ml
(salt)
BA
Buffer A > Buffer B
Stronger buffering capacity
• Amt of acid/salt higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff
• Higher buffer conc – Higher buffering capacity
Buffer A > BufferB
Stronger buffering capacity
• Amt of base/salt higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff
• Higher buffer conc – Higher buffering capacity
Which has greater buffering capacity? Which has greater buffering capacity?

10. Basic Buffer PreparationAcidic Buffer Preparation
Prepare Acidic Buffer pH = 5.2
• Choose pKa acid closest to pH 5.2
• pKa = 4.74 (ethanoic acid) chosen
• pH = pKa -lg [acid]
[salt]
• 5.2 = 4.74 – lg [acid]
[salt]
• [acid] = 0.35
[salt]
Ratio of [acid] = 0.35
[salt]
Prepare Basic Buffer at pH = 9.5 or pOH = 4.5
• Choose pKb base closest to pOH = 4.5
• pKb = 4.74 (NH3) chosen
• pOH = pKb -lg [base]
[salt]
• 4.5 = 4.74 – lg [base]
[salt]
• [base] = 1.74
[salt]
Ratio of [base] = 1.74
[salt]
Buffer solution
Buffering Capacity
2 2
3.5M, 10ml
(acid)
10M, 10ml
(salt)
0.35M, 10ml
(acid)
1M, 10ml
(salt)
BA
1.74M, 10ml
(base)
1M, 10ml
(salt)
0.174M, 10ml
(base)
0.1M, 10ml
(salt)
BA
Use same vol acid/salt but different conc ratio
BufferA BufferB
• 3.5M, 10ml (acid) = 0.35 or 0.35M, 10ml (acid) = 0.35
10M, 10ml (salt) 1M, 10ml (salt)
Use same vol base/salt but different conc ratio
BufferA BufferB
• 1.74M, 10ml (base) = 1.74 or 0.174M, 10ml (base) = 1.74
1M, 10ml (salt) 0.10M, 10ml (salt)
Which has greater buffering capacity? Which has greater buffering capacity?
Buffer A > Buffer B
Stronger buffering capacity
• Amt of acid/salt higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff
• Higher buffer conc – Higher buffering capacity
Buffer A > BufferB
Stronger buffering capacity
• Amt of base/salt higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff
• Higher buffer conc – Higher buffering capacity

11. Basic Buffer PreparationAcidic Buffer Preparation
Buffer solution
Buffering Capacity
3 3
0.35mol
(acid )
1mol
(salt)
0.035mol
(acid)
0.10mol
(salt)
BA
1.74mol
(base)
1mol
(salt)
0.174mol
(base)
0.1mol
(salt)
BA
Use fix vol, 1dm3 but diff mole ratio (acid/salt)
Buffer A Buffer B
• 0.35mol (acid) = 0.35 or 0.035mol (acid) = 0.35
1mol (salt) 0.1mol (salt)
1dm3 1dm3 1dm3 1dm3
Use fix vol, 1dm3 but diff mole ratio (base/salt)
Buffer A Buffer B
• 1.74mol (base) = 1.74 or 0.174mol (base) = 1.74
1mol (salt) 0.1mol (salt)
Which has greater buffering capacity? Which has greater buffering capacity?
Prepare Acidic Buffer pH = 5.2
• Choose pKa acid closest to pH 5.2
• pKa = 4.74 (ethanoic acid) chosen
• pH = pKa -lg [acid]
[salt]
• 5.2 = 4.74 – lg [acid]
[salt]
• [acid] = 0.35
[salt]
Ratio of [acid] = 0.35
[salt]
Prepare Basic Buffer at pH = 9.5 or pOH = 4.5
• Choose pKb base closest to pOH = 4.5
• pKb = 4.74 (NH3) chosen
• pOH = pKb -lg [base]
[salt]
• 4.5 = 4.74 – lg [base]
[salt]
• [base] = 1.74
[salt]
Ratio of [base] = 1.74
[salt]
Buffer A > Buffer B
Stronger buffering capacity
• Amt of acid/salt higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff
• Higher buffer conc – Higher buffering capacity
Buffer A > BufferB
Stronger buffering capacity
• Amt of base/salt higher to neutralise added H+ or OH-
• Ratio acid/salt same, pH buffer same but buffering capacity diff
• Higher buffer conc – Higher buffering capacity

12. Basic Buffer PreparationAcidic Buffer Preparation
Prepare Acidic Buffer at pH = 5.2
• Choose pKa acid closest to pH 5.2
• pKa = 4.74 (ethanoic acid) chosen
• pH = pKa -lg [acid]
[salt]
• 5.2 = 4.74 – lg [acid]
[salt]
• [acid] = 0.35
[salt]
Ratio of [acid] = 0.35
[salt]
Prepare Basic Buffer pH = 9.5 or pOH = 4.5
• Choose pKb base closest to pOH = 4.5
• pKb = 4.74 (NH3) chosen
• pOH = pKb -lg [base]
[salt]
• 4.5 = 4.74 – lg [base]
[salt]
• [base] = 1.74
[salt]
Ratio of [base] = 1.74
[salt]
Buffer solution
Buffering Capacity
4 4
Will pH change by adding water?
pH BufferA = pH Buffer B
• Same pH
• Adding water will not change pH
• Amt of acid/salt still the same
• Ratio conc acid/salt same, pH buffer same
0.35mol
(acid)
1mol
(salt )
0.35mol
(acid )
1mol
(salt)
BA
1.74mol
(base)
1mol
(salt)
1.74mol
(base)
1mol
(salt)
BA
Same mole ratio (acid/salt)but differenttotal volume
Buffer A BufferB
• 0.35mol (acid )= 0.35 in 1dm3 or 0.35mol (acid) = 0.35 in 2dm3
1mol (salt) 1mol (salt)
1dm3
2dm3
1dm3
Same mole ratio (base/salt)but differenttotal volume
Buffer A BufferB
• 1.74mol (base) = 1.74 in 1dm3 or 1.74mol (base) = 1.74 in 2dm3
1mol (salt) 1mol (salt)
2dm3
Add Water
Will pH change by adding water?
Add Water
pH BufferA = pH Buffer B
• Same pH
• Adding water will not change pH
• Amt of acid/salt still the same
• Ratio conc acid/salt same, pH buffer same
Weaker buffering capacity

13. Acidic Buffer PreparationAcidic Buffer Preparation
Prepare Acidic Buffer pH = 4.74
• Choose pKa acid closest to pH 4.74
• pKa = 4.74 (ethanoic acid) chosen
• pH = pKa -lg [acid]
[salt]
• 4.74 = 4.74 – lg [acid]
[salt]
• [acid] = 1.00
[salt]
Ratio of [acid] = 1.00
[salt]
Buffer solution
Buffering Capacity
5 5
Which has greater buffering capacity?
Buffer A > Buffer B
• Conc ratio [acid]/[salt] = 1
• Bufferhighest buffering capacity when pH = pKa
• Conc acid = Conc salt → highest buffering capacity
Concentration ratio
[acid]/[salt] = 1
1 mol
(acid)
1 mol
(salt)
A
1 mol
(salt)
B
Buffer A > Buffer B
• Further conc ratio [acid]/[salt] from 1
Same conc ratio (acid/salt)in 1dm3
Buffer A
• 1 mol (acid ) = 1.00
1 mol (salt)
1dm3 1dm3
Prepare Acidic Buffer at pH = 5.2
• Choose pKa acid closest to pH 5.2
• pKa = 4.74 (ethanoic acid) chosen
• pH = pKa -lg [acid]
[salt]
• 5.2 = 4.74 – lg [acid]
[salt]
• [acid] = 0.35
[salt]
Ratio of [acid] = 0.35
[salt]
Differentconc ratio (acid/salt)in 1dm3
Buffer B
• 0.35mol (acid ) = 0.35
1.00mol (salt)
Which has greater buffering capacity?
0.35mol
(acid)
Concentration ratio
[acid]/[salt] ratio < 1
Lower buffering capacity

14. No Salt Hydrolysis
Presence of ions from salt cause bonds in water to break
NEUTRALIZATION
HCI + NaOH → NaCI + H2O
Neutral salt
Strong acid and Strong base
NaCI – Ionize - Na+ and CI- ion
– Na+ doesn’t cause water hydrolysis
- No breaking bond in water.
Strong acid and Weak base Weak acid and Strong base
HCI + NH4OH → NH4CI + H2O CH3COOH + NaOH → CH3COONa + H2O
Acidic salt Basic salt
Salt Hydrolysis Salt Hydrolysis
No breaking
bond in water
NH4CI – Ionize - NH4
+ and CI- ion
- NH4
+ cause water hydrolysis
- Breaking bond in water
NH4
+ + H2O ↔ NH3 + H3O+
CH3COONa– Ionize - Na+ and CH3COO- ion
- CH3COO- causewater hydrolysis
- Breaking bond in water
CH3COO- + H2O ↔ CH3COOH + OH-
NH4
+ (Acid) - NH3 (Conjugate base)
lose H+ to produce H+ gain H+ to produce OH-
CH3COO- (Base) - CH3COOH (Conjugate acid)
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)

15. 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

16. 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

17. gain H+ to produce OH- - Basiclose H+ to produce H3O+ - Acidic
CH3COO- + H2O → CH3 COOH + OH-
Dissociation constant Ka and Kb
Weak acid and Weak base
CH3COOH + NH3 → CH3COONH4
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
Kb > Ka
BASIC
Amphoteric Ion
Ka = 4.7 x 10 -11 Kb = 2.3 x 10 -8
HCO3
- + H2O ↔ H3O+ + CO3
2- HCO3
- + H2O ↔ H2CO3 + OH-
Kb > Ka
BASIC
Solution of HCO3
- - Acidic or alkaline?
Solution of H2PO4
- - Acidic or alkaline?
H2PO4
- + H2O ↔ HPO4
2- + H3O+ H2PO4
- + H2O ↔ H3PO4 + OH-
lose H+ to produce H3O+ - Acidic
Ka = 6.2 x 10 -8
gain H+ to produce OH- - Basic
Kb = 1.4 x 10 -12
Ka > Kb
ACIDIC

18. IB QUESTIONS
Predict for each salt whether pH is <, >, = 7
1
HCI + Fe(OH)3 → FeCI3
strong acid + weak base → acidic salt
HNO3 + NH4OH → NH4NO3
NaNO3
strong acid + weak base → acidic salt
H2CO3 + NaOH → Na2CO3
Weak acid + strong base → basic salt
NH4NO3
FeCI3 Na2CO3
CH3COOLi KCN
HNO3 + NaOH → Na2CO3
strong acid + strong base → neutral salt
CH3COOH + LiOH → CH3COOLi HCN + KOH → KCN
2 3
pH < 7 pH > 7pH < 7
Predict for each salt whether pH is <, >, = 7
Weak acid + strong base → basic salt
pH > 7pH = 7
Weak acid + strong base → basic salt
pH > 7
Deduce the pH of solution
4 5 6
H2SO4 + NH3 → ? H3PO4 + KOH → ? HNO3 + Ba(OH)2 → ?7 8 9
strong acid + weak base → acidic salt
pH < 7
Weak acid + strong base → basic salt
pH > 7
strong acid + strong base → neutral salt
pH = 7

19. Acidic BufferCalculation
Find pH buffer - 0.20 mol CH3COONa(salt) add to 0.5dm3, 0.10M CH3COOH(acid)
Ka = 1.8 x 10-5
Conc CH3COO-
=Moles/volume
= 0.20/0.5
= 0.40M
Click here videos Khan Academy
Find conc of CH3COONa(salt) added to 1.0dm3 of 1.0M CH3COOH(acid)
Ka = 1.8 x 10-5M, pKa = 4.74 , pH 4.5
Find pH buffer - 0.10M CH3COOH(acid), 0.25M CH3COONa(salt)
Ka = 1.8 x 10-5
1st method (formula)
1
Convert Ka to pKa
2nd method (Ka)
2
1st method (formula) Convert Ka to pKa
2nd method (Ka)
3
1st method (formula)
Conc salt
2nd method (Ka)
Click here explanation from chem guide
14.5
]25.0[
]10.0[
lg74.4
][
][
lg



pH
pH
salt
acid
pKpH a
14.5
)102.7lg(
)lg(
102.7
10.0
))(25.0(
108.1
)(
))((
6
6
5
3
3












pH
pH
HpH
H
H
COOHCH
HCOOCH
Ka
34.5
]40.0[
]10.0[
lg74.4
][
][
lg



pH
pH
salt
acid
pKpH a
74.4
)108.1lg(
lg
108.1
5
5






a
a
aa
a
pK
pK
KpK
K
74.4
)108.1lg(
lg
108.1
5
5






a
a
aa
a
pK
pK
KpK
K
MCOOCH
COOCH
COOHCH
HCOOCH
Ka
0578.0
0.1
)1016.3)((
108.1
)(
))((
3
5
35
3
3








Msalt
salt
salt
salt
acid
pKpH a
0578.0][
24.0
][
]0.1[
lg
][
]0.1[
lg74.45.4
][
][
lg




34.5
)105.4lg(
)lg(
105.4
10.0
))(40.0(
108.1
)(
))((
6
6
5
3
3












pH
pH
HpH
H
H
COOHCH
HCOOCH
Ka
5
1016.3
)lg(5.4
)lg(






H
H
HpH
Conc [H+]

20. Find pH buffer - 0.50M NH3 (base), 0.32M NH4CI (salt)
Kb = 1.8 x 10-5
Basic BufferCalculation
Find pH buffer - 4.28g NH4CI (salt) add to 0.25dm3, 0.50NH3(base)
Kb = 1.8 x 10-5
Mole NH4CI = mass/RMM
= 4.28 / 53.5
= 0.08 mol
Conc NH4CI = moles/vol
= 0.08/0.25
= 0.32M
4
1st method (formula) 2nd method (Kb)
1st method (formula)
5
2nd method (Kb)
Conc salt
Find mass of CH3COONa added to 500ml, 0.10M CH3COOH(acid)
pH = 4.5, Ka = 1.8 x 10-5M, pKa = 4.74
Conc CH3COO- = 0.0578M → x RMM (82) → 4.74g in 1000ml
2.37g in 500ml
6
2nd method (Ka)1st method (formula)
Click here addition base to buffer
Click here addition acid to buffer
45.955.414
55.4
]32.0[
]50.0[
lg74.4
][
][
lg




pH
pOH
pOH
salt
base
pKpOH b
45.955.414
55.4
)1081.2lg(
)lg(
5






pH
pOH
pOH
OHpOH
5
5
3
4
423
1081.2
50.0
))(32.0(
108.1
)(
))((









OH
OH
NH
OHNH
K
OHNHOHNH
b
45.955.414
55.4
]32.0[
]50.0[
lg74.4
][
][
lg




pH
pOH
pOH
salt
base
pKpOH b
45.955.414
55.4
)1081.2lg(
)lg(
5






pH
pOH
pOH
OHpOH
5
5
3
4
423
1081.2
50.0
))(32.0(
108.1
)(
))((









OH
OH
NH
OHNH
K
OHNHOHNH
b
0578.0][
24.0
][
]10.0[
lg
][
]10.0[
lg74.45.4
][
][
lg
3
3
3
3








COOCH
COOCH
COOCH
COOCH
acid
pKpH a
5.4
10
)lg(5.4
)lg(






H
H
HpH
MCOOCH
COOCH
COOHCH
HCOOCH
K
HCOOCHCOOHCH
a
0578.0][
)10.0(
)10)((
108.1
)(
))((
3
5.4
35
3
3
33









Conc [H+]

21. Bicarbonate buffering system
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
][
][

baseConjugate
Acid
B + H2O ↔ BH+ + OH-
or
Ratio of acid
base
Dilution
Add water
pH buffer
pH will not change
Temperature
affect pH
pH change
Basic Bufferingsystem in blood
CO2 + H2O ↔ H2CO3 ↔ H+ + HCO3
-
Acid base homeostasis
- pH blood plasma constant
- buffer range 7.0 – 7.45
Increase CO2 – Shift right – More H+ – pH ↓ - Acidic
Decrease CO2 – Shift left – Less H+ - pH ↑ - Alkaline
H2CO3 ↔ HCO3
-
Weak acid Conjugate base
Exercise - release lactic acid H+/CO2
HCO3
- – base neutralize added acid
Respiratory acidosis (Hypoventilation)
Breathing too slowly – More CO2 in blood – pH ↓– Acidic
HCO3
- reabsorb/secretion by kidney, neutralize H+
Respiratory alkalosis (Hyperventilation)
Breathing too fast – Less CO2 in blood – pH ↑– Alkaline
Release of H+ by kidney to reduce pH ↓
HCO3
- secretion by kidney to reduce pH ↓
Altitude Sickness (Hyperventilation)
High altitude – [O2] ↓ – Hyperventilate ↑ – Less CO2 blood ↓ - pH ↑
Drug stimulate secretion HCO3
- / increase H+ secretion by kidney

22. Click here on pH calculation
Video on Acid/Base
Click here on pKa /pKb calculation How pH = pOH = 14 derived How Ka x Kb = Kw derived
Simulation on Acid/ Base
Click here on pH animation Click here to acid/base simulation
Click here on weak base simulation Click here strong acid ionization Click here on weak acid dissociation