ib chemistry on acid base buffers
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http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
Tutorial on Acid/Base Buffer solutions.
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
Distinguish bet strong and weak acid
Electrical conductivityRate of rxn pH
Strong acid
Strong acid → High ionization → High conc H+ → High conductivity → High rate rxn → Lower pH
Strong acid
OxoacidO atom > number ionizable proton
HNO3, H2SO4, HCIO3, HCIO4
Hydrohalic acidHI, HBr, HCI
Weak acid
Hydrohalic acidHF
OxoacidO atom ≥ number ionizable proton by 1
HCIO, HNO2, H3PO4
Carboxylic acid COOH
Strong base – contain OH- or O2-
LiOH, NaOH, CaO, K2O Ca(OH)2, Ba(OH)2
Weak base – contain electron rich nitrogen, NNH3, 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
HAHA
H+ A-H+
H+
A-
A-HA
HA
HA
HA
HA
HA
Ionizes partially
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-] = 0.0000001MpOH = -log [0.0000001]pOH = -log1010-7
pOH = 7pH + pOH = 14pH + 7 = 14pH = 7 (Neutral)
pH with Conc H+
pH = -log [H+][H+] = 0.0000001MpH = -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 measure of acidity of solution in logarithmic scale• pH = power of hydrogen or minus logarithm to base ten of hydrogen ion concentration
← Acidic – pH < 7 Alkaline – pH > 7 →
pOH with Conc OH-
pOH = -log [OH-][OH-] = 0.1MpOH = -log[0.1]pOH = 1pH + pOH = 14pH + 1 = 14
pH = 13 (Alkaline)
pH with Conc H+
pH = -log [H+][H+] = 0.01MpH = -log [0.01]pH = -log1010-2
pH = 2 (Acidic)
Easier pH scaleConc H+
Formula for acid/base calculation
[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/base calculation
Dissociation Constant for Weak Acid
pH = -log10[H+] pOH = -log10[OH-]pH + pOH = 14pH + 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
AHK a
HCOOCHCOOHCH 33
COOHCH
H
COOHCH
HCOOCHKa
3
2
3
3
Dissociation Constant for Weak Base
OHBHOHB 2
B
OHBHKb
OHNHOHNH 423
3
2
3
4
NH
OH
NH
OHNHKb
OHCOOCHOHCOOHCH 3323
OHCOOCHOHCOOHCH 3323
COOHCH
OHCOOCHKa
3
33
OHCOOHCHOHCOOCH 323
COOCH
OHCOOHCHKb
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
OHOHCOOCH
OHCOOHCH
COOHCH
OHCOOCH3
3
3
3
33
wba KKK
Formula for acid/base calculation
Ka /Kb measure equilibrium positionKa/Kb large ↑ – ↑ dissociation – shift to right – favour productKa/Kb large ↑ – pKa /pKb small ↓ – Stronger acid/base
Strong acid Large ↑ Ka
Weak acid Small ↓ Ka
Strong base Large ↑ Kb
Weak base Small ↓Kb
↑ Ka → ↓ pKa
Ka /Kb measure equilibrium positionKa /Kb small ↓ – ↓ dissociation – shift to left – reactant favourKa /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 dissociationconstant
Strong Acid Weak base
ba KK /
Str
on
g a
cid
Stro
ng
ba
se
Formula for acid/base calculation
[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/base calculation
Dissociation Constant for Weak Acid
pH = -log10[H+] pOH = -log10[OH-]pH + pOH = 14pH + 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
AHK a
HCOOCHCOOHCH 33
COOHCH
H
COOHCH
HCOOCHKa
3
2
3
3
Dissociation Constant for Weak Base
OHBHOHB 2
B
OHBHKb
OHNHOHNH 423
3
2
3
4
NH
OH
NH
OHNHKb
Dissociate partially ↔ 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
NH3 ↔ NH4+
Buffer Solution
Acid part
Neutralize
each other
Salt part
Base part
- NH3(weak base) + NH4CI (salt)- NH3 + H2O ↔ NH4
+ + OH− → NH3 molecule neutralise added H+
- NH4CI → NH4+ + CI− → NH4
+ neutralise added OH−
- Effective buffer 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/conjugate base
CH3COOH ↔ CH3COO-
Conjugate acid base pair
CH3COOH CH3COO-
Weak Acid Conjugate Base
BUFFER
Dissociate fully
HCOOCHCOOHCH 33
COOHCH 3 COONaCH 3
NaCOOCHCOONaCH 33
Dissociate partially
- CH3COOH (weak acid) + CH3COONa (salt)- CH3COOH ↔ CH3COO- + H+ → CH3COOH neutralise added OH−
- CH3COONa → CH3COO- + Na+ → CH3COO- neutralise added H+
- Effective buffer equal amt weak acid CH3COOH and base CH3COO-
COOHCH 3
COOCH 3BUFFER
Add acid H+Add alkaline OH-
Neutralize
each other
Basic buffer - weak base and its salt/conjugate acid
OHNHOHNH 423
NH3 + H2O ↔ NH4+ + OH-
NH3
Weak Base
NH4+
Conjugate acid
CINH 43NH
BUFFER
Conjugate acid base pair
Add acid H+ Add alkaline OH-
Neutralize
each other
Neutralize
each other
Dissociate partially
CINHCINH 44
3NH
4NH
Base partSalt part
Acid part
Dissociate fully
BUFFER
How to prepare acidic/ basic buffer
Acid Dissociation constantCH3COOH + H2O ↔ CH3COO- + H3O
+
Ka = (CH3COO-) (H3O+)
(CH3COOH)-lgKa = -lgH+ -lg (CH3COO-)
(CH3COOH)-lgH+ = -lg Ka + lg (CH3COO-)
(CH3COOH)pH = pKa + lg (CH3COO-)
(CH3COOH)
Acidic Buffer Formula• Mixture Weak acid + Salt/Conjugate base• CH3COOH ↔ CH3COO- + H+ (dissociate partially)• CH3COONa → CH3COO- + Na+ (dissociate fully)
Basic Buffer Formula• Mixture Weak base + Salt/Conjugate acid• NH3 + H2O ↔ NH4
+ + OH_ (dissociate partially)• NH4CI → NH4
+ + CI_ (dissociate fully)
pH = pKa - lg (acid)(salt)
pH = pKa + lg (salt)(acid)
Base Dissociation constantNH3 + 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 Buffer Acidic Buffer
salt salt
acid base
Henderson Hasselbalch Equation
multiply -lgboth sides
Henderson Hasselbalch Equation
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) = 0.351M, 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) = 1.741M, 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 remain same
Buffer solution
Buffer Preparation
1 1
2 2
3 Use fix vol, 1dm3 and use different mole ratio (Acid/salt) • 0.35 mole acid + 1 mole salt to 1 dm3 solvent = 0.35
Use fix vol, 1dm3 and use different mole 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
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 ratioBuffer A Buffer B
• 1M, 35ml (acid) = 0.35 or 0.1M, 35ml (acid) = 0.351M, 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 ratioBuffer A Buffer B
• 1M, 174ml (base) = 1.74 or 0.1M, 174ml (base) = 1.741M, 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 BStronger 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 > Buffer BStronger 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 ?
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 ratioBuffer A Buffer B
• 3.5M, 10ml (acid) = 0.35 or 0.35M, 10ml (acid) = 0.3510M, 10ml (salt) 1M, 10ml (salt)
Use same vol base/salt but different conc ratioBuffer A Buffer B
• 1.74M, 10ml (base) = 1.74 or 0.174M, 10ml (base) = 1.741M, 10ml (salt) 0.10M, 10ml (salt)
Which has greater buffering capacity ? Which has greater buffering capacity ?
Buffer A > Buffer BStronger 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 > Buffer BStronger 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
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) = 0.351mol (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) = 1.741mol (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 BStronger 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 > Buffer BStronger 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
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 Buffer A = 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)
BA1.74mol
(base)
1mol
(salt) 1.74mol
(base) 1mol(salt)
BA
Same mole ratio (acid/salt) but different total volumeBuffer A Buffer B
• 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 different total volumeBuffer A Buffer B
• 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 Buffer A = 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
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• Buffer highest 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]
Different conc 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
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- cause water 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)
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
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
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 = KbNEUTRAL
NH3 + HF → NH4F
salt
NH4F → NH4+ + F-
NH4+ + H2O → NH3 + H3O
+ F- + H2O → HF + OH-
cationanion
KaH3O
+ - Acidic KbOH- - Basic
Kb > KaBASIC
Amphoteric Ion
Ka = 4.7 x 10 -11 Kb = 2.3 x 10 -8
HCO3- + H2O ↔ H3O
+ + CO32- 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
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
NH4NO3FeCI3 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 > 7Deduce 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
Acidic Buffer Calculation
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
acidpKpH 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
HCOOCHK a
34.5
]40.0[
]10.0[lg74.4
][
][lg
pH
pH
salt
acidpKpH 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
HCOOCHK a
0578.0
0.1
)1016.3)((108.1
)(
))((
3
5
35
3
3
Msalt
salt
salt
salt
acidpKpH 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
HCOOCHK a
51016.3
)lg(5.4
)lg(
H
H
HpH
Conc [H+]
Find pH buffer - 0.50M NH3 (base), 0.32M NH4CI (salt)Kb = 1.8 x 10-5
Basic Buffer Calculation
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 1000ml2.37g 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
basepKpOH 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
OHNHK
OHNHOHNH
b
45.955.414
55.4
]32.0[
]50.0[lg74.4
][
][lg
pH
pOH
pOH
salt
basepKpOH 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
OHNHK
OHNHOHNH
b
0578.0][
24.0][
]10.0[lg
][
]10.0[lg74.45.4
][
][lg
3
3
3
3
COOCH
COOCH
COOCH
COOCH
acidpKpH a
5.410
)lg(5.4
)lg(
H
H
HpH
MCOOCH
COOCH
COOHCH
HCOOCHK
HCOOCHCOOHCH
a
0578.0][
)10.0(
)10)((108.1
)(
))((
3
5.4
35
3
3
33
Conc [H+]
Bicarbonate buffering system
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][
][
baseConjugate
Acid
B + H2O ↔ BH+ + OH-
or
Ratio of acidbase
DilutionAdd water
pH buffer
pH will not change
Temperature affect pH
pH change
Basic Buffering system 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
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
Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/http://4photos.net/en/image:44-225901-Water_droplets_on_blue_backdrop__images
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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