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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