Chemical calculations used in medicine

part 2

Pavla Baliacutenovaacute

Acid ndash base theories

bull Arrhenius theory

acid = substance that is able to lose H+ (CH3COOH harr CH3COO- + H+)

base = substance that is able to lose OH-

bull Broumlnsted ndash Lowry theory

acid = chemical species (molecule or ion) that is able to lose H+

base = chemical species (molecule or ion) with the ability to accept H+

HCl + H2O harr H3O+ + Cl-

acid 1 base 2 acid 2 base 1

conjugated pair 1

Water H2O

The purest water is not all H2O rarr about 1 molecule in 500 million transfers a proton H+ to another H2O molecule giving a hydronium ion H3O+ and a hydroxide ion OH- H2O + H2O harr H3O+ + OH- = dissociation of water

The concentration of H3O+ in pure water is 0000 0001 M or 110-7 MThe concentration of OH- is also 110-7 M Pure water is a neutral solution without an excess of either H3O+ and OH- ionsEquilibrium constant of water Keq = [H3O+] [OH-] [H2O]2

Ion product of water Kw Kw = [H3O+] [OH-] Kw = 110-7 110-7 = 110-14

Ion product of water

Kw = [H3O+] [OH-]

10ndash14 = [H3O+] [OH-] log log(a b) = log a + log b

log 10ndash14 = log ([H3O+] [OH-])

log 10ndash14 = log [H3O+] + log [OH-]

-14 = log [H3O+] + log [OH-] ( -1 )

14 = - log [H3O+] - log [OH-]

- log KW pH pOH - log KW = pKW

pKW = pH + pOH = 14

Addition of an acid to pure water rarr increasing H3O+

and OH- will fall until the product equals 110-14Addition of a base to pure water rarr increasing OH- and H3O+ will fall until the product equals 110-14

ExampleLemon juice has a [H3O+] of 001 M What is the [OH-] [H3O+ ] [OH-] = 110-14

110-2 [OH-] = 110-14 110-2

[OH-] = 110-12

pH scale

Exponential numbers express the often minute actual concentration of H3O+ and OH- ions

In 1909 S P L Soslashrensen proposed that only the number in the exponent be used to express acidity Soslashrensenacutes scale came to be known as the pH scale (bdquopower of hydrogenldquo)

pH = log 1[H3O+] = - log [H3O+]

e g The pH of a solution whose [H3O+ ] is 110-4 equals 4

Representative pH values

90 ndash 100hand soap

735 ndash 745blood

70pure water

60urine

50coffee

29vinegar

15 ndash 20gastric juice

05lead-acid battery

pH Substance

pH of strong acidsStrong acids are those that react completely with water to form H3O+ and anion (HCl H2SO4 HClO4 HNO3hellip)Generally HA rarr H+ + A-

e g HCl + H2O harr H3O+ + Cl-

pH = - log cH+ = - log cHA

Calculations 1) 01 M HCl pH =

2) Strong acids pH a) 16 c = molL b) 30 c = molL

3) 008 M H2SO4 pH =

4) Dilution of a strong acid c1 = 01 M rarr c2 = 001 M ∆ pH =

pH of weak acids

Weak acids react only to a slight extent with water to form relatively few H3O+ ions Most of the molecules of the weak acids remain in the molecular form (uncharged)HA rarr H+ + A-

Kdis = [H+][A-] [H+] = [A-] [HA] [HA] = cHA

Kdis = [H+]2

cHA Kdis = KHA

KHA cHA = [H+]2 log log (KHA cHA ) = 2 log [H+] log KHA + log cHA = 2 log [H+] frac12 frac12 log KHA + frac12 log cHA = log [H+] middot (-1)-frac12 log KHA - frac12 log cHA = - log [H+] - log KHA = pKHA

frac12 pKHA - frac12 log cHA = pH rarr pH = frac12 pKHA - frac12 log cHA

pH = frac12 (pKHA ndash log cHA)

pH of weak acids - calculations

1) 001 M acetic acid Kdis = 18 x 10-5 pH =

2) 01 M lactic acid pH = 24 Kdis =

3) dilution of a weak acid c1 = 01 M rarr c2 = 001 M

∆ pH =

pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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• Slide 8
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• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14

Acid ndash base theories

bull Arrhenius theory

acid = substance that is able to lose H+ (CH3COOH harr CH3COO- + H+)

base = substance that is able to lose OH-

bull Broumlnsted ndash Lowry theory

acid = chemical species (molecule or ion) that is able to lose H+

base = chemical species (molecule or ion) with the ability to accept H+

HCl + H2O harr H3O+ + Cl-

acid 1 base 2 acid 2 base 1

conjugated pair 1

Water H2O

The purest water is not all H2O rarr about 1 molecule in 500 million transfers a proton H+ to another H2O molecule giving a hydronium ion H3O+ and a hydroxide ion OH- H2O + H2O harr H3O+ + OH- = dissociation of water

The concentration of H3O+ in pure water is 0000 0001 M or 110-7 MThe concentration of OH- is also 110-7 M Pure water is a neutral solution without an excess of either H3O+ and OH- ionsEquilibrium constant of water Keq = [H3O+] [OH-] [H2O]2

Ion product of water Kw Kw = [H3O+] [OH-] Kw = 110-7 110-7 = 110-14

Ion product of water

Kw = [H3O+] [OH-]

10ndash14 = [H3O+] [OH-] log log(a b) = log a + log b

log 10ndash14 = log ([H3O+] [OH-])

log 10ndash14 = log [H3O+] + log [OH-]

-14 = log [H3O+] + log [OH-] ( -1 )

14 = - log [H3O+] - log [OH-]

- log KW pH pOH - log KW = pKW

pKW = pH + pOH = 14

Addition of an acid to pure water rarr increasing H3O+

and OH- will fall until the product equals 110-14Addition of a base to pure water rarr increasing OH- and H3O+ will fall until the product equals 110-14

ExampleLemon juice has a [H3O+] of 001 M What is the [OH-] [H3O+ ] [OH-] = 110-14

110-2 [OH-] = 110-14 110-2

[OH-] = 110-12

pH scale

Exponential numbers express the often minute actual concentration of H3O+ and OH- ions

In 1909 S P L Soslashrensen proposed that only the number in the exponent be used to express acidity Soslashrensenacutes scale came to be known as the pH scale (bdquopower of hydrogenldquo)

pH = log 1[H3O+] = - log [H3O+]

e g The pH of a solution whose [H3O+ ] is 110-4 equals 4

Representative pH values

90 ndash 100hand soap

735 ndash 745blood

70pure water

60urine

50coffee

29vinegar

15 ndash 20gastric juice

05lead-acid battery

pH Substance

pH of strong acidsStrong acids are those that react completely with water to form H3O+ and anion (HCl H2SO4 HClO4 HNO3hellip)Generally HA rarr H+ + A-

e g HCl + H2O harr H3O+ + Cl-

pH = - log cH+ = - log cHA

Calculations 1) 01 M HCl pH =

2) Strong acids pH a) 16 c = molL b) 30 c = molL

3) 008 M H2SO4 pH =

4) Dilution of a strong acid c1 = 01 M rarr c2 = 001 M ∆ pH =

pH of weak acids

Weak acids react only to a slight extent with water to form relatively few H3O+ ions Most of the molecules of the weak acids remain in the molecular form (uncharged)HA rarr H+ + A-

Kdis = [H+][A-] [H+] = [A-] [HA] [HA] = cHA

Kdis = [H+]2

cHA Kdis = KHA

KHA cHA = [H+]2 log log (KHA cHA ) = 2 log [H+] log KHA + log cHA = 2 log [H+] frac12 frac12 log KHA + frac12 log cHA = log [H+] middot (-1)-frac12 log KHA - frac12 log cHA = - log [H+] - log KHA = pKHA

frac12 pKHA - frac12 log cHA = pH rarr pH = frac12 pKHA - frac12 log cHA

pH = frac12 (pKHA ndash log cHA)

pH of weak acids - calculations

1) 001 M acetic acid Kdis = 18 x 10-5 pH =

2) 01 M lactic acid pH = 24 Kdis =

3) dilution of a weak acid c1 = 01 M rarr c2 = 001 M

∆ pH =

pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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• Slide 14

Water H2O

The purest water is not all H2O rarr about 1 molecule in 500 million transfers a proton H+ to another H2O molecule giving a hydronium ion H3O+ and a hydroxide ion OH- H2O + H2O harr H3O+ + OH- = dissociation of water

The concentration of H3O+ in pure water is 0000 0001 M or 110-7 MThe concentration of OH- is also 110-7 M Pure water is a neutral solution without an excess of either H3O+ and OH- ionsEquilibrium constant of water Keq = [H3O+] [OH-] [H2O]2

Ion product of water Kw Kw = [H3O+] [OH-] Kw = 110-7 110-7 = 110-14

Ion product of water

Kw = [H3O+] [OH-]

10ndash14 = [H3O+] [OH-] log log(a b) = log a + log b

log 10ndash14 = log ([H3O+] [OH-])

log 10ndash14 = log [H3O+] + log [OH-]

-14 = log [H3O+] + log [OH-] ( -1 )

14 = - log [H3O+] - log [OH-]

- log KW pH pOH - log KW = pKW

pKW = pH + pOH = 14

Addition of an acid to pure water rarr increasing H3O+

and OH- will fall until the product equals 110-14Addition of a base to pure water rarr increasing OH- and H3O+ will fall until the product equals 110-14

ExampleLemon juice has a [H3O+] of 001 M What is the [OH-] [H3O+ ] [OH-] = 110-14

110-2 [OH-] = 110-14 110-2

[OH-] = 110-12

pH scale

Exponential numbers express the often minute actual concentration of H3O+ and OH- ions

In 1909 S P L Soslashrensen proposed that only the number in the exponent be used to express acidity Soslashrensenacutes scale came to be known as the pH scale (bdquopower of hydrogenldquo)

pH = log 1[H3O+] = - log [H3O+]

e g The pH of a solution whose [H3O+ ] is 110-4 equals 4

Representative pH values

90 ndash 100hand soap

735 ndash 745blood

70pure water

60urine

50coffee

29vinegar

15 ndash 20gastric juice

05lead-acid battery

pH Substance

pH of strong acidsStrong acids are those that react completely with water to form H3O+ and anion (HCl H2SO4 HClO4 HNO3hellip)Generally HA rarr H+ + A-

e g HCl + H2O harr H3O+ + Cl-

pH = - log cH+ = - log cHA

Calculations 1) 01 M HCl pH =

2) Strong acids pH a) 16 c = molL b) 30 c = molL

3) 008 M H2SO4 pH =

4) Dilution of a strong acid c1 = 01 M rarr c2 = 001 M ∆ pH =

pH of weak acids

Weak acids react only to a slight extent with water to form relatively few H3O+ ions Most of the molecules of the weak acids remain in the molecular form (uncharged)HA rarr H+ + A-

Kdis = [H+][A-] [H+] = [A-] [HA] [HA] = cHA

Kdis = [H+]2

cHA Kdis = KHA

KHA cHA = [H+]2 log log (KHA cHA ) = 2 log [H+] log KHA + log cHA = 2 log [H+] frac12 frac12 log KHA + frac12 log cHA = log [H+] middot (-1)-frac12 log KHA - frac12 log cHA = - log [H+] - log KHA = pKHA

frac12 pKHA - frac12 log cHA = pH rarr pH = frac12 pKHA - frac12 log cHA

pH = frac12 (pKHA ndash log cHA)

pH of weak acids - calculations

1) 001 M acetic acid Kdis = 18 x 10-5 pH =

2) 01 M lactic acid pH = 24 Kdis =

3) dilution of a weak acid c1 = 01 M rarr c2 = 001 M

∆ pH =

pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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• Slide 2
• Slide 3
• Slide 4
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• Slide 14

Ion product of water

Kw = [H3O+] [OH-]

10ndash14 = [H3O+] [OH-] log log(a b) = log a + log b

log 10ndash14 = log ([H3O+] [OH-])

log 10ndash14 = log [H3O+] + log [OH-]

-14 = log [H3O+] + log [OH-] ( -1 )

14 = - log [H3O+] - log [OH-]

- log KW pH pOH - log KW = pKW

pKW = pH + pOH = 14

Addition of an acid to pure water rarr increasing H3O+

and OH- will fall until the product equals 110-14Addition of a base to pure water rarr increasing OH- and H3O+ will fall until the product equals 110-14

ExampleLemon juice has a [H3O+] of 001 M What is the [OH-] [H3O+ ] [OH-] = 110-14

110-2 [OH-] = 110-14 110-2

[OH-] = 110-12

pH scale

Exponential numbers express the often minute actual concentration of H3O+ and OH- ions

In 1909 S P L Soslashrensen proposed that only the number in the exponent be used to express acidity Soslashrensenacutes scale came to be known as the pH scale (bdquopower of hydrogenldquo)

pH = log 1[H3O+] = - log [H3O+]

e g The pH of a solution whose [H3O+ ] is 110-4 equals 4

Representative pH values

90 ndash 100hand soap

735 ndash 745blood

70pure water

60urine

50coffee

29vinegar

15 ndash 20gastric juice

05lead-acid battery

pH Substance

pH of strong acidsStrong acids are those that react completely with water to form H3O+ and anion (HCl H2SO4 HClO4 HNO3hellip)Generally HA rarr H+ + A-

e g HCl + H2O harr H3O+ + Cl-

pH = - log cH+ = - log cHA

Calculations 1) 01 M HCl pH =

2) Strong acids pH a) 16 c = molL b) 30 c = molL

3) 008 M H2SO4 pH =

4) Dilution of a strong acid c1 = 01 M rarr c2 = 001 M ∆ pH =

pH of weak acids

Weak acids react only to a slight extent with water to form relatively few H3O+ ions Most of the molecules of the weak acids remain in the molecular form (uncharged)HA rarr H+ + A-

Kdis = [H+][A-] [H+] = [A-] [HA] [HA] = cHA

Kdis = [H+]2

cHA Kdis = KHA

KHA cHA = [H+]2 log log (KHA cHA ) = 2 log [H+] log KHA + log cHA = 2 log [H+] frac12 frac12 log KHA + frac12 log cHA = log [H+] middot (-1)-frac12 log KHA - frac12 log cHA = - log [H+] - log KHA = pKHA

frac12 pKHA - frac12 log cHA = pH rarr pH = frac12 pKHA - frac12 log cHA

pH = frac12 (pKHA ndash log cHA)

pH of weak acids - calculations

1) 001 M acetic acid Kdis = 18 x 10-5 pH =

2) 01 M lactic acid pH = 24 Kdis =

3) dilution of a weak acid c1 = 01 M rarr c2 = 001 M

∆ pH =

pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14

Addition of an acid to pure water rarr increasing H3O+

and OH- will fall until the product equals 110-14Addition of a base to pure water rarr increasing OH- and H3O+ will fall until the product equals 110-14

ExampleLemon juice has a [H3O+] of 001 M What is the [OH-] [H3O+ ] [OH-] = 110-14

110-2 [OH-] = 110-14 110-2

[OH-] = 110-12

pH scale

Exponential numbers express the often minute actual concentration of H3O+ and OH- ions

In 1909 S P L Soslashrensen proposed that only the number in the exponent be used to express acidity Soslashrensenacutes scale came to be known as the pH scale (bdquopower of hydrogenldquo)

pH = log 1[H3O+] = - log [H3O+]

e g The pH of a solution whose [H3O+ ] is 110-4 equals 4

Representative pH values

90 ndash 100hand soap

735 ndash 745blood

70pure water

60urine

50coffee

29vinegar

15 ndash 20gastric juice

05lead-acid battery

pH Substance

pH of strong acidsStrong acids are those that react completely with water to form H3O+ and anion (HCl H2SO4 HClO4 HNO3hellip)Generally HA rarr H+ + A-

e g HCl + H2O harr H3O+ + Cl-

pH = - log cH+ = - log cHA

Calculations 1) 01 M HCl pH =

2) Strong acids pH a) 16 c = molL b) 30 c = molL

3) 008 M H2SO4 pH =

4) Dilution of a strong acid c1 = 01 M rarr c2 = 001 M ∆ pH =

pH of weak acids

Weak acids react only to a slight extent with water to form relatively few H3O+ ions Most of the molecules of the weak acids remain in the molecular form (uncharged)HA rarr H+ + A-

Kdis = [H+][A-] [H+] = [A-] [HA] [HA] = cHA

Kdis = [H+]2

cHA Kdis = KHA

KHA cHA = [H+]2 log log (KHA cHA ) = 2 log [H+] log KHA + log cHA = 2 log [H+] frac12 frac12 log KHA + frac12 log cHA = log [H+] middot (-1)-frac12 log KHA - frac12 log cHA = - log [H+] - log KHA = pKHA

frac12 pKHA - frac12 log cHA = pH rarr pH = frac12 pKHA - frac12 log cHA

pH = frac12 (pKHA ndash log cHA)

pH of weak acids - calculations

1) 001 M acetic acid Kdis = 18 x 10-5 pH =

2) 01 M lactic acid pH = 24 Kdis =

3) dilution of a weak acid c1 = 01 M rarr c2 = 001 M

∆ pH =

pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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pH scale

Exponential numbers express the often minute actual concentration of H3O+ and OH- ions

In 1909 S P L Soslashrensen proposed that only the number in the exponent be used to express acidity Soslashrensenacutes scale came to be known as the pH scale (bdquopower of hydrogenldquo)

pH = log 1[H3O+] = - log [H3O+]

e g The pH of a solution whose [H3O+ ] is 110-4 equals 4

Representative pH values

90 ndash 100hand soap

735 ndash 745blood

70pure water

60urine

50coffee

29vinegar

15 ndash 20gastric juice

05lead-acid battery

pH Substance

pH of strong acidsStrong acids are those that react completely with water to form H3O+ and anion (HCl H2SO4 HClO4 HNO3hellip)Generally HA rarr H+ + A-

e g HCl + H2O harr H3O+ + Cl-

pH = - log cH+ = - log cHA

Calculations 1) 01 M HCl pH =

2) Strong acids pH a) 16 c = molL b) 30 c = molL

3) 008 M H2SO4 pH =

4) Dilution of a strong acid c1 = 01 M rarr c2 = 001 M ∆ pH =

pH of weak acids

Weak acids react only to a slight extent with water to form relatively few H3O+ ions Most of the molecules of the weak acids remain in the molecular form (uncharged)HA rarr H+ + A-

Kdis = [H+][A-] [H+] = [A-] [HA] [HA] = cHA

Kdis = [H+]2

cHA Kdis = KHA

KHA cHA = [H+]2 log log (KHA cHA ) = 2 log [H+] log KHA + log cHA = 2 log [H+] frac12 frac12 log KHA + frac12 log cHA = log [H+] middot (-1)-frac12 log KHA - frac12 log cHA = - log [H+] - log KHA = pKHA

frac12 pKHA - frac12 log cHA = pH rarr pH = frac12 pKHA - frac12 log cHA

pH = frac12 (pKHA ndash log cHA)

pH of weak acids - calculations

1) 001 M acetic acid Kdis = 18 x 10-5 pH =

2) 01 M lactic acid pH = 24 Kdis =

3) dilution of a weak acid c1 = 01 M rarr c2 = 001 M

∆ pH =

pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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Representative pH values

90 ndash 100hand soap

735 ndash 745blood

70pure water

60urine

50coffee

29vinegar

15 ndash 20gastric juice

05lead-acid battery

pH Substance

pH of strong acidsStrong acids are those that react completely with water to form H3O+ and anion (HCl H2SO4 HClO4 HNO3hellip)Generally HA rarr H+ + A-

e g HCl + H2O harr H3O+ + Cl-

pH = - log cH+ = - log cHA

Calculations 1) 01 M HCl pH =

2) Strong acids pH a) 16 c = molL b) 30 c = molL

3) 008 M H2SO4 pH =

4) Dilution of a strong acid c1 = 01 M rarr c2 = 001 M ∆ pH =

pH of weak acids

Weak acids react only to a slight extent with water to form relatively few H3O+ ions Most of the molecules of the weak acids remain in the molecular form (uncharged)HA rarr H+ + A-

Kdis = [H+][A-] [H+] = [A-] [HA] [HA] = cHA

Kdis = [H+]2

cHA Kdis = KHA

KHA cHA = [H+]2 log log (KHA cHA ) = 2 log [H+] log KHA + log cHA = 2 log [H+] frac12 frac12 log KHA + frac12 log cHA = log [H+] middot (-1)-frac12 log KHA - frac12 log cHA = - log [H+] - log KHA = pKHA

frac12 pKHA - frac12 log cHA = pH rarr pH = frac12 pKHA - frac12 log cHA

pH = frac12 (pKHA ndash log cHA)

pH of weak acids - calculations

1) 001 M acetic acid Kdis = 18 x 10-5 pH =

2) 01 M lactic acid pH = 24 Kdis =

3) dilution of a weak acid c1 = 01 M rarr c2 = 001 M

∆ pH =

pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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pH of strong acidsStrong acids are those that react completely with water to form H3O+ and anion (HCl H2SO4 HClO4 HNO3hellip)Generally HA rarr H+ + A-

e g HCl + H2O harr H3O+ + Cl-

pH = - log cH+ = - log cHA

Calculations 1) 01 M HCl pH =

2) Strong acids pH a) 16 c = molL b) 30 c = molL

3) 008 M H2SO4 pH =

4) Dilution of a strong acid c1 = 01 M rarr c2 = 001 M ∆ pH =

pH of weak acids

Weak acids react only to a slight extent with water to form relatively few H3O+ ions Most of the molecules of the weak acids remain in the molecular form (uncharged)HA rarr H+ + A-

Kdis = [H+][A-] [H+] = [A-] [HA] [HA] = cHA

Kdis = [H+]2

cHA Kdis = KHA

KHA cHA = [H+]2 log log (KHA cHA ) = 2 log [H+] log KHA + log cHA = 2 log [H+] frac12 frac12 log KHA + frac12 log cHA = log [H+] middot (-1)-frac12 log KHA - frac12 log cHA = - log [H+] - log KHA = pKHA

frac12 pKHA - frac12 log cHA = pH rarr pH = frac12 pKHA - frac12 log cHA

pH = frac12 (pKHA ndash log cHA)

pH of weak acids - calculations

1) 001 M acetic acid Kdis = 18 x 10-5 pH =

2) 01 M lactic acid pH = 24 Kdis =

3) dilution of a weak acid c1 = 01 M rarr c2 = 001 M

∆ pH =

pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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pH of weak acids

Weak acids react only to a slight extent with water to form relatively few H3O+ ions Most of the molecules of the weak acids remain in the molecular form (uncharged)HA rarr H+ + A-

Kdis = [H+][A-] [H+] = [A-] [HA] [HA] = cHA

Kdis = [H+]2

cHA Kdis = KHA

KHA cHA = [H+]2 log log (KHA cHA ) = 2 log [H+] log KHA + log cHA = 2 log [H+] frac12 frac12 log KHA + frac12 log cHA = log [H+] middot (-1)-frac12 log KHA - frac12 log cHA = - log [H+] - log KHA = pKHA

frac12 pKHA - frac12 log cHA = pH rarr pH = frac12 pKHA - frac12 log cHA

pH = frac12 (pKHA ndash log cHA)

pH of weak acids - calculations

1) 001 M acetic acid Kdis = 18 x 10-5 pH =

2) 01 M lactic acid pH = 24 Kdis =

3) dilution of a weak acid c1 = 01 M rarr c2 = 001 M

∆ pH =

pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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pH of weak acids - calculations

1) 001 M acetic acid Kdis = 18 x 10-5 pH =

2) 01 M lactic acid pH = 24 Kdis =

3) dilution of a weak acid c1 = 01 M rarr c2 = 001 M

∆ pH =

pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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pH of strong basesStrong bases are those that are ionized completelyGenerally BOH rarr B+ + OH-

ie NaOH harr Na+ + OH-

Other examples KOH LiOH Ba(OH)2 Ca(OH)2

pOH = - log cBOH pH = 14 - pOH

Calculations1) 001 M KOH pH = 2) Strong bases pH a) 11 c = b) 103 c = 3) 01 M Ba(OH)2 pH = 4) 50 mL of a solution contains 4 mg of NaOH Mr (NaOH) = 40 pH =

pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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pH of weak bases

Weak bases are those that react with water only to a slight extent producing relatively few OH- ions The weak base remains in the molecular formGenerally BOH harr B+ + OH-

Kdis = [B+] x [OH-] pKBOH = - log Kdis

[BOH] e g NH3 + H2O harr NH4

+ + OH-

pOH = frac12 (pKBOH ndash log cBOH) pH = 14 ndash pOH

Calculations1) 02 M NH4OH pK = 474 pH =

2) 006 M dimethylamine pK = 327 pH =

pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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pH of buffers

Buffers are solutions that keep the pH value nearly constant when acid or base is addedBuffer solution containsbull weak acid + its salt (ie CH3COOH + CH3COONa)bull weak base + its salt (ie NH4OH + NH4Cl)bull two salts of an oxoacid (ie HPO4

2- + H2PO41-)

bull amphoteric compounds (ie aminoacids and proteins)

Henderson ndash Hasselbalch equationpH = pKA + log (cS x VS cA x VA) rarr for acidic buffer

pOH = pKB + log (cS x VS cB x VB) pH = 14 ndash pOH rarr for basic buffer

pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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pH of buffers - calculations

1) 200 mL 05 M acetic acid + 100 mL 05 M sodium acetate rarr bufferpKA = 476 pH =

2) 20 mL 005 M NH4Cl + 27 mL 02 M NH4OH rarr buffer

K = 185 x 10-5 pH =

3) The principal buffer system of blood is a bicarbonate buffer (HCO3

- H2CO3) Calculate a ratio of HCO3- H2CO3 components if

the pH is 738 and pK(H2CO3) = 61

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