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 conductivity Rate of rxn pH

Strong acid

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

Hydrohalic acid HI, HBr, HCI

Weak acid

Hydrohalic acid HF

Oxoacid O 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, 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

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 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-] = 0.1M 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 scale Conc H+

Conc [H+] = 1 x 10-12 pH = -lg[H+] pH = -lg[10-12] pH = 12

Conc [OH-]= 1 x 10-2 pOH = -log10[OH-]

pOH = -log1010-2 = pOH = 2 pH + pOH = 14 pH + 2 = 14 pH = 12

Conc [H+] = 1 x 10-2 pH = -lg[H+] pH = -lg[10-2] pH = 2

Alkaline

Alkaline

Acidic

Acidic

Kw - Ionic product constant water

Using conc [H+] pH = -log10[H+]

pH = -log10[H+] pOH = -log10[OH-] pH + pOH = 14 Kw = [H+][OH-]

Using conc [OH-] pOH = -log10[OH-]

Conc [OH-]= 1 x 10-12 pOH = -log10[OH-]

pOH= -log1010-12 =pOH = 12 pH + pOH = 14 pH + 12 = 14 pH = 2

Formula for acid/base calculation

OH

OHOHK c

2

3

OHOHOHKc 32

OHOHKw 3

OHOH3

14100.1

7714 101101100.1

7101 OH

OHOHOHOH 322

H2O dissociate forming H3O+ and OH- (equilibrium exist)

14100.1 wK

Dissociation water small [H2O] is constant

Kw = 1.0 x 10-14 Ionic Product constant water at -25C

Kc - Dissociation constant water

7

3 101 OH

Number sig fig in log calculation Significant number in log calculation

log10(3575)=3.55327 = 3.5532

log10(3.000x104) = 4.477121 = 4.4771 log10(3.3 x 104) = 4.5185 = 4.51 Calculation involve pH = -log10[H+]

Conc H+ = 1.9 x 10-4

pH= -log10[1.9 x 10-4] = 3.721 = 3.72

Measurement scale not linear • Simple average CANNOT be used • Average of pH 7, pH 8, pH 9 pH scale is logarithmic, pH = -log[H+] Correct average = convert to H+ conc pH 7 = -log10[H+] → H+ = 10-7

pH 8 = -log10[H+] → H+ = 10-8

pH 9 = -log10[H+] → H+ = 10-9

pH pH= -lg10H+ Conc H+

0 0 = -lg10100 1.0

1 1 = -lg1010-1 0.1

2 2 = -lg1010-2 0.01

3 3 = -lg1010-3 0.001

4 4 = -lg1010-4 0.0001

5 5 = -lg1010-5 0.00001

6 6 = -lg1010-6 0.000001

7 7 = -lg1010-7 0.0000001

8 8 = -lg1010-8 0.00000001

9 9 = -lg1010-9 0.000000001

10 10= -lg1010-10 0.0000000001

11 11= -lg1010-11 0.00000000001

12 12= -lg1010-12 0.000000000001

13 13= -lg1010-13 0.0000000000001

14 14= -lg1010-14 0.00000000000001

Easier using pH scale than Conc [H+] • Low pH – High H+ conc – More acidic • High pH – Low H+ conc – Less acidic • pH 3 (10x) more acidic > than pH 4 • 1 unit change in pH is 10 fold change in Conc [H+]

Relationship between pH and Conc H+

Uncertainty involving pH

83

987

Average

Uncertainty involving pH

4 sig fig 5 sig fig/4 decimal place

4 sig fig 5 sig fig/4 decimal place

Conc H+ = 3.2 x 10-5 M

pH = - log10[3.2 x 10-5]= 4.4948 = 4.49

2 sig fig 3 sig fig/2 decimal place

2 sig fig 3 sig fig/2 decimal place

2 sig fig 3 sig fig/2 decimal place

2 sig fig 3 sig fig

2 sig fig 3 sig fig

2 sig fig 3 sig fig

pH solution = 7.40. Cal conc of H+ ions

7.40 = -log10 [H+] [H+] = 10-7.40

= 4.0 x 10-8

3 sig fig 2 sig fig

2 sig fig

4.7

]107.3lg[

107.3

3

101010

8

8

987

pH

pH

Average

Average

pH weak acid at various concentration

OHCOOCHOHCOOHCH 3323

Extend of dissociation depend on initial concentration acid

Conc of acid Observed pH CH3COOH Calculated pH HCI

0.10 2.7 1.0

0.010 3.0 2.0

0.0010 3.5 3.0

0.00010 4.2 4.0

CIHHCI

Weak acid Strong acid

Dissociate partially Dissociate completely

At same acid concentration • HCI has HIGHER [H+] > CH3COOH

• HCI has LOWER pH < CH3COOH • HCI dissociate completely- Strong acid • CH3COOH dissociate partially- Weak acid

At decreasing acid concentration • Extend of dissociation for CH3COOH increase • pH weak acid closer to strong acid • Dilution increase the extend of dissociation

Conc decrease

OHCOOCHOHCOOHCH 3323

Trends

Addition Water

Dilution shift equilibrium to right

Decrease conc of CH3COOH, CH3COO- and H+

Conc on left side is more effected due to CH3COO- and H+

Equilibrium shift to right to increase conc of CH3COO- and H+ again

Extend of dissociation for acid increase (shift to right)

О

О

Concept Map

[H+] [OH-]

pH pOH

Kw = [H+] x [OH-] = 1 x 10-14

pH + pOH = 14

pH = -lg [H+] [H+] = 10-pH pOH = -lg [OH-] [OH-] = 10-pOH

OHOH3

14100.1 7714 101101100.1

OHOHOHOH 322

H2O dissociate forming H3O+ and OH- (equilibrium exist)

14100.1 wK

Kw = 1.0 x 10-14 Ionic Product constant water at -25C

Kc – Ionic Product Constant Water

H+ OH-

Ionic Product Water, Kw, is Temperature dependent

Temp/C Kw [H+] [OH-] pH

0 1.5 x 10-15 0.39 x 10-7 0.39 x 10-7 7.47

10 3.0 x 10-15 0.55 x 10-7 0.55 x 10-7 7.27

20 6.8 x 10-15 0.82 x 10-7 0.82 x 10-7 7.08

25 1.0 x 10-14 1.00 x 10-7 1.00 x 10-7 7.00

30 1.5 x 10-14 1.22 x 10-7 1.22 x 10-7 6.92

40 3.0 x 10-14 1.73 x 10-7 1.73 x 10-7 6.77

50 5.5 x 10-14 2.35 x 10-7 2.35 x 10-7 6.63

OHOHOHOH 322

OHOHKw 3

molkJH /57 Temp increase ↑ → Equilibrium shift right → Reduce Temp ↓ → More ion form Kw increase ↑

Temp ↑ - shift right – more H+/OH- – Kw ↑ Temp ↑ - Kw ↑ – H+ ion ↑ - pH ↓

At 25C, Kw - 1.0 x 10-14

Conc [H+] = [OH−]= 1.0 x 10-7

Neutral pH = 7

At 50C, Kw - 5.5 x 10-14

Conc [H+]= [OH−]= 2.35 x 10-7

Neutral pH = 6.63

OHOHKw 3

At 25C, Kw - 1.0 x 10-14

•Kw = [H+][OH−] • 1.0 x 10-14 = [H+][OH−] • [H+][OH−] = [10-7][10-7] • pH = -lg[H+] • pH = -lg [1.0 x 10-7] • Neutral pH = 7

At 50C, Kw - 9.3 x 10-14

•Kw = [H+][OH−] •9.3 x 10-14 = [H+][OH−] •[H+]2 = 9.3 x 10-14 •[H+] = 3.05 x 10-7 • pH = -lg[3.05 x 10-7] • Neutral pH = 6.5

Amount same

Amount same

Ionic Product Water, Kw, is Temperature dependent

Temp/C

Kw [H+] [OH-] pH

0 1.5 x 10-15 0.39 x 10-7 0.39 x 10-7 7.47

10 3.0 x 10-15 0.55 x 10-7 0.55 x 10-7 7.27

20 6.8 x 10-15 0.82 x 10-7 0.82 x 10-7 7.08

25 1.0 x 10-14 1.00 x 10-7 1.00 x 10-7 7.00

30 1.5 x 10-14 1.22 x 10-7 1.22 x 10-7 6.92

40 3.0 x 10-14 1.73 x 10-7 1.73 x 10-7 6.77

50 5.5 x 10-14 2.35 x 10-7 2.35 x 10-7 6.63

OHOHOHOH 322

OHOHKw 3

molkJH /57

Temp increase ↑→ Equilibrium shift right → Reduce Temp ↓→ More ion form Kw increase ↑

Temp ↑ - shift right – more H+/OH- – Kw ↑ Temp ↑ - Kw ↑ – H+ ion ↑ - pH ↓

At 25C, Kw - 1.0 x 10-14

Conc [H+] = [OH−]= 1.0 x 10-7

Neutral pH = 7

At 50C, Kw - 5.5 x 10-14

Conc [H+]= [OH−]= 2.35 x 10-7

Neutral pH = 6.63

Kc ionization water = 1.80 x 10-16. Based on magnitude of Kc which direction does it lies? Calculate Kw for water assume [H2O] is constant = 55.6 mol/dm3

OH

OHHKc

2

Kw = 1.0 x 10-14 Ionic Product constant water at -25C

• Direction to the left

• Mostly undissociated water molecules

treac

productKc

tan

14

16

16

100.1

1080.155

551080.1

OHH

OHH

OHH

14100.1 wK

OHHOH2

OHHOHKK cw 2

Fraction of ionized = Amt ionized = 1.00 x 10-7 = 18 x 10-10 Initial amt 55.6

7714 101101100.1

Kc small

18 molecule ionized in 10 000 000 000

Amount same Amount same

Formula for acid/base calculation

[OH-] [H+] Kw = [H+] x [OH-] = 1 x 10-14

[OH-] = 10-pOH pOH = -lg [OH-]

pOH pH

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 = 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

AHKa

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 position Ka/Kb large ↑ – ↑ dissociation – shift to right – favour product Ka/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 position 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 Base Conjugate acid base pair

Small dissociation constant

Strong Acid Weak base

ba KK /

Str

ong a

cid

Stro

ng b

ase

Formula for acid/base calculation

[OH-] [H+] Kw = [H+] x [OH-] = 1 x 10-14

[OH-] = 10-pOH pOH = -lg [OH-]

pOH pH

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 = 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

AHKa

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

What is pH for [H+] = 1 x 10-12 M pH = -lg [10-12]

pH = 12

What is conc of H+ of pH 3.20?

3.20 = -lg [H+] [H+] = 10 –2.20

[H+] = 6.3 x 10-4

pH = -log10[H+] pOH = -log10[OH-] pH + pOH = 14 Kw = [H+][OH-]

Formula acid/base calculation

2 sig fig 1 sig fig 3 sig fig 2 sig fig

What is pH for [OH-] = 0.15M pOH = -lg [0.15] pOH = 0.823 pH + pOH = 14 pH = 14 – 0.823 = 13.2

pOH = -log[OH-]

3 sig fig 2 sig fig

Calculate conc of H+, OH- and pH for 0.001 M HCI.

1 2 3

4

CIHHCI0.001 ↔ 0.001 0.001

OHHOH2

HCI H2O

OHHKwAssuming H+ all from HCI = 0.0010

)()( 2OHHHCIHH

= 0.001 Negligible / too little

OHH14100.1

0.3

001.0log

log

10

10

pH

pH

HpH

0.31114

11

101001.0

100.1

001.0100.1

1114

14

pH

pOH

OH

OH

or

Cal conc OH-/pH when 3.o x 10-4 H+ add water

HCI H2O

CIHHCI OHHOH2

OHHKw

OHH14100.1

11

4

14

414

103.3100.3

100.1

100.3100.1

OH

OH

3x10-4 ↔ 3x10-4

52.3

100.3log

log

4

10

10

pH

pH

HpH

5

11

10loglog

101001.0

100.1

001.0100.1

11

1114

14

pH

HpH

H

H

00.1

10.0log

log

10

10

pH

pH

HpH

Cal pH of 0.10 M HCI

H2O

Assuming H+ all from HCI = 0.10

)()( 2OHHHCIHH

CIHHCI0.10 mol 0.10 mol

= 0.10

Strong Acid/Base calculation

Strong acid • 100% dissociation (complete)

Strong base • 100% dissociation (complete)

CIHHCI OHKKOHShift right Shift right

2 sig fig 3 sig fig

Cal pH of 0.10M H2SO4

H2O

2

442 2 SOHSOH

0.10 mol 0.20 mol

Assuming H+ all from H2SO4 = 0.20

700.0

20.0log

log

10

10

pH

pH

HpH

)()( 242 OHHSOHHH

= 0.20

2 sig fig 3 sig fig

OHKKOH0.001 mol 0.001 mol

Cal pH of 0.001 M KOH

H2O

Assume OH- from KOH = 0.10

)()( 2OHOHKOHOHOH

OHHKw

= 0.001

11,3

001.0log

log

pHpOH

pOH

OHpOH

OHCaOHCa 2)( 2

20.001 mol 0.002 mol

Cal pH of 0.001M Ca(OH)2

H2O

Assume OH- from Ca(OH)2 = 0.002

)()( 2OHOHKOHOHOH

= 0.002

3.11,7.2

002.0log

log

pHpOH

pOH

OHpOH

OHHKw

3.11

105log

105002.0

100.1

002.0100.1

12

1214

14

pH

pH

H

H

0.2

01.0log

log

10

10

pH

pH

HpH

OH

OHOHKc

2

3

OHOHKOHK wc 32

OHOH3

14100.1 7714 101101100.1

OHOHOHOH 322

H2O dissociate forming H3O+ and OH- (equilibrium exist)

14100.1 wK

Dissociation water small [H2O] is constant

Kw - Ionic product constant water Kw = 1.0 x 10-14 Ionic Product constant water at -25C

Kc - Dissociation constant water

Cal conc of H+ ,OH- and pH of water Cal conc of H+ ,OH- and pH of 0.01M HCI

OHHOH2

OHHKw

OHH14100.1

7714 101101100.1

7101 H

H2O H2O

HCI

OHHOH2

H2O

OHHKw

OHH14100.1

Assuming H+ all from HCI = 0.01

)()( 2OHHHCIHH

H+ = 0.01 + 1.0x10-12

= 0.01 + 0.000000000001 ≈ 0.01

OHHOH2

H+ = 1x10-12 OH- = 1x10-12

CIHHCI0.01 mol 0.01 mol 0.01

1 mol ↔ 1 mol 1mol

0.000000000001

0.000000000001

H+ OH-

= 0.01 = 0.000000000001

or

0.7

101log 7

10

pH

pH

0.21214

12

100.101.0

100.1

01.0100.1

1214

14

pH

pOH

OH

OH

OH

OHOHKc

2

3

OHOHKOHK wc 32

OHOH3

14100.1 7714 101101100.1

OHOHOHOH 322

H2O dissociate forming H3O+ and OH- (equilibrium exist)

14100.1 wK

Dissociation water small [H2O] is constant

Kw - Ionic product constant water Kw = 1.0 x 10-14 Ionic Product constant water at -25C

Kc - Dissociation constant water

Cal conc of H+ ,OH- and pH of 0.01M KOH Cal conc of H+ ,OH- and pH of 0.1M H2SO4

7.0

2.0log

log

10

10

pH

pH

HpH

H2O

KOH

H2SO4

OHHOH2

H2O

OHHKw

OHH14100.1

Assuming H+ all from H2SO4 = 0.2

7.03.1314

3.13

100.52.0

100.1

2.0100.1

1414

14

pH

pOH

OH

OH

)()( 242 OHHSOHHH

H+ = 0.2 + 5 x 10-14

= 0.2 + 0.000000000000005 ≈ 0.2

OHHOH2

H+ = 5x10-14 OH- = 5x10-14

2

442 2 SOHSOH

0.1 mol 0.2 mol

0.2

1 mol ↔ 1 mol 1 mol

0.00000000000005

0.0000000000005

H+ OH-

= 0.2 = 0.00000000000005

OHHOH2

1 mol ↔ 1 mol 1 mol

OHHOH2

OHHKw

OHKKOH

0.01

Assuming OH- all from KOH = 0.01

)()( 2OHOHKOHOHOH

= 0.01 = 0.000000000001

12

10log

log

10101.0

100.1

01.0100.1

12

10

10

1214

14

pH

pH

HpH

H

H

or

H+ = 1x10-12 OH- = 1x10-12

0.01 mol 0.01 mol

Approximation and Assumption

Ka very small < 10-5 Not much change acid conc Approximation is VALID Ionization make no diff to conc HA

SMALLKa SMALLKa Find pH of 0.10 M HA, weak acid Ka - 1.8 x 10-5

- Little product form

- Initial conc reactant unchanged

Using

approximation

0.10 – x ≈ 0.10

HA ↔ H+ + A-

Initial conc 0.10 0 0

Change 0.10 - x +x +x

Eq Conc 0.10 – x +x +x

HA ↔ H+ + A-

HA

AHKa

xx

10.0108.1

2

5

10.0

108.1

2

5 x

31034.1 x

[HA] = (0.10 – 0.00134) = 0.098 ≈ 0.10

[HA]initial ≈ [HA]eq

Calculation Weak Acid (Using ICE Method)

87.2

1034.1log

log

3

10

10

pH

pH

HpH

HA ↔ H+ + A-

Find Ka of 0.02 M HA, weak acid, [H]+ = 0.0012M

HA ↔ H+ + A-

Initial conc 0.02 0 0

Change 0.02 – 0.0012 +0.0012 +0.0012

Eq Conc ≈ 0.02 +0.0012 +0.0012

HA

AHKa

02.0

0012.00012.0aK

Using

approximation

0.02 – 0.0012 ≈ 0.02

5102.7 aK

Find Ka of 0.01M HA, weak acid, pH = 5.0

HA ↔ H+ + A-

HA ↔ H+ + A-

Initial conc 0.01 0 0

Change 0.01 – 1x10-5 +1x10-5 +1x10-5

Eq Conc ≈ 0.01 +1x10-5 +1x10-5

5

10

101

log0.5

H

H

HA

AHKa

01.0

101101 55 aK

Using

approximation

0.01 – 1x10-5 ≈ 0.01

8108.1 aK

< 5% rule

%3.1%10010.0

1034.1

%5

3

concInitial

x

Approximation and Assumption

SMALLKa SMALLKa

HA ↔ H+ + A-

HA

AHKa

HA

H2

6101.4

4104.2 HA

Calculation Weak Acid (Using ICE Method)

5101.3

log50.4

log

H

H

HpH

Find pH of 0.100M HA, weak acid, pKa = 4.20

HA

AHKa

600.2

1051.2log 3

pH

pH

Find Conc HA, weak acid, pH = 4.50, Ka = 4.1 x 10-6

HA

256 101.3

101.4

HA ↔ H+ + A-

100.0

1031.6

2

5

H

51031.6

log2.4

log

a

a

aa

K

K

KpK 31051.2 H

3 sig fig 4 sig fig

Find Ka of 0.01M CH3COOH, pH = 3.4

CH3COOH ↔ CH3COO- + H+

Initial conc 0.01 0 0

Change 0.01 – 0.0004 +0.0004 +0.0004

Eq Conc ≈ 0.01 +0.0004 +0.0004

0004.001.0

10424

3

2

3

3

COOHCH

H

COOHCH

HCOOCHKa

4104

log4.3

log

H

H

HpH

5106.1 aK

01.0

10424

aK

Using

approximation

0.01 – 0.0004 ≈ 0.01

Ka very small < 10-5 Not much change acid conc Approximation VALID

Ionization make no diff to conc acid

Find pH of 0.75M CH3COOH, Ka = 1.8 x 10 -5

CH3COOH ↔ CH3COO- + H+

Initial conc 0.75 0 0

Change 0.75 - x +x +x

Eq Conc ≈ 0.75 +x +x

x

x

COOHCH

HCOOCHKa

75.0

2

3

3

Using

approximation

0.75 – x ≈ 0.75

75.0

108.1

2

5

H

3107.3 H

40.2

107.3log 3

pH

pH

Approximation and Assumption

Kb very small < 10-5 Not much change reactant conc Approximation is VALID Ionization make no diff to conc B

SMALLKb SMALLKb Find pH of 0.010M B, weak base Kb - 1.8 x 10-5

- Little product form

- Initial conc reactant unchanged

Using

approximation

0.01 – x ≈ 0.01

B + H2O ↔ BH+ + OH-

Initial conc 0.01 0 0

Change 0.01 - x +x +x

Eq Conc 0.01 – x +x +x

B + H2O ↔ BH+ + OH-

B

OHBHKb

xx

010.0108.1

2

5

4102.4 x

[B] = (0.01 – 0.00042) ≈ 0.01

[B]initial ≈ [B]eq

Calculation Weak Base (Using ICE Method)

6.10

37.314

37.3

102.4log

log

4

pH

pH

pOH

pOH

OHpOH

Find Kb of 0.030M B, weak base, pH = 10.0

Using

approximation

0.03 – 0.0001 ≈ 0.03

7103.3 bK

010.0

108.1

2

5 x

B + H2O ↔ BH+ + OH-

B + H2O ↔ BH+ + OH-

Initial conc 0.03 0 0

Change 0.03 - x +x +x

Eq Conc 0.03 – x +x +x

B

OHBHKb

xx

Kb

030.0

2

4100.1

log4

log

1014

OH

OH

OHpOH

pOH

030.0

100.124

bK

Find conc of B, weak base, pH 10.8, Kb - 4.36 x 10-4

B + H2O ↔ BH+ + OH-

B

OHBHKb

B

22.34 100.1

1036.4

2.3100.1

log2.3

log

8.1014

OH

OH

OHpOH

pOH

4101.9 B

< 5% rule

%2.4%10001.0

102.4

%5

4

concInitial

x

x

x

NH

OH

20.0

2

3

2

20.0

108.1

2

5 x

Approximation and Assumption

SMALLKb SMALLKb

Calculation Weak Base (Using ICE Method)

60.11400.214

400.2

1046.4log

log

3

pH

pOH

pOH

OHpOH

31046.4 OH

3

4

NH

OHNHKb

3109.1 x

Using

approximation

0.20 – 0.0019 ≈ 0.20

4101.9 B

Find pH of 0.0500M B, weak base pKb - 3.40

B + H2O ↔ BH+ + OH-

B

OHBHKb

0500.0

1098.3

2

4

OH

41098.3

log40.3

log

b

b

bb

K

K

KpK

Find conc of B, weak base pH - 10.8, pKa = 10.64

B + H2O ↔ BH+ + OH-

B

OHBHKb

B

OH2

41036.4

41036.4

log36.3

log

36.3

64.1014

14

b

b

bb

b

b

ba

K

K

KpK

pK

pK

pKpK

4103.6

log2.3

log

2.38.1014

OH

OH

OHpOH

pOH B

244 103.6

1036.4

NH3 + H2O ↔ NH4+ + OH-

Initial conc 0.20 0 0

Change 0.20 – x +x +x

Eq Conc ≈ 0.20 +x +x

Find pH of 0.20M NH3 - Kb - 1.8 x 10-5

28.11

72.214

72.2

109.1log

log

3

pH

pH

pOH

pOH

OHpOH

Find Kb of 0.10M CH3NH2 , pH = 11.8

CH3NH2 + H2O ↔ CH3NH3+ + OH-

Initial conc 0.10 0 0

Change 0.10 – x +x +x

Eq Conc 0.10 - x +x +x

23

33

NHCH

OHNHCHKb

xx

10.0

2

3103.6

log20.2

20.28.1114

14

OH

OH

pOH

pOHpH

3

3

103.610.0

103.6

bK

4102.4 bK

> 5% rule

Approximation

%7.6%10010.0

103.6

%5

3

concInitial

x

H+ = 1 x 10-8 + 9.5 x 10-8

H+ = 10.5 x 10-8

980.6

105.10log

log

8

10

10

pH

pH

HpH

H2O HCI

OHHOH2

H2O

OHHKw

xx 814 101100.1 Assuming H+ from HCI and H2O

)()( 2OHHHCIHH

OHHOH2

CIHHCIx

H+ = 1 x 10-8 + x

8105.9 x

Cal pH of 0.10M HCI

CIHHCI0.10 mol 0.10 mol

Assuming H+ all from HCI = 0.10

)()( 2OHHHCIHH = 0.10

00.1

10.0log

log

10

10

pH

pH

HpH

2 sig fig 3 sig fig

Cal pH of 1 x 10-8M HCI

1 x 10-8 Assuming dissociation

H+ ions from water = x 1 x 10-8

pH of very STRONG CONC acid

Strong acid • 100% dissociation (complete)

CIHHCI

Shift right

H+ ions from water is negligible

Assume all H+ ions come from ACID

pH of very STRONG DILUTED acid

H+ ions from water is SIGNIFICANT

All H+ ions come from ACID and H2O

Assuming H+ from HCI = 1 x 10-8

0.8

101log

log

8

10

10

pH

pH

HpH

ALKALINE!!!!!!!

Click here to view

Table for Ka/Kb Expt acid/base (RSC)

Click here to view

1 x 10-8

Formula for acid/base calculation

[OH-] [H+] Kw = [H+] x [OH-] = 1 x 10-14

[OH-] = 10-pOH pOH = -lg [OH-]

pOH pH

pH = -lg [H+] [H+] = 10-pH

pH + pOH = 14

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

HF + H2O ↔ F- + H3O+

STRONG BASE

WEAK BASE

Kb increase

pKb increase pKb decrease

Kb decrease

STRONG ACID

WEAK ACID

Ka increase

pKa decrease

Ka decrease

pKa increase

Ka pKa

Kb pKb

pKa = -lg [Ka]

pKb = -lg [Kb]

Ka = 10-pKa

Kb = 10-pKb

Ka x Kb = Kw

Ka x Kb = 1 x 10-14 pKa + pKb = 14

pKa + pKb = pKw

Find Kb for F- , Ka HF - 6.8 x 10-4

HF (acid) - F- (conjugate base)

4

4

14

14

1098.3

108.6

101

101

)()(

b

b

a

b

wba

K

K

KK

KFKHFK

3

4

2

4

2

442

4243

POHHPO

HPOHPOH

POHHPOH

Successive acid dissociation constant

3

443 3 POHPOH

Polyprotic acid – dissociate releasing 1 proton each time

13

3

8

2

3

1

108.4

102.6

105.7

K

K

K

3

443 3 POHPOH

+ Less acidic

Increasing difficulty

removing H+ from

negatively charged ion

Most acidic

wba KKK

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