IB Chemistry on Acid Base Dissociation Constant and Ionic Product Water

1. Strong/Weak Acid and Base
Strong Acid/Weak Acid
Strong acid - HI, HBr, HCI, HNO3, H2SO4, HCIO3, HCIO4
Weak Acid - CH3COOH, HF, HCN, H2CO3, H3BO3, H3PO4
Strong Base/ Weak Base
Strong base - LiOH, KOH, NaOH, CsOH, Ca(OH)2
Weak Base - NH3, C2H5NH2, (CH3)2NH, C3H5O2NH2
Distinguishbet strong and weak acid
ElectricalconductivityRate of rxn pH
Strongacid
Strong acid → High ionization → High conc H+ → High conductivity→ High rate rxn → Lower pH
Strong acid
Oxoacid
O atom > number ionizable proton
HNO3, H2SO4, HCIO3,HCIO4
Hydrohalicacid
HI, HBr, HCI
Weak acid
Hydrohalicacid
HF
Oxoacid
O atom ≥ number ionizable protonby 1
HCIO, HNO2, H3PO4
Carboxylicacid
COOH
Strong base – containOH- or O2-
LiOH, NaOH, CaO, K2O Ca(OH)2, Ba(OH)2
Weak base – contain 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

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

3. 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/basecalculation
  
 OH
OHOH
Kc
2
3


    
 OHOHOHKc 32
  
 OHOHKw 3
  
 OHOH3
14
100.1
  7714
101101100.1 

  7
101 
OH
  
 OHOHOHOH 322
H2O dissociateforming H3O+ and OH-
(equilibriumexist)
14
100.1 
wK
Dissociation water small [H2O] is constant
Kw = 1.0 x 10-14 Ionic Product constantwater at -25C
Kc - Dissociation constant water
  7
3 101 
OH

4. 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
8
3
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 fig3 sig fig/2 decimal place
2 sig fig3 sig fig/2 decimal place
2 sig fig 3 sig fig
2 sig fig3 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

5. pH weak acid at variousconcentration

 OHCOOCHOHCOOHCH 3323
Extendof dissociationdependon initialconcentrationacid
Conc of acid Observed pH CH3COOH CalculatedpH 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 dissociatepartially- 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
Decreaseconc of CH3COOH,CH3COO- andH+
Conc on left side is more effecteddue to CH3COO- and H+
Equilibrium shift to right to increase
conc of CH3COO- andH+ 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

6.   
 OHOH3
14
100.1   7714
101101100.1 


 OHOHOHOH 322
H2O dissociate forming H3O+ and OH-
(equilibrium exist)
14
100.1 
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

7. 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
OHH
Kc
2


Kw = 1.0 x 10-14 Ionic Product constant water at -25C
• Direction to the left
• Mostly undissociated water molecules
 
 treac
product
Kc
tan

  
 
  
   14
16
16
100.1
1080.155
55
1080.1







OHH
OHH
OHH
14
100.1 
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

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

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

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

11. 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-]
Formulaacid/basecalculation
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

 CIHHCI
0.001 ↔ 0.001 0.001

 OHHOH2
HCIH2O
  
 OHHKw
Assuming H+ all from HCI = 0.0010
)()( 2OHHHCIHH 

= 0.001 Negligible / too little
  
 OHH14
100.1
 
 
0.3
001.0log
log
10
10


 
pH
pH
HpH
  
 
0.31114
11
101
001.0
100.1
001.0100.1
11
14
14










pH
pOH
OH
OH
or
Cal conc OH-
/pH when3.o x 10-4
H+
add water
HCI
H2O

 CIHHCI
 OHHOH2
  
 OHHKw
  
 OHH14
100.1
  
  11
4
14
414
103.3
100.3
100.1
100.3100.1










OH
OH
3x10-4 ↔ 3x10-4
 
 
52.3
100.3log
log
4
10
10





pH
pH
HpH
5

12.   
 
   
11
10loglog
101
001.0
100.1
001.0100.1
11
11
14
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 


 CIHHCI
0.10 mol 0.10 mol
= 0.10
StrongAcid/Base calculation
Strong acid
• 100% dissociation (complete)
Strongbase
• 100% dissociation (complete)

 CIHHCI 
 OHKKOHShift right
Shift right
2 sig fig3 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 fig3 sig fig

 OHKKOH
0.001 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
2
0.001 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
105
002.0
100.1
002.0100.1
12
12
14
14











pH
pH
H
H

13.  
 
0.2
01.0log
log
10
10


 
pH
pH
HpH
  
 OH
OHOH
Kc
2
3


    
 OHOHKOHK wc 32
  
 OHOH3
14
100.1   7714
101101100.1 


 OHOHOHOH 322
H2O dissociate forming H3O+ and OH-
(equilibrium exist)
14
100.1 
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
  
 OHH14
100.1
  7714
101101100.1 

  7
101 
H
H2O H2O
HCI

 OHHOH2
H2O
  
 OHHKw
  
 OHH14
100.1
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

 CIHHCI
0.01 mol 0.01 mol0.01
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
100.1
01.0
100.1
01.0100.1
12
14
14










pH
pOH
OH
OH

14.   
 OH
OHOH
Kc
2
3


    
 OHOHKOHK wc 32
  
 OHOH3
14
100.1   7714
101101100.1 


 OHOHOHOH 322
H2O dissociate forming H3O+ and OH-
(equilibrium exist)
14
100.1 
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
  
 OHH14
100.1
Assuming H+ all from H2SO4 = 0.2
  
 
7.03.1314
3.13
100.5
2.0
100.1
2.0100.1
14
14
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
101
01.0
100.1
01.0100.1
12
10
10
12
14
14













pH
pH
HpH
H
H
or
H+ = 1x10-12 OH- = 1x10-12
0.01 mol 0.01 mol

15. Approximationand Assumption
Ka very small < 10-5
Not much change acid conc
Approximation is VALID
Ionizationmake 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
AH
Ka

  
 x
x

 
10.0
108.1
2
5
 
 10.0
108.1
2
5 x
 
3
1034.1 
x
[HA] = (0.10 – 0.00134)
= 0.098 ≈ 0.10
[HA]initial ≈ [HA]eq
CalculationWeak Acid (UsingICE 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
AH
Ka


  
 02.0
0012.00012.0
aK
Using
approximation
0.02 – 0.0012 ≈ 0.02
5
102.7 
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
AH
Ka


  
 01.0
101101 55 

aK
Using
approximation
0.01 – 1x10-5 ≈ 0.01
8
108.1 
aK
< 5% rule
%3.1%100
10.0
1034.1
%5
3




concInitial
x

16. Approximation and Assumption
SMALLKa  SMALLKa 
HA ↔ H+ + A-
  
 HA
AH
Ka

  
 HA
H
2
6
101.4



  4
104.2 
HA
CalculationWeak Acid (UsingICE Method)
 
 
  5
101.3
log50.4
log






H
H
HpH
Find pH of 0.100M HA, weak acid, pKa = 4.20
  
 HA
AH
Ka


 
600.2
1051.2log 3

 
pH
pH
Find Conc HA, weak acid, pH = 4.50, Ka = 4.1 x 10-6
 
 HA
25
6 101.3
101.4

 

HA ↔ H+ + A-
 
 100.0
1031.6
2
5



H
 
 
5
1031.6
log2.4
log




a
a
aa
K
K
KpK   3
1051.2 
H
3 sig fig4 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
104
24
3
2
3
3




COOHCH
H
COOHCH
HCOOCH
Ka
 
 
  4
104
log4.3
log






H
H
HpH
5
106.1 
aK
 
 01.0
104
24

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
HCOOCH
Ka



75.0
2
3
3
Using
approximation
0.75 – x ≈ 0.75
 
 75.0
108.1
2
5



H
  3
107.3 
H
 
40.2
107.3log 3

 
pH
pH

17. Approximationand Assumption
Kb very small < 10-5
Not much change reactant conc
Approximation is VALID
Ionizationmake 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
OHBH
Kb


 
 x
x

 
010.0
108.1
2
5
4
102.4 
x
[B] = (0.01 – 0.00042)
≈ 0.01
[B]initial ≈ [B]eq
CalculationWeak 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
7
103.3 
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
OHBH
Kb


 
 x
x
Kb


030.0
2
 
 
  4
100.1
log4
log
1014







OH
OH
OHpOH
pOH
 
 030.0
100.1
24

bK
Find conc of B, weak base, pH 10.8, Kb - 4.36 x 10-4
B + H2O ↔ BH+ + OH-
  
 B
OHBH
Kb


 
 B
22.3
4 100.1
1036.4

 

 
 
  2.3
100.1
log2.3
log
8.1014







OH
OH
OHpOH
pOH
  4
101.9 
B
< 5% rule
%2.4%100
01.0
102.4
%5
4




concInitial
x

18.  
 
 
 x
x
NH
OH



20.0
2
3
2
 
 20.0
108.1
2
5 x
 
Approximationand Assumption
SMALLKb  SMALLKb 
CalculationWeak Base (Using ICE Method)
 
 
60.11400.214
400.2
1046.4log
log
3






pH
pOH
pOH
OHpOH
  3
1046.4 
OH
  
 3
4
NH
OHNH
Kb


3
109.1 
x
Using
approximation
0.20 – 0.0019 ≈ 0.20
  4
101.9 
B
Find pH of 0.0500M B, weak base pKb - 3.40
B + H2O ↔ BH+ + OH-
  
 B
OHBH
Kb


 
 0500.0
1098.3
2
4



OH
 
 
4
1098.3
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
OHBH
Kb


 
 B
OH
2
4
1036.4



 
 
4
1036.4
log36.3
log
36.3
64.1014
14







b
b
bb
b
b
ba
K
K
KpK
pK
pK
pKpK
 
 
  4
103.6
log2.3
log
2.38.1014







OH
OH
OHpOH
pOH 
 B
24
4 103.6
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
OHNHCH
Kb


 
 x
x


10.0
2
 
  3
103.6
log20.2
20.28.1114
14






OH
OH
pOH
pOHpH
 
 3
3
103.610.0
103.6




bK
4
102.4 
bK
> 5% rule
Approximation
%7.6%100
10.0
103.6
%5
3




concInitial
x

19. 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

 CIHHCI
x
H+ = 1 x 10-8 + x
  8
105.9 
x
Cal pH of 0.10M HCI

 CIHHCI
0.10 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 fig3 sig fig
Cal pH of 1 x 10-8M HCI
1 x 10-8Assuming 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/KbExpt acid/base (RSC)
Click here to view
1 x 10-8

20. 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
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 increasepKb 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
K
K
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 

21. 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