IB Chemistry on Acid Base Dissociation Constant and Ionic Product Water

1. http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
Tutorial on Acid/Base Dissociation Constant, Ka,
pKa, pKb and pH

2. 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
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

3. 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 scaleConc H+

4. 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
OHOH
Kc
2
3


    
 OHOHOHKc 32
  
 OHOHKw 3
  
 OHOH3
14
100.1
  7714
101101100.1 

  7
101 
OH
  
 OHOHOHOH 322
H2O dissociate forming H3O+ and OH-
(equilibrium exist)
14
100.1 
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

5. 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

6. 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

7.   
 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

8. 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

9. 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 = 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


Dissociation Constant 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 

10. 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 BaseConjugate acid base pair
Small dissociation
constant
Strong Acid Weak base
ba KK /
Strongacid
Strongbase

11. 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 = 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


Dissociation Constant for Weak Base

 OHBHOHB 2
  
 B
OHBH
Kb



 OHNHOHNH 423
  
 
 
 3
2
3
4
NH
OH
NH
OHNH
Kb


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

12. 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

 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 when 3.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

13.   
 
   
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
Strong Acid/Base calculation
Strong acid
• 100% dissociation (complete)
Strong base
• 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

14.  
 
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

15.   
 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

16. 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
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
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
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

17. Approximation and Assumption
SMALLKa  SMALLKa 
HA ↔ H+ + A-
  
 HA
AH
Ka

  
 HA
H
2
6
101.4



  4
104.2 
HA
Calculation Weak Acid (Using ICE 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

18. 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
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
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
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

19.  
 
 
 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
  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

20. 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

21. 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 

22. 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

23. Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://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 tutorial
http://lawrencekok.blogspot.com