http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
Tutorial on Acid/Base Dissociation Constant, Ka,
pKa, pKb and pH
Strong/Weak Acid and Base
Strong Acid/Weak Acid
Strong acid - HI, HBr, HCI, HNO3, H2SO4, HCIO3, HCIO4
Weak Acid - CH3COOH, HF, HCN, H2CO3, H3BO3, H3PO4
Strong Base/ Weak Base
Strong base - LiOH, KOH, NaOH, CsOH, Ca(OH)2
Weak Base - NH3, C2H5NH2, (CH3)2NH, C3H5O2NH2
Distinguish bet strong and weak acid
Electrical conductivityRate of rxn pH
Strong acid
Strong acid → High ionization → High conc H+ → High conductivity → High rate rxn → Lower pH
Strong acid
OxoacidO atom > number ionizable proton
HNO3, H2SO4, HCIO3, HCIO4
Hydrohalic acidHI, HBr, HCI
Weak acid
Hydrohalic acidHF
OxoacidO atom ≥ number ionizable proton by 1
HCIO, HNO2, H3PO4
Carboxylic acid COOH
Strong base – contain OH- or O2-
LiOH, NaOH, CaO, K2O Ca(OH)2, Ba(OH)2
Weak base – contain electron rich nitrogen, NNH3, C2H5NH2, (CH3)2NH, C3H5O2NH2
Strong base Weak base
1 2 3
Weak acid
0.1 M HCI 0.1 M CH3COOH
H+ 0.1 mole 0.0013 mole
pH 1 (Low) 2.87 (High)
Electrical conductivity High (Ionize completely) Low (Ionize partially)
Rate with magnesium Fast Slow
Rate with calcium carbonate
Fast Slow
Weaker acid → Low ionization → Low conc H+ → Low conductivity → Low rate rxn → High pH
Strong acid
HA A-H+
H+ H+
H+
H+ H+
H+
H+A-
A-
A-
A- A-
A-
Ionizes completely
Weak acid
HAHA
H+ A-H+
H+
A-
A-HA
HA
HA
HA
HA
HA
Ionizes partially
Easier using pH scale than Conc [H+]• Conc H+ increase 10x from 0.0001(10-4) to 0.001(10-3) - pH change by 1 unit from pH 4 to 3• pH 3 is (10x) more acidic than pH 4• 1 unit change in pH is 10 fold change in Conc [H+]
Conc OH- increase ↑ by 10x
pH increase ↑ by 1 unit
pOH with Conc OH-
pOH = -log [OH-][OH-] = 0.0000001MpOH = -log [0.0000001]pOH = -log1010-7
pOH = 7pH + pOH = 14pH + 7 = 14pH = 7 (Neutral)
pH with Conc H+
pH = -log [H+][H+] = 0.0000001MpH = -log [0.0000001]pH = -log1010-7
pH = 7 (Neutral)
Conc H+ increase ↑ by 10x
pH decrease ↓ by 1 unit
pH measurement of Acidity of solution
• pH is the measure of acidity of solution in logarithmic scale• pH = power of hydrogen or minus logarithm to base ten of hydrogen ion concentration
← Acidic – pH < 7 Alkaline – pH > 7 →
pOH with Conc OH-
pOH = -log [OH-][OH-] = 0.1MpOH = -log[0.1]pOH = 1pH + pOH = 14pH + 1 = 14
pH = 13 (Alkaline)
pH with Conc H+
pH = -log [H+][H+] = 0.01MpH = -log [0.01]pH = -log1010-2
pH = 2 (Acidic)
Easier pH scaleConc H+
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 = 2pH + pOH = 14pH + 2 = 14pH = 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 = 12pH + pOH = 14pH + 12 = 14pH = 2
Formula for acid/base calculation
OH
OHOHK c
2
3
OHOHOHKc 32
OHOHKw 3
OHOH 3
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 calculationSignificant 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+ concpH 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 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
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 increaseconc 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
OHOH 3
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 formKw 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 formKw 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
productK c
tan
14
16
16
100.1
1080.155
551080.1
OHH
OHH
OHH
14100.1 wK
OHHOH 2
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-pOHpOH = -lg [OH-]
pOHpH
pH = -lg [H+] [H+] = 10-pH
pH + pOH = 14
Formula for acid/base calculation
Dissociation Constant for Weak Acid
pH = -log10[H+] pOH = -log10[OH-]pH + pOH = 14pH + pOH = pKw
Kw = [H+][OH-]Ka x Kb = Kw
Ka x Kb = 1 x 10-14
pKa = - lg10Ka
pKb = - lg10Kb
pKa + pKb = pKw
pKa + pKb = 14
AHHA
HA
AHK a
HCOOCHCOOHCH 33
COOHCH
H
COOHCH
HCOOCHKa
3
2
3
3
Dissociation Constant for Weak Base
OHBHOHB 2
B
OHBHKb
OHNHOHNH 423
3
2
3
4
NH
OH
NH
OHNHKb
OHCOOCHOHCOOHCH 3323
OHCOOCHOHCOOHCH 3323
COOHCH
OHCOOCHKa
3
33
OHCOOHCHOHCOOCH 323
COOCH
OHCOOHCHKb
3
3
Derive Ka x Kb = Kw
Relationship bet Weak acid and its conjugate base
Weak acid Conjugate Base
COOCH
OHCOOHCH
COOHCH
OHCOOCH
3
3
3
33
OHOHCOOCH
OHCOOHCH
COOHCH
OHCOOCH3
3
3
3
33
wba KKK
Formula for acid/base calculation
Ka /Kb measure equilibrium positionKa/Kb large ↑ – ↑ dissociation – shift to right – favour productKa/Kb large ↑ – pKa /pKb small ↓ – Stronger acid/base
Strong acid Large ↑ Ka
Weak acid Small ↓ Ka
Strong base Large ↑ Kb
Weak base Small ↓Kb
↑ Ka → ↓ pKa
Ka /Kb measure equilibrium positionKa /Kb small ↓ – ↓ dissociation – shift to left – reactant favourKa /Kb small ↓ – pKa /pKb high ↑– Weak acid/base
↑ Kb → ↓ pKb
↓ Ka → ↑ pKa
↓ Kb →↑ pKb
For weak acid/ base
CIHHCI OHNHOHNH 423
Shift right Shift left
CH3COOH + H2O ↔ CH3COO- + H3O+
CH3COOH CH3COO-CH3COOH ↔ CH3COO-
Strong Acid Weak conjugate BaseConjugate acid base pair
Small dissociationconstant
Strong Acid Weak base
ba KK /
Str
on
g a
cid
Stro
ng
ba
se
Formula for acid/base calculation
[OH-][H+]Kw = [H+] x [OH-] = 1 x 10-14
[OH-] = 10-pOHpOH = -lg [OH-]
pOHpH
pH = -lg [H+] [H+] = 10-pH
pH + pOH = 14
Formula for acid/base calculation
Dissociation Constant for Weak Acid
pH = -log10[H+] pOH = -log10[OH-]pH + pOH = 14pH + pOH = pKw
Kw = [H+][OH-]Ka x Kb = Kw
Ka x Kb = 1 x 10-14
pKa = - lg10Ka
pKb = - lg10Kb
pKa + pKb = pKw
pKa + pKb = 14
AHHA
HA
AHK a
HCOOCHCOOHCH 33
COOHCH
H
COOHCH
HCOOCHKa
3
2
3
3
Dissociation Constant for Weak Base
OHBHOHB 2
B
OHBHKb
OHNHOHNH 423
3
2
3
4
NH
OH
NH
OHNHKb
Dissociate partially ↔ used
Weak acid/base
Ka /Kb value pKa /pKb value easier!
Click here weak acid dissociation Click here weak acid dissociation Click here CH3COOH dissociation Click here strong acid ionization
Weak acid/base Animation
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 = 14pH = 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
OHHOH 2
HCIH2O
OHHK wAssuming 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
HCIH2O
CIHHCI OHHOH 2
OHHK w
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 rightShift 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
OHKKOH0.001 mol 0.001 mol
Cal pH of 0.001 M KOH
H2O
Assume OH- from KOH = 0.10
)()( 2OHOHKOHOHOH
OHHK w
= 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
OHHK w
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
OHOHK c
2
3
OHOHKOHK wc 32
OHOH 3
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 waterKw = 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
OHHOH 2
OHHK w
OHH14100.1
7714 101101100.1
7101 H
H2O H2O
HCI
OHHOH 2
H2O
OHHK w
OHH14100.1
Assuming H+ all from HCI = 0.01
)()( 2OHHHCIHH
H+ = 0.01 + 1.0x10-12
= 0.01 + 0.000000000001≈ 0.01
OHHOH 2
H+ = 1x10-12 OH- = 1x10-12
CIHHCI0.01 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
100.101.0
100.1
01.0100.1
1214
14
pH
pOH
OH
OH
OH
OHOHK c
2
3
OHOHKOHK wc 32
OHOH 3
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 waterKw = 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
OHHOH 2
H2O
OHHK w
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
OHHOH 2
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
OHHOH 2
1 mol ↔ 1 mol 1 mol
OHHOH 2
OHHK w
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 concApproximation 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 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
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 concApproximation 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 concApproximation 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
OHHOH 2
H2O
OHHK w
xx 814 101100.1Assuming H+ from HCI and H2O
)()( 2OHHHCIHH
OHHOH 2
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 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
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 = 14pH + pOH = pKw
Kw = [H+][OH-]Ka x Kb = Kw
Ka x Kb = 1 x 10-14
pKa = - lg10Ka
pKb = - lg10Kb
pKa + pKb = pKw
pKa + pKb = 14
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
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
Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/http://4photos.net/en/image:44-225901-Water_droplets_on_blue_backdrop__images
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com
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