1. ACIDS AND BASES

2. REVISE Brønsted-Lowry acids and bases Amphoteric substances
Conjugate acid base pairs
Neutralisation
Neutral: pH = 7 ([H+] = [OH-])
Acidic: pH < 7 ([H+] > [OH-])
Basic: pH > 7 ([H+] < [OH-])
pH = -log [H+]
pOH = -log [OH-]
pH + pOH = at 25oC
Kw = [OH-][H+]
Kw = at 25oC
Kw = KaKb

3. STRONG ACIDS AND BASES
Strong acids and bases  react nearly “completely” to produce H+ and OH-
 equilibrium constants are large
e.g.: HCl H+ + Cl-
Complete dissociation:
large
small
Common strong acids:
HCl, HBr, HI, H2SO4, HNO3, HClO4
(Why is HF not a strong acid?)
Common strong bases:
LiOH, NaOH, KOH, RbOH, CsOH, R4NOH

4. LiOH  Li+ + OH- Example: Calculate the pH of 0.1M LiOH. 0.1M 0.1M
Strong base 
Dissociates completely
LiOH  Li OH-
0.1M
Start:
0.1M
0.1M
Complete rxn:
pOH = -log [OH-]
= -log (0.1) = 1
 pH = = 13

5. IMPOSSIBLE!!!!!! Problem: What is the pH of 1x10-8 M KOH? As before:
pOH = -log (1x10-8) = 8
pH = 14 – 8 = 6
BUT
pH 6  acidic conditions and KOH is a strong base
IMPOSSIBLE!!!!!!

6. We do this by systematic treatment of equilibrium.
Since the concentration of KOH is so low (1x10-8 M), we need to take the ionisation of water into account.
In pure water [OH-] = 1x10-7 M, which is greater than the concentration of OH- from KOH.
We do this by systematic treatment of equilibrium.
Charge balance: [K+] + [H+] = [OH-]
Mass balance: [K+] = 1x10-8 M
Equilibria: [H+][OH-] = Kw = 1x10-14 M
3 equations + 3 unknowns  solve simultaneously
Find: pH = 7.02
Hint: You end up with a quadratic equation which you solve using the formula.

7. Also note that:
Only pure water produces 1x10-7 M H+ and OH-.
If there is say 1x10-4 M HBr in solution,
pH = 4 and [OH-] = 1x10-10 M
But the only source of OH- is from the dissociation of water.
 if water produces 1x10-10 M OH- 
it can only produce 1x10-10 M H+ due to the dissociation of water.
 pH in this case is due mainly to the dissociation of HBr and not the dissociation of water.
It is thus important to look at the concentration
of acid and bases present.

8. Some guidelines regarding the concentrations of acids and bases:
When conc > 1x10-6 M
 calculate pH as usual
When conc < 1x10-8 M
 pH = 7
(there is not enough acid or base to affect the pH of water)
When conc  1x x10-6 M
 Effect of water ionisation and added acid and bases are comparable, thus:
use the systematic treatment of equilibrium approach.

9. WEAK ACIDS AND BASES
Weak acids and bases  react only “partially” to produce H+ and OH-
 equilibrium constants are small
HA H+ + A-
Partial dissociation
small
large
Acid dissociation constant
Common weak acids:
carboxylic acids
(e.g. acetic acid = CH3COOH)
ammonium ions
(e.g. RNH3+, R2NH2+, R3NH+)
Common weak bases:
carboxylate anions
(e.g. acetate = CH3COO-)
amines
(e.g. RNH2, R2NH, R3N)

10. base hydrolysis constant/ base “dissociation” constant
B + H2O BH+ + OH-
Weak base
partial dissociation
Kb small
base hydrolysis constant/ base “dissociation” constant
NOTE:
pKa = -log Ka
pKb = -log Kb
As K increases, its p-function decreases and vice versa.

11. Problem:
Find the pH of a solution of formic acid given that the formal concentration is 2 M and Ka = 1.80x10-4.
HCOOH H+ + HCOO-
H2O H+ + OH-
Systematic treatment of equilibria:
Charge balance: [H+] = [HCOO-] + [OH-]
Mass balance: 2 M = [HCOOH] + [HCOO-]
Equilibria:
4 equations  4 unknowns
 difficult to solve

12.  [HCOO-] >> [OH-]
Make an assumption:
[H+] due to acid dissociation  [H+] due to water dissociation
Produces HCOO-
Produces OH-
[HCOO-] large
[OH-] small
 [HCOO-] >> [OH-]
 Charge balance: [H+]  [HCOO-]

13. Charge balance: [H+]  [HCOO-]
Mass balance: 2 M = [HCOOH] + [H+]
Equilibria:
Let [H+] = [HCOO-] = x
Or x =
 No negative conc’s
[H+] = [HCOO-] = M
 pH = 1.7

14. OR since [HCOOH] > 1x10-6, we can calculate pH as usual
Weak acid
equilibrium conditions
HCOOH H HCOO- 2M
Start:
2-x x x
Equilibrium:
Solve as before

15. FRACTION OF DISSOCIATION, 
Fraction of acid in the form A-
For the above problem:
 =
[HCOO-] F =
0.019 M
2 M =
 Acid is 0.95% dissociated at 2 M formal concentration
Weak electrolytes dissociate more as they are diluted.

16. FRACTION OF ASSOCIATION
WEAK BASE EQUILIBRIA
B + H2O BH+ + OH-
Charge balance: [BH+] = [OH-]
Mass balance: F = [B] + [BH+]
Equilibria:
Let [BH+] = [OH-] = x
FRACTION OF ASSOCIATION

17. CONJUGATE ACIDS AND BASES
Relationship between Ka and Kb for a conjugate acid- base pair:
Ka.Kb = Kw = 1x at 25oC
If Ka is very large (strong acid)
Then Kb must be very small (weak conjugate base)
And vice versa
Base so weak it is not
a base at all in water
If Ka is very small, say 1x10-6 (weak acid)
Then Kb must be small, 1x10-8 (weak conjugate base)
Greater acid strength, weaker conjugate base strength, and vice versa.

18. Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for ammonia.
Problem:
Calculate the pH of 0.1 M NH3, given that pKa = for ammonia. Kb
NH3 + H2O NH4+ + OH-
base
acid
pKa = -log Ka
Ka = 5.70x10-10
Kb = Kw Ka =
1x10-14
5 .70x10-10
= x10-5

19. Solve for x using the quadratic equation
Problem:
Calculate the pH of 0.1 M NH3, given that pKa = for ammonia.
Kb = x2
F - x
Where x = [OH-] = [NH+] = x2
0.1 - x
= x10-5
Solve for x using the quadratic equation
Negative value discarded
Find x = 1.31x10-3 M = [OH-]
pOH = -log [OH-]
= 2.88
pH = 14 – 2.88 = 11.12

20. BUFFERS Mixture of an acid and its conjugate base.
Buffer solution  resists change in pH when acids or bases are added or when dilution occurs.
Mix:
A moles of weak acid + B moles of conjugate base
Find:
moles of acid remains close to A, and
moles of base remains close to B
 Very little reaction
HA H+ + A-
Le Chatelier’s principle

21. HENDERSON-HASSELBALCH EQUATION
For acids:
When [A-] = [HA], pH = pKa
For bases:
pKa applies to this acid Kb 
B + H2O BH+ + OH-  Ka
base
acid
acid
base

22. Derivation:
HA H+ + A-

23. The acid or base is consumed by A- or HA respectively
Why does a buffer resist change in pH when small amounts of strong acid or bases is added? ?
The acid or base is consumed by A- or HA respectively
A buffer has a maximum capacity to resist change to pH.
Buffer capacity, :
 Measure of how well solution resists change in pH when strong acid/base is added.
Larger   more resistance to pH change

24. A buffer is most effective in resisting changes in pH when:
pH = pKa
i.e.: [HA] = [A-]
 Choose buffer whose pKa is as close as possible to the desired pH.
pKa  1 pH unit

25. Problem:
Calculate the pH of a solution containing M NH3 and M NH4Cl given that the acid dissociation constant for NH4+ is 5.7x10-10.
NH3 + H2O NH4+ + OH-
acid
base Ka
pKa = 9.244
pKa applies to this acid
pH = log
(0.200)
(0.300)
pH = 9.07

26. POLYPROTIC ACIDS AND BASES
Can donate or accept more than one proton.
In general:
Diprotic acid:
H2L HL- + H+ Ka1  K1
HL L2- + H+ Ka2  K2
Diprotic base:
L2- + H2O HL- + OH- Kb1
HL- + H2O H2L + OH- Kb2
Relationships between Ka’s and Kb’s:
Ka1. Kb2 = Kw
Ka2. Kb1 = Kw

27. Using pKa values and mass balance equations, the fraction of each species can be determined at a given pH.

29. ACID-BASE TITRATIONS
We will construct graphs to see how pH changes as titrant is added.
Start by:
writing chemical reaction between titrant and analyte
using the reaction to calculate the composition and pH after each addition of titrant

30. TITRATION OF STRONG BASE WITH STRONG ACID
Example:
Titrate ml of M KOH with M HBr.
HBr + KOH KBr + H2O
What is of interest to us in an acid-base titration:
H+ + OH- H2O
Mix strong acid and strong base
 reaction goes to completion
H+ + OH-  H2O

31. * Calculate volume of HBr needed to reach the equivalence point, Veq:
Example:
Titrate ml of M KOH with M HBr.
* Calculate volume of HBr needed to reach the equivalence point, Veq:
C1V1
C2V2 n1 n2 =
But n1 = n2 = 1
 CHBrVeq = CKOHVKOH
( M)Veq = ( M)(50.00 ml)
Veq = ml

32. There are 3 parts to the titration curve:1
Before reaching the equivalence point
 excess OH- present
At the equivalence point
 [H+] = [OH-] 2 3
After reaching the equivalence point
 excess H+ present

33. Before reaching the equivalence point  excess OH- present
HBr + KOH KBr + H2O
Say 2.00 ml HBr has been added.
Starting nOH-
= (0.02 M)(0.050 L)
= 1x10-3 mol
COH- =
nunreacted
Vtotal
nH+ added
= (0.1 M)(0.002 L)
= 2x10-4 mol
COH- = M
Vtotal = mL
= 52 mL
= L
nOH- unreacted
= 8x10-4 mol
Kw = [H+][OH-]
1x10-14 = [H+]( M)
[H+] = 6.500x10-13 M
 pH = 12.19

34. At the equivalence point
 nH+ = nOH-
pH is determined by dissociation of H2O:
H2O H+ + OH- x x
Kw = [H+][OH-]
1x10-14 = x2
x = 1x10-7 M  [H+] = 1x10-7 M
 pH = 7
pH = 7 at the equivalence point ONLY for strong acid – strong base titrations!!

35. After reaching the equivalence point  excess H+ present
HBr + KOH KBr + H2O
Say ml HBr has been added.
Starting nOH-
= 1x10-3 mol
CH+ =
nexcess
Vtotal
nH+ added
= (0.1 M)( L)
= 1.010x10-3 mol
CH+ = 1.664x10-4 M
pH = 3.78
nH+ excess
= 1x10-5 mol
Vtotal = mL
= 60.1 mL
= L

36. Note:
A rapid change in pH near the equivalence point occurs.
Equivalence point where:
slope is greatest
second derivative is
zero (point of inflection)

37. Calculate titration curve by calculating pH values after a number of additions of HBr.

38. TITRATION OF WEAK ACID WITH STRONG BASE
Example:
Titrate ml of M formic acid with M NaOH.
HCO2H + NaOH HCO2Na + H2O
OR HCO2H + OH- HCO2- + H2O HA A-
pKa = 3.745
Equilibrium constant so large  reaction “goes to completion” after each addition of OH-
Ka = 1.80x10-4
Kb = 5.56x10-11
Strong and weak react completely

39. * Calculate volume of NaOH needed to reach the equivalence point, Veq:
Example:
Titrate ml of M formic acid with M NaOH.
HCO2H + OH- HCO2- + H2O
* Calculate volume of NaOH needed to reach the equivalence point, Veq:
C1V1
C2V2 n1 n2 =
But n1 = n2 = 1
 CNaOHVeq = CFAVFA
( M)Veq = ( M)(50.00 ml)
Veq = ml

40. There are 4 parts to the titration curve:
Before base is added  HA and H2O present. HA weak acid,  pH determined by equilibrium:
HA H+ + A- Ka
From first addition of NaOH to immediately before equivalence point  mixture of unreacted HA and A-
HA + OH- A- + H2O
BUFFER!!  use Henderson-Hasselbalch eqn for pH 2 1

41. At the equivalence point
 all HA converted to A-. A- is a weak base whose pH is determined by reaction:
A- + H2O HA + OH- Kb 4 3
4) Beyond the equivalence point
 excess OH- added to A-.
Good approx: pH determined by strong base (neglect small effect from A-)

42. Before base is added
 HA and H2O present. HA = weak acid.
Ka = 1.80x10-4
HA H+ + A-
F- x x
1.80x10-4 = x2 x
x x10-4x – 3.60x10-6 = 0
x = 1.81x10-3
 [H+] = 1.81x10-3
 pH = 2.47

43. Say 2.00 ml NaOH has been added.
From first addition of NaOH to immediately before equivalence point  mixture of unreacted HA and A- . BUFFER!!
HCO2H + OH- HCO2- + H2O
Say 2.00 ml NaOH has been added.
Starting nHA
= (0.02 M)(0.05 L)
= 1x10-3 mol
nOH- added
= (0.1 M)(0.002 L)
= 2x10-4 mol
HA + OH A- + H2O
1x10-3
2x10-4
8x10-4
Start -
End
pH = log
2x10-4
8x10-4
 pH = 3.14

44. But VA = VHA = VTot

45. Special condition:
When volume of titrant = ½ Veq
pH = pKa
Since:
nHA = nA-

46. At the equivalence point
 all HA converted to A-. A- = weak base (nHA = nNaOH)
Starting nHA
= 1x10-3 mol
HA + OH A- + H2O
1x10-3
Start -
End
 nOH-
= 1x10-3 mol
Solution contains just A-  a solution of weak base

47. A- + H2O HA + OH- Kb = 5.56x10-11 F- x x x FA- = nA- V = 1x10-3 mol
Vtotal = mL
= 60 mL = L
= M
5.56x10-11 = x2 x
[OH-] = 9.63x10-7 M
x x10-11x – 9.27x10-13 = 0
pOH = 6.02
x = 9.63x10-7
pH = 7.98
pH is slightly basic at equivalence point for strong base-weak acid titrations

48. CALCULATED TITRATION CURVE

49. Titration curve depends on Ka of HA.
As HA becomes a weaker acid the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect
 not practical to titrate an acid or base that is too weak.

50. Titration curve depends on extent of dilution of HA.
As HA becomes a more dilute the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect
 not practical to titrate a very dilute acid or base.

51. TITRATION OF WEAK BASE WITH STRONG ACID
This is the reverse of the titration of weak base with strong acid.
The titration reaction is:
B + H+ BH+
Recall:
Strong and weak react completely

52. There are 4 parts to the titration curve:
Before acid is added
 B and H2O present.
B weak base  pH determined by equilibrium:
B + H2O BH+ + OH- Kb
F-x x

53. From first addition of acid to immediately before equivalence point
 mixture of unreacted B and BH+
B + H BH+
BUFFER!!
 use Henderson-Hasselbalch equation for pH
pKa applies to this acid

54. At the equivalence point  all B converted to BH+.
BH+ is a weak acid  determined pH by reaction:
BH B + H+ Ka
F-x x
FBH+ = n
Vtotal
Take dilution into account
pH is slightly acidic (pH below 7) for strong acid-weak base titrations

55. 4) Beyond the equivalence point
 excess H+ added to BH+.
Good approx: pH determined by strong acid (neglect small effect from BH+)
Example:
50.00 ml of 0.05 M NaCN is titrated with 0.1 M HCl. Ka for NaCN = 6.20x1010
Draw the titration curve by calculating pH at various volumes of HCl.

56. TITRATION CURVE OF WEAK BASE WITH STRONG ACID

57. TITRATIONS IN DIPROTIC SYSTEMS
Example - a base that is dibasic:
B + H+  BH+
BH+ + H+  BH22+
pKb1 = 4.00
pKb2 = 9.00
With corresponding reactions:
Two end points will be observed.

59. FINDING END POINTS WITH A pH ELECTRODE
After each small addition of titrant the pH is recorded and a titration curve is plotted.
2 ways of determining end points from this:
using derivatives
using a Gran plot

60. Setup

61. But there are autotitrators!
Titrando from Metrohm

62. USING DERIVATIVES End point is taken where the slope is greatest
Or where the 2nd derivative is zero

63. USING A GRAN PLOT A problem with using derivatives
 titration data is most to obtain near the end point
Example – titration of a weak acid, HA
Ka =
[H+]H+[A-] A-
[HA] HA
HA H+ + A-
NOTE:
pH electrode responds to hydrogen ion ACTIVITY, not concentration

64. Say we titrated Va ml of HA (formal conc = Fa) with Vb ml of NaOH (formal conc = Fb) :
HA + OH- A- + H2O
[A-] =
nOH(titrated)
VTotal =
VbFb
Va + Vb
[HA] =
nHA(initial) – nOH(titrated)
VTotal =
VaFa- VbFb
Va + Vb
Substitute into the equilibrium constant:
Ka =
[H+]H+[A-] A-
[HA] HA

65. Rearrange:
VaFa Fb
- Vb
[H+]H+ = 10-pH
= Ve - Vb

66. Gran plot equation:
Gran plot  Graph of Vb10-pH vs Vb
If is constant, then:
A-
HA
Slope = -Ka
A-
HA
and
x-intercept = Ve
Use data taken before end point to find end point
Can determine Ka from slope

67. Use only linear portion of graph
Extrapolate graph to get Ve

68. FINDING END POINTS WITH INDICATORS
Acid-base indicator  acid or base itself
Various protonated species have different colours
HIn H+ + In-

69. Choose indicator whose colour change is as close as possible to the pH of the end point
Indicators transition range overlaps the steepest part of the titration curve

70. Indicator error: difference between the observed end point (colour change) and the true equivalence point.
Systematic error
Random error
Visual uncertainty associated with distinguishing the colour of the indicator reproducibly
Why do we only add a few drops of indicator?
Indicator is an acid/base itself  will react with analyte/titrant
Few drops  neglible relative to amount of analyte

71. Good luck with the test and don’t panic!

72. HOW LONG DOES IT TAKE YOU TO FIND THE MAN’S HEAD?