This document provides an overview of acid-base theories and properties. It covers the Bronsted-Lowry and Lewis theories of acids and bases. It defines strong and weak acids and bases, and how their strength affects properties like conductivity and reaction rate. It also introduces the pH scale and explains how pH is determined by the concentration of hydrogen ions in solution.

1. Main
Menu
Lesson 1
Theories of Acids and Bases

2. Lesson 1: Theories of Acids/Bases
Main
 Objectives:
 Reflect on prior knowledge of acids and bases
 Understand the Bronsted-Lowry theory of acidity and
identify Bronsted-Lowry acids and bases
 Understand the Lewis theory of acidity and identify Lewis
acids and bases

3. Solubility of Acids and Akalis
 Most acids are soluble or react strongly with water
 Some bases are soluble, some are insoluble
 Soluble bases are called ALKALIS
Main

4. Bronsted-Lowry Acids and Bases
+(aq) + OH-(aq)
Main
 It’s all about protons (H+)
 Acid: Proton donor
 HCl(aq)  H+(aq) + Cl-(aq)
 H2SO4(aq)  2H+(aq) + SO4
2-(aq)
 Base: Proton acceptor
 NH3(aq) + H2O(l)  NH4
 OH-(aq)* + H+(aq)  H2O(l)
*From any soluble hydroxide or other alkali
 If we mention acid/base without mentioning the type, we
generally mean a Bronsted-Lowry one.

5. Conjugate Acids and Bases
 A conjugate acid/base pair are two species that differ by
a single proton.
 A conjugate base is a species that has one less proton
 A conjugate acid is a species that has one more proton
+ is its conjugate acid
Main
 For example:
 Hydrochloric acid, HCl
 HCl is the acid, Cl- is its conjugate base
 The HCl can donate a proton…it is an acid
 The Cl- could accept a proton….it is a base
 Ammonia, NH3
 NH3 is the base, NH4
 The NH3 can accept a proton….it is a base
 The NH4
+ could donate a proton….it is an acid What is the link?

6. Lewis Acids and Bases
 Acid: electron pair acceptor
 Species with an incomplete octet/outer-shell
 Base: electron pair donor
Main
 Species with a lone pair
 For example:
Gilbert Lewis

7. Main
Key Points
 Bronsted-Lowry Theory
 Acid is a proton donor
 Base is a proton acceptor
 Lewis Theory
 Acid is electron pair acceptor
 Base is electron pair donor

8. Main
Menu
Lesson 2
Properties of Acids and Bases

9. Lesson 2: Properties of Acids and Bases
Main
 Objectives:
 Understand the effect of acids/alkalis on indicators
 Understand that acids neutralises bases (and vice versa)
 Understand the reactions of acids with:
 Metals
 Carbonates
 Hydrogen carbonates
 Conduct experiments to confirm the above properties

10. Main
Indicators
 Universal (actually a mixture of indicators):
 Litmus:
 Acid is Red / Base is Blue
 Phenolpthalein:
 Acid is Colourless / Alkali is Pink
 This is sorcery due to some clever chemistry that we will meet in the
HL part of the course.

11. Main
Neutralisation
ACID + ALKALI  SALT + WATER
hydrochloric acid + sodium hydroxide  sodium
chloride + water
sulfuric acid + ammonia  ammonium sulfate + water
 This is always an exothermic reaction.
 It is called neutralisation but won’t always
lead to a perfectly neutral solution

12. Main
Acids and Metals
ACID + METAL  SALT + HYDROGEN
phosphoric acid + magnesium  magnesium
phosphate + hydrogen
ethanoic acid + sodium  sodium ethanoate +
hydrogen
 Very reactive metals may do this
explosively

13. Acids and Carbonates
ACID + CARBONATE  SALT + CARBON DIOXIDE + WATER
hydrochloric acid + calcium carbonate  calcium chloride + carbon
dioxide + water
nitric acid + sodium carbonate  sodium nitrate + carbon dioxide +
water
Main

14. Acids and Hydrogencarbonates
ACID + HYDROGENCARBONATE  SALT + CARBON DIOXIDE + WATER
hydrochloric acid + calcium hydrogencarbonate  calcium chloride + carbon dioxide +
water
nitric acid + sodium hydrogencarbonate  sodium nitrate + carbon dioxide +
water
Main

15. Main
Key Points
 Acids react in similar ways to each other
 Bases react in similar ways to each other
 Reactions tend to neutralise acid/base properties

16. Main
Menu
Lesson 3
The pH Scale

17. Lesson 3: The pH Scale
Main
 Objectives:
 Understand how pH relates to acidity/bascicity
 Understand how pH relates to changes in hydrogen ion
concentration
 Make a pH colour chart by diluting acids/alkalis

18. Main
The pH Scale
 Runs from 1 for most acid up to 14 for most alkali?
 Nonsense, can go below 0 for very strong acids and
above 14 for very strong alkalis.

19. What is pH?.....the ‘Power of Hydrogen’
 pH is determined by the concentration of H+ in a solution.
Main
 HL only: pH = -log10[H+]
 Each one step increase in pH corresponds to a 10 fold decrease in
the concentration of H+
 pH 0 : [H+] = 1.0x100 mol dm-3 (i.e. 1.0)
 pH 1 : [H+] = 1.0x10-1 mol dm-3 (i.e. 0.1)
 pH 2 : [H+] = 1.0x10-2 mol dm-3 (i.e. 0.01)
 Follow this pattern to work out the H+ concentration required for:
 pH 3
 pH 5
 pH 7
 pH 9
 pH 14

20. So a really really weak solution of an acid is
actually an alkali?
THIS IS ONLY NEEDED AT HL…MORE DETAIL LATER
 No!.... Pure water exists in an equilibrium as follows:
H2O  H+ + OH-
 The position of the equilibrium is way over to the left.
 However there is always a certain concentration of H+, even in pure
Main
water.
 In pure water: [H+] = 1.00x10-7 mol dm-3
 You can only get a lower concentration than this by shifting the
equilibrium to the left by adding OH-

21. Main
Key Points
 pH is determined by the concentration of H+ ions
 Every one step increase in pH corresponds to a ten-fold
decrease in H+ concentration

22. Main
Menu
Lesson 4
Strong and Weak Acids and Bases

23. Lesson 4: Strong and Weak Acids and Bases
Main
 Objectives:
 Understand the difference between strong and weak
acids and bases
 Complete an experiment to explore the difference in
properties of strong/weak acids and bases

24. Strong and Weak Acids
Main
 Strong Acids:
 HA(aq)  H+(aq) + A-(aq) …..the acid fully dissociates into ions
 For example: HCl(aq)  H+(aq) + Cl-(aq)
 Includes: hydrochloric, sulfuric, phosphoric, nitric
 Strong acids have weak conjugate bases
 Weak Acids:
 HA(aq)  H+(aq) + A-(aq) …..the acid only partially dissociates into ions
 For example: HF(aq)  H+(aq) + F-(aq)
 Includes: hydrofluoric, ethanoic, carbonic
 Weak acids have strong conjugate bases

25. Strong and Weak Bases
Main
 Strong Bases:
 BOH(aq)  B+(aq) + OH-(aq) …..the base fully dissociates into ions
 For example: NaOH(aq)  Na+(aq) + OH-(aq)
 Includes: group (I) hydroxides, barium hydroxide
 Strong bases have weak conjugate acids
 Weak Bases:
 BOH(aq)  B+(aq) + OH-(aq) …..the base only partially dissociates into ions
 For example: NH3(aq) + H2O(l)  NH4
+(aq) + OH-(aq)
 Includes: ammonia, amines
 Weak bases have strong conjugate acids

26. Main
So what?
 The equilibrium has a profound effect on the
properties of the acid/base
 Compared with strong acids of the same
concentration, weak acids:
 Have lower electrical conductivity
 React more slowly
 pH is higher (less acid)
 Change pH more slowly when diluted
 However, they neutralise the same volume of alkali
 Weak bases follow a similar pattern

27. Main
Key Points
 Strong acids/bases dissociate fully into ions
 Weak acids/bases only partially dissociate, forming
an equilibrium
 The strong/weak character has a significant effect on
the chemical properties

28. Main
Menu
Lesson 5
HL Only
pH, pOH and the Ionic Product of Water

29. Lesson 5: pH, pOH and the Ionic Product of Water
Main
 Objectives:
 Understand the ionic product of water and use it calculate
H+ and OH- concentrations
 Calculate pH
 Calculate pOH

30. The ionic product of water, Kw
 Water can be both an acid and a base, this leads to
the following equilibrium:
 2 H2O  H3O+ + OH- ……or more simply:
 H2O  H+ + OH-
 The equilibrium for this reaction is called the ionic
product of water and has the symbol Kw:
[ ][ ] 1 10 14    K  H OH   W
Main

31. Calculating [H+] and [OH-] of pure water:
 Kw = [H+][OH-] = 1.00x10-14 mol2 dm-6
 Since when pure, [H+] = [OH-]
Main
 [H+]2 = 1.00x10-14
 [H+] = √1.00x10-14 = 1.00x10-7 mol dm-3
 Kw varies with temperature:
 At 273 K, Kw = 1.14x10-15, calculate [H+]
 At 373 K, Kw = 5.13x10-13, calculate [OH-]
 Do these changes mean the self-dissociation of water is
endothermic or exothermic? Justify your answer.

32. Calculating [H+] and [OH-] of strong acids
Main
 Must use the equilibrium
 For example, what is the concentration of OH- in 2.00 mol
dm-3 sulphuric acid solution?
 Calculate H+ from the data given in the question, there is a
small effect from the equilibrium but it can be ignored.
 Since H2SO4 produces two protons:
 [H+] = 2 x 2.00 = 4.00 mol dm-3
 Now we use the ionic product of water:
 [H+][OH-] = 1.00x10-14
 [OH-] = 1.00x10-14 / [H+] = 1.00x10-14 / 4.00
= 2.50x10-15 mol dm-3

33. Calculating [H+] and [OH-] of strong bases
Main
 Must use the equilibrium
 For example, what is the concentration of H+ in
0.150 mol dm-3 sodium hydroxide acid solution?
 Calculate OH- from the data given in the question, there is a
small effect from the equilibrium but it can be ignored.
 Since NaOH only produces one hydroxide:
 [OH-] = 0.150 mol dm-3
 Now we use the ionic product of water:
 [H+][OH-] = 1.00x10-14
 [H+] = 1.00x10-14 / [OH-] = 1.00x10-14 / 0.150
= 6.67x10-14 mol dm-3

34. Main
pH and pOH
 You are familiar with the pH scale, based on [H+]:
 0 (very strong acid)  7 (neutral)  14 (very strong
alkali)
 There is an analogue called pOH based on [OH-]:
 0 (very strong alkali)  7 (neutral)  14 (very strong
acid)

35. Calculating pH and pOH of strong acids
pH = -log10[H+] AND pOH = -log10[OH-]
 For example, what is the concentration of pH and pOH of 2.00 mol dm-3
Main
sulphuric acid solution?
 pH = -log10[H+] = -log10(4.00) = -0.602
 pOH = -log10[OH-] = -log10(2.5x10-15)= 14.6
 Note: pH + pOH = 14*…..this allows us to take a short cut:
 pOH = 14 – pH = 14 – (-0.602) = 14.6
 pH = 14 – pOH = 14 – 14.6 = -0.602
*This ‘14’ is known as
pKw, i.e. –log10(1.00x10-
14)

36. Calculating pH and pOH of strong bases
pH = -log10[H+] AND pOH = -log10[OH-]
 For example, what is the concentration of pH and pOH of 0.150 mol dm-3
Main
sodium hydroxide acid solution?
 pH = -log10[H+] = -log10(6.67x10-14) = 13.2
 pOH = -log10[OH-] = -log10(0.150)= 0.824
 Note: pH + pOH = 14*…..this allows us to take a short cut:
 pOH = 14 – pH = 14 – (13.2) = 0.824
 pH = 14 – pOH = 14 – 0.824 = 13.2
*This ‘14’ is known as
pKw, i.e. –log10(1.00x10-
14)

37. Calculating [H+] and [OH-]
Main
 Very simple:
 [H+] = 10-pH
 For example; solution of pH 6.2
 [H+] = 10-6.2 = 6.3x10-7 mol dm-3
 [OH-] = 10-pOH
 For example; solution of pH 6.2
 pOH = 14 - 6.2 = 7.8
 [OH-] = 10-7.8 = 1.6x10-8 mol dm-3

38. Main
Key Points
 Kw = [H+][OH-] = 1.00x10-14
 pH = -log10[H+] = 14 - pOH
 pOH = -log10[OH-] = 14 - pH

39. Main
Menu
Lesson 6
HL Only
Ka/pKa and Kb/pKb

40. Lesson 6: Ka/pKa and Kb/pKb
Main
 Objectives:
 Understand the concepts of Ka and Kb
 Understand the concepts of pKa and pKb

41. Weak Acids: Ka and pKa
 Weak acids dissociate to form an equilibrium
 
H A
[ ][ ]
HA
[ ]

Main
 HA(aq)  H+(aq) + A-(aq)
 This has the equilibrium constant (aka acid dissociation
constant), Ka
Ka
 The values for Ka are often very small, so we use ‘pKa’ to
make them easier to handle:
pKa = -log10(Ka)

42. Acid Ka pKa
Hydronium ion, H3O+ 1.00 0.00
Oxalic acid, HO2CCO2H 5.9x10-2 1.23
Hydrofluoric, HF 7.2x10-4 3.14
Methanoic, CHOOH 1.77x10-4 3.75
Ethanoic, CH3COOH 1.76x10-5 4.75
Phenol, C6H5OH 1.6x10-10 9.80
Main
Ka and pKa in action
 In order of decreasing acid strength:
 Smaller Ka
 weaker acid
 Smaller pKa
 stronger acid

43. Weak Bases: Kb and pKb
 Weak bases dissociate to form an equilibrium
 BOH(aq)  B+(aq) + OH-(aq)
 This has the equilibrium constant (aka base dissociation
 
B OH
[ ][ ]
BOH
[ ]

Main
constant), Kb
Kb
 The values for Kb are often very small, so we use ‘pKb’ to
make them easier to handle:
pKb = -log10(Kb)

44. Base Kb pKb
Diethylamine 1.3x10-3 2.89
Ethylamine 5.6x10-4 3.25
Methylamine 4.4x10-4 3.36
Ammonia 1.8x10-5 4.74
Main
Kb and pKb in action
 In order of decreasing base strength:
 Smaller Kb
 weaker base
 Smaller pKb
 stronger base

45. Measuring Ka/pKa and Kb/pKb
 pKa and pKb can be determined experimentally
 At the point of half neutralisation:
Main
 pH = pKa
 pOH = pKb
 This is a convenient artefact of the mathematics
 Follow the instructions here

46. Main
Key Points
 For a weak acid:
 
H A
[ ][ ]
Ka

 AND pKa = -log10(Ka)
HA
[ ]
 For a weak base:
 
B OH
[ ][ ]
Kb

 AND pKb = -log10(Kb)
BOH
[ ]
 At the point of half-neutralisation:
 pKa = pH AND pKb = pOH

47. Main
Menu
Lesson 7
HL Only
Weak acid/base calculations

48. Lesson 7: Weak acid/base calculations
Main
 Objectives:
 Perform calculations involving weak acids
 Perform calculations involving weak bases

49. Main
Recap
WEAK ACIDS
 Ka = [H+][A-]/[HA]
 pKa = -log10(Ka)
 Smaller Ka
 weaker acid
 Smaller pKa
 stronger acid
Acid Ka pKa
Hydronium ion, H3O+ 1.00 0.00
Oxalic acid, HO2CCO2H 5.9x10-2 1.23
Hydrofluoric, HF 7.2x10-4 3.14
Methanoic, CHOOH 1.77x10-
4
3.75
Ethanoic, CH3COOH 1.76x10-
5
4.75
Phenol, C6H5OH 1.6x10-10 9.80
WEAK BASES
 Kb = [B+][OH-]/[BOH]
 pKb = -log10(Kb)
 Smaller Kb
 weaker base
 Smaller pKb
 stronger base
Base Kb pKb
Diethylamine,
(C2H5)2NH
1.3x10-3 2.89
Ethylamine, C2H5NH2 5.6x10-4 3.25
Methylamine, CH3NH2 4.4x10-4 3.36
Ammonia, NH3 1.8x10-5 4.74

50. We need to be able to solve problems such
as:
 What is the pH of an 1.50 M solution of weak acid,
X?
 What is the [OH-] of a solution of weak base, Y?
 50 cm3 of a 0.1 M solution of acid X reacts with 25
cm3 of a 0.1 M solution of base Y, what is the
resulting pH?
 The pH of a 0.250 M solution of weak acid Z is 5.4,
what is it’s Ka and pKa?
Main

51. We will need to use a variety of equations:
Kb

Main
 New(ish) today:
 Ka × Kb = Kw = 1.00x10-14
 pKa + pKb = pKw = 14.0
 pH + pOH = pKw = 14.0
 And from previous lessons:
 
B OH
[ ][ ]
BOH
[ ]
 
H A
[ ][ ]
HA
[ ]
Ka

 pKa = -log10(Ka) pKb = -log10(Kb)
 pH = -log10[H+] pOH = -log10[OH-]

52. Example 1: Calculation of [OH-]
 What is the concentration of OH- ions in a 0.500 mol dm-3 solution of
ammonia (Kb = 1.8x10-5)? What % of the NH3 molecules have
dissociated?
 Since it is a weak base and the equilibrium is to the right we assume
that at equilibrium, [NH3] is the same as stated in the question. So:
Main
 Kb = [NH4
+][OH-]/[NH3] Sub all known values into
equation
 1.8x10-5 = [NH4
+][OH-]/ 0.500 Looks like there are 2 unknowns
 However, since [NH4+] = [OH-]:
 1.8x10-5 = [OH-]2 / 0.500 Rearrange to make [OH-] the
subject
 [OH-] = √(1.8x10-5 x 0.500) Perform calculation
 [OH-] = 0.0030 mol dm-3
 % Dissociation = 0.0030 / 0.500 x 100 = 0.60%

53. Example 2: Calculating pH
 What is the pH of a 0.225 mol dm-3 solution of oxalic acid (HOOCCOOH, Ka
= 5.9x10-2), and how does the pH change on ten-fold dilution?
 Again, assume [HA] is as stated in the question
 Ka = [H+][A-] / [HA] Sub in all known values
 5.9x10-2 = [H+][A-] / 0.225 Looks like two unknowns, but isn’t really
 This time, since oxalic acid produces two protons, [A-] = ½ [H+] so the
Main
expression becomes:
 5.9x10-2 = ½ [H+]2 / 0.225 Rearrange to
make [H+] subject
 [H+] = √((5.9x10-2 x 0.225) / 2) = 0.0815 mol dm-3
 pH = -log10[H+] = -log10(0.0815) = 1.09
 Now with the ten-fold dilution
 [H+] = √((5.9x10-2 x 0.0225) / 2) = 0.0257 mol dm-3
 pH = -log10[H+] = -log10(0.0257) = 1.59….i.e. ten-fold dilution increased pH by 0.50

54. Example 3: Calculating Kb from pH
 A 0.0350 mol dm-3 solution of methylamine (CH3NH2) has a pH of
11.59. Determine Kb of methylamine and Ka of the methylammonium
ion (CH3NH3
Main
+).
 Since pH = 11.59
 pOH = 14 – 11.59 = 2.41
 [OH-] = 10-2.41 = 3.92x10-3 mol dm-3
 Remember:
To calculate Ka of the conjugate acid
use:
Ka x Kb = Kw
Ka = Kw / Kb
= 1.00x10-14 / 4.39x10-4
= 2.78x10-11
 [BOH] at equilibrium is same as stated in question
 [OH-] = [B+]
 Kb = [B+][OH-]/[BOH] Known values subbed in
 Kb = (3.92x10-3).(3.92x10-3)/0.0350
 Kb = 4.39x10-4

55. Example 4: Calculating Ka, pKa, Kb and pKb
from each other
 The pKa of benzoic acid is 4.20. Calculate Ka, Kb and
pKb
 Ka = 10-pKa = 10-4.20 = 6.31x10-5
 Kb x Ka = Kw
Kb = Kw / Ka = 1.00x10-14 / 6.31x10-5 = 1.58x10-10
 pKb = -log10(Kb) = -log10(1.58x10-10) = 9.80
Main
 OR
 Since pKa + pKb = pKw
 pKb = 14 – pKa = 14 – 4.20 = 9.80

56. Main
Key Points
 Calculations rely on two key assumptions:
 Concentration of HA or BOH at equlibrium is the same as
given in the question
 Reasonable as equilibrium effects mean dissociation is often
1% or less
 [H+]/[OH-] and [A-]/[B+] are not separate variables but are
related to each other
 Expressing [A-]/[B+] in terms of [H+]/OH-] is a key step

57. Main
Menu
Lesson 8
HL Only
Buffers

58. Main
Lesson 8: Buffers
 Objectives:
 Describe and explain the function of buffers
 Calculate the pH of buffer solutions
 Make an acid and an alkaline buffer solution and observe
its buffering activity.

59. Main
Buffers
 Buffers are solutions that resist changes to
their pH
 Acid buffers:
 Weak acid and salt of it’s conjugate base
 E.g. Ethanoic acid and sodium ethanoate
 Alkaline buffer:
 Weak base and a salt of its conjugate acid
 E.g. ammonia and ammonium chloride
 Common in nature, blood being the best
example: pH kept at 7.35-7.45

60. How they work: acidic buffers
 Buffer is a mixture of HA and A- solutions, establishing the following
Main
equilibrium:
HA  H+ + A-
 The addition of extra A- ions forces the equilibrium to the left due to
Le Chatelier’s principle. This creates a large reservoir of un-dissociated
HA
 Adding acid:
 Equilibrium shifts to the left, reducing the increase in H+
 Adding base:
 Equilibrium shifts to the right, reversing the decrease in H+
 Adding water:
 Both sides affected equally so pH unchanged

61. How they work: basic buffers
 Buffer is a mixture of BOH and B+ solutions, establishing the
Main
following equilibrium:
BOH  B+ + OH-
 The addition of extra B+ ions forces the equilibrium the left due to Le
Chatelier’s principle. This creates a large reservoir of un-dissociated
BOH
 Adding base:
 Equilibrium shifts to the left, reducing the increase in OH-
 Adding acid:
 Equilibrium shifts to the right, reversing the decrease in OH-
 Adding water:
 Both sides affected equally so pH unchanged

62. Calculating pH of a Buffer
 What is the pH of a buffer comprising a 0.450 mol dm-3 ammonia
(NH3, Kb = 1.8x10-5) and 0.200 mol dm-3 ammonium chloride?
NH OH
[ ][ ]

NH
[ ]
NH
[ ]
K
 We need to make two assumptions:
Main
 [NH4

4
+] is equal to that stated in the question…since there will be very little
coming from dissociated NH3
 [NH3] is equal to that stated in the question…since very little NH3 has
dissociated







 


 
[ ]
log
3
10
3
4
NH
pOH pK
b
b
4.39
5

14.0 4.39 9.61
0.200
0.450
log (1.8 10 ) log10
10
pOH
pOH
  






    
pH

63. Calculating the proportions for a buffer
 What mass of sodium ethanoate (Mr = 82.0) should be added to 0.500 dm3
of 0.400 mol dm-3 ethanoic acid solution (pKa = 4.76) to make a buffer of ph
5.70?
 
H CH COO
[ ][ ]
3
CH COOH
[ ]
 
H CH COO
[ ][ ]
CH COOH
[ ]
K CH COOH
[ ]

H
[ ]
a


10 
0.400
K
a
CH COO
CH COO mol dm
5.7
3
10

Main
 Similar to before:
 Two assumptions:
3
Ka

 CH3COO- will be only due to added sodium ethanoate
 [CH3COOH] is as stated in the question
Note: Ka= 10-pKa
4.76


[ ]
3

[ ]
3
3
3


 Mass sodium ethanoate required:
3
3.48


 Mass = moles x molar mass = (3.48 x 0.500) x 82.0 = 143 g

64. Main
Key Points
 Calculations are similar to those for weak acids/bases but
need re-expressing as:

 


 
B
10 BOH

 
B OH
[ ][ ]
BOH
[ ]
K
b
pOH pK
 


[ ]
[ ]
log
b

H A
[ ][ ]
HA
[ ]
K
 Calculations make two key assumptions:


A
[ ]
 Concentration of the weak acid/base is as stated in the
question
 Concentration of the conjugate base/acid is as stated in the
question
 These assumptions are valid as equilibrium effects mean very
little of the weak acid/base dissociates
 

 

 

 
[ ]
log
10 HA
pH pK
a
a

65. Main
Menu
Lesson 9
HL Only
Salt Hydrolysis

66. Lesson 9: Salt Hydrolysis
Main
 Objectives:
 Understand why some salts do not form neutral solutions
 Predict whether a given salt will form an acidic, neutral or
alkaline solution
 Complete an experiment to determine any trends in
acidity/basicity of salt solutions

67. Acid Salt Hydrolysis
 When you dissolve sodium chloride in water it forms
a neutral solution of Na+ and Cl- ions:
NaCl(s)  Na+(aq) + Cl-(aq)
 When you dissolve ammonium chloride in water it
forms a weakly acidic solution of NH4
Main
+ and Cl- ions
 Why?....discuss

68. +  NH3 + H+
Main
Why acidic?
 Ammonium chloride contains the NH4
+ ion
 NH4
+ is the conjugate acid of the weak base NH3:
+ + OH-
NH3 + H2O  NH4
 The NH4
+ ion therefore is weakly acidic and will establish
the following equilibrium:
NH4
 Thus, ammonium chloride will form a weakly acidic
solution

69. Basic Salt Hydrolysis
 When you dissolve sodium chloride in water it forms
a neutral solution of Na+ and Cl- ions:
NaCl  Na+(aq) + Cl-(aq)
 When you dissolve sodium ethanoate in water it
forms a weakly basic solution of Na+ and CH3COO-ions
Main
 Why?....discuss

70. Recap: Conjugate Acids and Bases
 A conjugate acid/base pair are two species that differ by
a single proton.
 A conjugate base is a species with has one less proton
 A conjugate acid is a species with one more proton
+ is its conjugate acid
Main
 For example:
 Hydrochloric acid, HCl
 HCl is the acid, Cl- is a conjugate base
 The HCl can donate a proton…it is an acid
 The Cl- could accept a proton….it is a base
 Ammonia, NH3
 NH3 is the base, NH4
 The NH3 can accept a proton….it is a base
 The NH4
+ could donate a proton….it is an acid

71. Main
Why basic?
 Sodium ethanoate contains the CH3COO- ion
 CH3COO- is the conjugate base of the weak acid
CH3COOH:
CH3COOH  H+ + CH3COO-
 The CH3COO- ion therefore is weakly basic and will
establish the following equilibrium:
CH3COO- + H2O  CH3COOH + OH-
 Thus, sodium ethanoate will form a weakly basic solution

72. Main
Rules of Thumb
 Conjugate bases:
 Conjugate bases of weak acids will form basic solutions
 Conjugate bases of strong acids will form neutral
solutions
 Conjugate acids:
 Conjugate acids of weak bases will form acidic solutions
 Conjugate acids of strong bases will form neutral
solutions
 The final pH of an individual salt will depend on the
relative acidity of both the conjugate acid and
conjugate base is formed from

73. Main
Key Points
 Salts of a weak acid and strong base form basic solutions:
 Due to the conjugate base of the weak acid being basic
 Salts of a weak base and a strong acid form acidic solutions:
 Due to the conjugate acid of the weak base being acidic
 Salts of a strong acid and strong base form neutral solutions
 Neither the conjugate base of the acid or the conjugate acid of the
base form acid/base equilibria
 Salts of a weak acid and weak base can be acidic, basic or
neutral
 Depends on the relative basicity of the conjugate base and acidity
of the conjugate acid

74. Main
Menu
Lesson 10
HL Only
Acid-Base Titrations

75. Lesson 10: Acid-Base Titrations
Main
 Objectives:
 Complete titrations of each of the four possible
combinations of acid and base, recording data with a
data-logger
 Use software to produce a graph of the results of your
titrations
 Explain in full the characteristic shape of each graph

76. Anatomy of an acid-base titration curve:
In this case weak-acid/strong-base
Half-equivalence
Main
point:
i.e. Half the
volume added at
the equivalence
point
The pH of this
point is equal to
the pKa of the
weak acid.
Equivalence
point:
i.e. The point of
inflection (where
the gradient starts
to drop again).
Represents the
point at which the
acid has just been
neutralised.
pH Intercept:
Higher pH if a
weaker acid used
Buffer Region:
A lot of base has
to be added to
result in only a
small change of
pH

77. Adding acid to base
Main
 Your graphs for
adding base to
acid will have
the same shape
but be mirror
images.

78. Main
Key Points
 Acid-base titration curves have a characteristic
shape
 The shape depends on the combination of
strong/weak acid-strong/weak base used
 The pH changes very fast at the point of equivalence
due to the logarithmic nature of the pH scale
 The ‘flatter’ parts of the weak acid/base curves are
due to equilibrium effects

79. Main
Menu
Lesson 11
HL Only
Indicators

80. Lesson 11: Indicators
Main
 Objectives:
 Understand how indicators work
 Determine suitable indicators for a reaction
 Make indicators from a range of natural products

81. Main
Indicators
 An indicator is a compound whose colour depends on pH
 For example:
 Phenolphthalein
 pH< 8: COLOURLESS pH>10: PINK
 Methyl orange
 pH<3.2: RED pH>4.4: ORANGE
 Bromothymol blue:
 pH<6.0: YELLOW pH>7.6: BLUE
 Most indicators change colour only once (sometimes twice). The obvious
exception is universal indicator which is actually a mixture of several
indicators with different colour change ranges.
 Note: the pH generally changes over a small range, but the logarithmic
nature of pH means often equates to a single drop.

82. How Indicators Work (what you need):
 Indicators are weak acids/bases in their own right
 In solution indicators form an equilibrium:
In + H+ ⇌ InH+
COULOUR 1 COLOUR 2
Main
 Where: ‘In’ stands for indicator
 As [H+] changes, the equilibrium moves to the left or right, thus changing the
colour
 The structure of indicators change depending on the pH:
 Higher pH can cause weak-acid groups to deprotonate
 Low pH can cause weak-base groups to protonate
 This can have knock-on effects on the structure, of the indicator molecule which
changes its colour

83. How Indicators Work (in detail):
 Colour in many organic compounds comes from having overlapping π-
systems, with many delocalised electrons
 + 2H+
colourless pink
Main
 For example: phenolphthalein
 The colour derives from conjugated π-systems which are radically altered by
the changes in structure
 The double bond on the central carbon atom allows the π-systems in the three
benzene rings to interact with each other, leading to the pink colour
 This is more detail than required by the IB!!!

84. Main
pH range and pKa
 Most indicators change range within ±1.0 of their pKa
 pKa data for indicators can be found in your data
booklet
Indicator pKa
pH
Range
Colour Change
Acid Alkali
methyl orange 3.46 3.2–4.4 Red Yellow
bromophenol blue 4.10 3.0–4.6 Yellow Blue
bromocresol green 4.90 3.8–5.4 Yellow Blue
methyl red 5.00 4.8–6.0 Red Yellow
bromothymol blue 7.30 6.0–7.6 Yellow Blue
phenol red 8.00 6.6–8.0 Yellow Red
phenolphthalein 9.50 8.2–
10.0
Colourless Pink

85. Main
Using Indicators
 Indicators are NOT USED to measure pH
 Colour change is not an accurate measurement
 Range of colour change limits the measurements that
could be taken
 pH probes are very accurate
 Indicators ARE USED:
 To determine the end-point of reactions in titrations
 To give a ‘rough and ready’ idea of pH

86. Main
Key Points
 Indicators change colour over a narrow pH range
 The colour change generally occurs at a pH within ±
1.0 of the indicator’s pKa
 Indicators should be chosen with a pH range that
matches the expected equivalence point as closely
as possible