1. ________
Unit IV
Chemical Equilibrium
Acid-Base Review

2. Closed Systems at Equilibrium
A chemical system that is separate from it’s surroundings where no matter can enter or leave is called a closed system.
Closed Systems at Equilibrium
Evidence from many chemical reactions occurring in a closed system has shown us that after some reactions appear to have stopped, there is a mixture of reactants and products present.
Na2SO4(aq) + CaCl2(aq) CaSO4(s) + 2 NaCl(aq)
forward
reverse
We assume that any closed system with no observable changes occurring is in a state of dynamic equilibrium.
The forward reaction (collisions between reactants to form products) and the reverse reaction (collisions between products to form reactants) are occurring simultaneously and at the same rate.

3. Chemical Reaction Equilibrium

4. 1 – start with reactants only
2 – start with a mixture of reactants and product
3 – start with products only
No matter the starting conditions, the system reaches a state of dynamic equilibrium each time.

5. The rate of the forward reaction decreases as the number of reactant molecules decreases (fewer collisions).
The rate of the reverse reaction in increases as the number of product molecules increases (more collisions).
Dynamic equilibrium is reached when the rate of the forward reaction is equal to the rate of the reverse reaction.

6. Percent Yield
Percent yield provides a way to refer to the chemicals present in equilibrium systems.
78%
The maximum possible yield is calculated using stoichiometry, assuming all of the reactant molecules are used up to form products.

7. Based on percent yield, equilibrium systems fall under one of the above classifications.
ICE Tables
I – initial
ICE tables are used for quantitative calculations involving chemical equilibrium systems that are not quantitative (i.e. < 99.9% yield).
C – change
E – equilibrium

8. The Equilibrium Constant, Kc
Consider the following generic reaction equation for a system at equilibrium:
A, B, C, D – chemical formulas
a, b, c, d – coefficients
The equilibrium law expression allows us to calculate the value of the equilibrium constant, Kc.
products
reactants
The greater the value of the equilibrium constant, the more the products are favoured at equilibrium.
If Kc > 1, then the products are favoured at equilibrium.
If Kc < 1, then the reactants are favoured at equilibrium.

9. First, write the balanced equation with whole-number coefficients.
Ions in solution must be represented as single entities. Equilibrium constant expressions are always written from the net ionic form of reaction equations. Spectator ions are not included.

10. Kc = 40 (given) Kc = 0.025 formation decomposition reciprocal

11. Equilibrium law expressions do NOT include solids or liquids because their concentrations are fixed – the chemical amount (number of moles) per unit volume is a constant value.

12. Henri Louis Le Châtelier
According to Le Châtelier’s principle, when a chemical system at equilibrium is disturbed by a change in property of the system, the system always appears to react to oppose the change, until a new equilibrium is reached.
We will examine the effects of changing the
Henri Louis Le Châtelier
(1850 –1936)
concentrations of reactants or products
temperature of the system
gas volume (or pressure)

13. Le Châtelier’s Principle and Concentration Changes
The addition of a reactant to a system at equilibrium produces an equilibrium shift forward (to the right). The forward reaction produces more product molecules to oppose the change introduced.
The sudden upward “spike” in [HF(g)] indicates that hydrogen fluoride is added.
The equilibrium shift to the right is indicated by the gradual decrease in concentration of the reactant and the gradual increase in concentration of the products.

14. The removal of a product will also shift the equilibrium forward, producing more product molecules to oppose the change introduced.
The sudden downward “spike” in [HCl(g)] indicates that hydrogen chloride is removed.
The equilibrium shift to the right is again indicated by the gradual decrease in concentration of the reactant and the gradual increase in concentration of the products.
Changing concentration has no effect on the value of the equilibrium constant, Kc.

15. Collision-Reaction Theory and Concentration Changes
When we increase the concentration of a reactant, we assume the collisions between reactants are much more frequent, significantly increasing the rate of the forward reaction.
As the concentration of the products increases, so does the rate of the reverse reaction, and the rate of the forward reaction decreases as reactant molecules are used up.
Eventually the rates become equal and a new dynamic equilibrium is reached.

16. Le Châtelier’s Principle and Temperature Changes
The heat energy in an equation is treated as though it were a reactant or a product.
reactants + energy products
(endothermic)
reactants products + energy
(exothermic)
Heating or cooling is like adding or removing heat energy from the system.
This reaction is endothermic, so:
heating shifts the equilibrium to the right
cooling shifts the equilibrium to the left

17. ΔrH = – 198 kJ
The negative enthalpy change indicates the reaction is exothermic, so:
heating shifts the equilibrium to the left
cooling shifts the equilibrium to the right
The gradual increase in concentration of the product with a gradual decrease in reactants indicates a shift to the right.
The system has been cooled.
This shift to the products increases the value of the equilibrium constant, Kc.

18. Collision-Reaction Theory and Energy Changes
ΔrH = – 198 kJ
Cooling the system decrease the rates of both the forward and the reverse reaction because the particles move slower and collide less often.
But, the reverse rate decreases more than the forward rate, resulting the production of more product and the release of more heat energy.

19. Le Châtelier’s Principle and Gas Volume Changes
Recall that the pressure and volume of a gas are inversely proportional.
An increase in volume (decreasing pressure) will cause a shift toward the side with the larger number of moles of gaseous entities.
An decrease in volume (increasing pressure) will cause a shift toward the side with the smaller number of moles of gaseous entities.
3 moles of gas
2 moles of gas
A system with equal amount of gas on each side is not affected by a change in volume. The same for systems with liquid or solid entities.
Adding or removing gasses that are NOT involved in the reaction may change the pressure, but does not affect the equilibrium.

20. Collision-Reaction Theory and Gas Volume Changes
A volume (pressure) change will produce a “spike” in concentrations of all gaseous entities, followed by a gradual equilibrium shift.
The sudden increase in concentrations of all gases indicates that the volume has been decreased (pressure increased).
Changing volume (pressure) has no effect on the value of the equilibrium constant, Kc.

21. Temperature is the only change that will affect the value of the equilibrium constant, Kc.

22. The Water Ionization Constant, Kw
Water is in a liquid state, so we eliminate it from the equilibrium expression.

23. Kw is the ionization constant for water.
In a neutral solution, [H3O+(aq)] = [OH–(aq)], so:
[ H3O+(aq) ] = 1×10–7 mol/L
neutral solution:
[ H3O+(aq) ] > 1×10–7 mol/L
acidic solution:
[ H3O+(aq) ] < 1×10–7 mol/L
basic solution:

24. Communicating Concentrations: pH and pOH
pH = −log [ H3O+(aq) ]
[ H3O+(aq) ] = 10−pH
pOH = −log [ OH−(aq) ]
[ OH−(aq) ] = 10−pOH
pOH + pH = 14

25. Acid Strength as an Equilibrium Position
A strong acid is explained as an acid that reacts quantitatively with water to form hydronium ions.
e.g. hydrochloric acid
HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)
>99%
A single arrow is used instead of:
A weak acid is explained as an acid that reacts partially with water to form hydronium ions (most less than 50%).
e.g. acetic acid
1.3%

26. HA(aq) + H2O(l) → H3O+(aq) + A–(aq)
Percent Ionization:
Percent ionization is a way to express acid strength quantitatively.
HA(aq) + H2O(l) → H3O+(aq) + A–(aq)
It relates the concentration of hydronium ions at equilibrium to the original concentration of the acid.

27. HX(s/l/g) → H+(aq) + X–(aq)
Arrhenius Theory
Acids ionize in water to produce hydrogen ions plus an anion.
HX(s/l/g) → H+(aq) + X–(aq)
Bases dissociate in water to produce hydroxide ions plus a cation.
Svante Arrhenius
(1859 – 1927)
MOH(s) → M+(aq) + OH–(aq)
Limitation:
It failed to predict the acidic or basic properties of some compounds.

28. e.g. HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)
Modified Arrhenius Theory
Acids are substances that react with water to produce hydronium ions.
e.g. HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)
Bases are substances that react with water to produce hyrdoxide ions.
Svante Arrhenius
(1859 – 1927)
e.g. NH3(aq) + H2O(l) → OH–(aq) + NH4+(aq)
Limitations:
There is no provision for reactions that do not occur in aqueous solution.
Some substances have both acidic and basic properties: sodium bicarbonate (NaHCO3(s)), for example. These substances are called amphoteric.

29. The Proton Transfer Concept
The Brønsted–Lowry concept does away with defining a substance as being an acid or a base.
An entity is referred to as acting as an acid or acting as a base in the context of a specific reaction.
Johannes Brønsted
(1879 – 1947)
A Brønsted–Lowry acid is a proton donor in a specific reaction.
A Brønsted–Lowry base is a proton acceptor in a specific reaction.
Thomas Lowry
(1874 – 1936)

30. HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)
A Brønsted–Lowry reaction involves a single proton transfer from one entity (the acid) to another (the base).
1) hydrogen chloride in water H+
HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)
acid
base
2) ammonia in water H+
base
acid
Water has the ability to either accept or donate an proton, making it amphiprotic.

31. water as an acid water as a base HCO3–(aq) HPO42–(aq) HSO4–(aq)
Other examples include:
HCO3–(aq)
HPO42–(aq)
HSO4–(aq)
H2PO4–(aq)

32. H3O+(aq) + NH3(aq) → H2O(aq) + NH4+(aq)
3) hydrochloric acid and ammonia solution H+
H3O+(aq) + NH3(aq) → H2O(aq) + NH4+(aq)
acid
base
4) ammonia gas and hydrogen chloride gas H+
HCl(g) + NH3(g) → NH4Cl(s)
acid
base
Water does not have to be present!

33. The hydrogen carbonate ion is amphiprotic because it can either . . .
a) accept a proton H+
base
acid
b) donate a proton H+
acid
base
The presence of the hydrogen carbonate ion makes sodium bicarbonate (baking soda) an amphoteric substance.

34. Bicarbonate ions can . . .
1) raise the pH of a strong acid solution H+
base
acid
2) lower the pH of a strong base solution H+
acid
base

35. Conjugate Acids and Bases
A Brønsted–Lowry reaction results in an acid–base dynamic equilibrium, where the forward and reverse reactions occur at the same time and at the same rate. H+
base
acid
acid
base H+
A pair of substances that differ only by a proton is called a conjugate acid–base pair.
conjugate pair
conjugate pair

36. The weaker an acid, the stronger it’s conjugate base.
Acetic acid has a strong attraction for its own proton (i.e. doesn’t donate it very readily), so it is a weak acid.
It’s conjugate base, the acetate ion (CH3COO–(aq)), is a stronger base than water. It has a greater attraction for protons.
The weaker an acid, the stronger it’s conjugate base.

37. HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)
A hydrogen chloride molecule has a much weaker attraction for its own proton than water. This makes HCl(aq) a strong acid.
The stronger an acid, the less it attracts its own proton and the stronger a base, the more it attracts another proton.
The stronger an acid, the weaker it’s conjugate base.

39. Predicting Acid–Base Equilibria
In a system that contains several different possible acid-base reactions, the only significant reaction is a proton transfer from the strongest acid present to the strongest base present. H+
strongest acid
strongest base
For an aqueous solution system, we represent all entities as they exist in solution.
Strong Electrolytes: write in dissociated form
ionic salts
e.g. NaCl(aq) → Na+(aq) + Cl–(aq)
strong acids
e.g. HNO3(aq) + H2O(l) → H3O+(aq) + NO3–(aq)
strong bases
e.g. NaOH(aq) → Na+(aq) + OH–(aq)

40. Weak Electrolytes: write as is
weak acids
e.g. CH3COOH (aq)
weak bases
e.g. NH3(aq)

41. Steps For Predicting the Predominant Acid–Base Reaction
1) List all entities as they appear in solution, including H2O(l).
2) Label all possible aqueous acids and bases.
3) Label the strongest acid (SA) and strongest base (SB) using the table on pages 8 and 9 in the data booklet.
4) Write an equation showing the transfer of one proton from the strongest acid to the strongest base, with the products being the conjugate base and acid of the reactants.
5) Predict the position of the equilibrium, based on the fact that the side that is opposite the strongest acid is favoured.
>50%
<50%
products favoured
reactants favoured

42. S A A
Na+(aq)
OH–(aq)
CH3COOH(aq)
H2O(l) S B B
metal ions are treated as spectators
>50% A S A
NH3(aq)
H2O(l)
H3O+(aq) S B B
The reaction is quantitative.

43. Another Example
Write the balanced acid–base equilibrium equation when aqueous potassium hydrogen carbonate (KHCO3(aq)) is mixed with aqueous sodium hydrogen phosphate (Na2HPO4(aq)). S A A A
K+(aq)
Na+(aq)
HCO3–(aq)
HPO42–(aq)
H2O(l) B S B B
<50%
Strongest acid, so reactants are favoured.

44. Perchloric acid is one of the six strong acids, so it forms H3O+(aq).
Yet Another Example!
Write the balanced acid–base equilibrium equation when perchloric acid (HClO4(aq)) is mixed with aqueous calcium hydroxide (Ca(OH)2(aq)).
Perchloric acid is one of the six strong acids, so it forms H3O+(aq). S A A
ClO4–(aq)
H3O+(aq)
Ca+(aq)
OH–(aq)
H2O(l) B S B B
>99%

45. The first six acids in the table from the data booklet are strong acids because they all react quantitatively with water to form hydronium ions.

46. Most other acids are weak, in that they react with water to a much lesser extent.
For example, hydrofluoric acid:
undissociated molecules
dissociated molecules
The acid ionization constant (Ka) indicates the extent to which an acid will react with water. It is a ratio of the dissociated form of the acid to the undissociated form.

47. Calculating [H3O+(aq)] from Ka
The Rule Of 1000
The value of x in the denominator can be omitted whenever the original concentration of the acid is at least 1000 times the numerical value of the Ka.
For any weak acid:

48. Base Strength and the Ionization Constant, Kb
All ionic hydroxides completely dissociate upon dissolving, so they are considered to be strong bases.

49. Most other bases are weak, in that they react with water to a much lesser extent.
For example, citrate ions:
The base ionization constant (Kb) indicates the extent to which a base will react with water.

50. The Ka–Kb Relationship for Conjugate Acid–Base Pairs
Calculating Kb from Amount Concentrations
The Ka–Kb Relationship for Conjugate Acid–Base Pairs
Consider the case of a general weak acid of the form HX(aq) and its reaction with water:

51. Ka × Kb = Kw For the conjugate base X–(aq):
Notice what happens when we multiply these two equilibrium expressions:
Ka × Kb = Kw
(recall: Kw = 1.0 × 10–14)

52. The Effect of Amphoteric Entities
The Rule Of 1000
The value of x in the denominator can be omitted whenever the original concentration of the base is at least 1000 times the numerical value of the Kb.
For any weak base:
The Effect of Amphoteric Entities
Remember, amphoteric entities can act as an acid or as a base.
To decide which one, compare the values of the Ka and Kb:
If Ka > Kb, then it acts as an acid.
If Ka < Kb, then it acts as a base.

53. Buffering Regions, Endpoints and Indicators
The endpoint is the observable colour change.
The equivalence point is when the amount of acid and base are precisely chemically equal.
NO!!

54. Before any acid titrant is added, the sample solution is mostly water molecules and excess hydroxide ions.
Buffering occurs because initially, any acid added immediately reacts with the excess hydroxide and is converted to water.
NO!!

55. The equivalence point is approached as the excess hydroxide ions in the sample solution are almost all reacted with the added acid.
Now the solution consists of water molecules and excess hydronium ions.
NO!!

56. Bromothymol blue is a good indicator to use for a strong acid–strong base titration, because it changes colour very close to the equivalence point.

57. Acid–Base Indicator Equilibrium
An indicator is a Brønstead–Lowry conjugate weak acid–base pair formed when an indicator dye dissolves in water.
conjugate pair
Acid
(red litmus)
Base
(blue litmus)
An increase in hydronium ions present causes an equilibrium shift to the left, producing more of the red–coloured acid form.
The presence of excess hydroxide, the hydroxide removes hydronium, producing an equilibrium shift to the right (blue colour).

58. Polyprotic Entities and Sequential Reactions
Na2CO3(aq) → 2 Na+(aq) + CO32–(aq)
The carbonate ion is a diprotic base because it can accept two protons.
Hydrochloric acid is a strong acid, so it is like adding hydronium ions (H3O+(aq)).

59. H3O+(aq) + CO32–(aq) → H2O(l) + HCO3–(aq)
H3O+(aq) + HCO3–(aq) → H2O(l) + H2CO3(aq)

60. H2SO4(aq) — HSO4–(aq) — SO42–(aq)
Nitric acid is a monoprotic acid:
HNO3(aq) — NO3–(aq)
1 hydrogen!
Sulfuric acid is a diprotic acid:
H2SO4(aq) — HSO4–(aq) — SO42–(aq)
2 hydrogens!
Remember: for every proton transferred by a polyprotic entity, the strength of the new acid or base entity formed greatly decreases.

61. H3PO4(aq) — H2PO4–(aq) — HPO42–(aq) — PO43–(aq)
OH–(aq) + H2PO4– (aq) → H2O(l) + HPO42– (aq)
OH–(aq) + H3PO4(aq) → H2O(l) + H2PO4– (aq)
As a general rule, only quantitative reactions (i.e. 100% reaction) produce detectable equivalence points in an acid–base titration.

63. pH Curve Shape versus Acid and Base Strength
Weak acid–weak base titrations do not have a detectable equivalence point, so pH curves are normally not done for those reactions.

64. Strong Base/Strong Acid
H3O+(aq) + OH–(aq) → 2 H2O(l)
equivalence point is always at a pH of 7
large vertical portion
a wide range of indicators would be useful

65. H3O+(aq) + NH3(aq) → H2O(l) + NH4+(aq)
Strong Acid/Weak Base
For example, hydrochloric acid is added to aqueous ammonia:
H3O+(aq) + NH3(aq) → H2O(l) + NH4+(aq)
equivalence point is always less than a pH of 7
equivalence point pH = 4.6
choose an indicator that changes colour at a lower pH value e.g. bromocresol green

66. CH3COOH(aq) + OH–(aq) → H2O(l) + CH3COO –(aq)
Strong Base/Weak Acid
For example, barium hydroxide is added to acetic acid:
CH3COOH(aq) + OH–(aq) → H2O(l) + CH3COO –(aq)
equivalence point pH = 9.2
equivalence point is always greater than a pH of 7
choose an indicator that changes colour at a higher pH value e.g. phenolphthalein

69. pH Curve Buffering Regions and Buffering Solutions
A buffer is a combination of any weak acid with its conjugate base, in the same solution.

70. (a) OH–(aq) + CH3COOH(aq) → H2O(l) + CH3COO–(aq)
(b) H3O+(aq) + CH3COO–(aq) → H2O(l) + CH3COOH(aq)
Buffer capacity is the limit of the ability of a buffer to maintain a pH level.

71. Blood plasma is buffered because the chemical reactions that take place in our bodies must be in a narrow pH range.