1. Equilibrium, Acids & Bases
Most of the reactions that we looked at in thermochemistry and electrochemistry have started out as reactants and ended up as products. Up until this point in chemistry, the common assumption has been that all reactions proceed to completion. This means that all reactants are completely used up in a reaction and completely converted to products as in quantitative stoichiometric reactions. -

2. Chemical Equilibrium
100%
Not all reactions are quantitative (reactants  products)
Evidence: For many reactions reactants are present even after the reaction appears to have stopped
Recall the conditions necessary for a chemical reaction:
Particles must collide with the correct orientation and have sufficient energy
If product particles can collide effectively also, a reaction is said to be reversible
Rate of reaction depends on temperature, surface area and concentration
Equilibrium is a balance point. It occurs is a closed system, which means that no matter enters or leaves, only energy does (closed system). It occurs at a constant temperature and pressure.
Equilibrium is reached under the following conditions:
The rate of forward reaction is equal to the rate of the reverse reaction
Concentrations remain constant; this does not necessarily mean equal concentrations of reactants and products
All measurable properties are constant, meaning there are no macroscopic changes; concentration, volumes, temperature, pressure and color remain constant

3. - Opposing changes occur simultaneously at the same rate
Chemical Equilibrium
Consider the following reversible reaction:
The final state of this chemical system can be explained as a competition between:
We assume this system is closed (so the reactants and products cannot escape) and will eventually reach a:
DYNAMIC EQUILBRIUM
- Opposing changes occur simultaneously at the same rate
The collisions of reactants
to form products The collisions of products to re-form reactants
Dynamic= constantly changing
- When two opposing processes proceed at the same rate. This type of equilibrium takes time to establish.

4. Chemical Equilibrium DYNAMIC EQUILIBRIUM
Consider the following hypothetical system:
AB + CD  AD + BC forward reaction, therefore
AD + BC  AB + CD reverse reaction
Initially, only AB and CD are present. The forward reaction is occurring exclusively at its highest rate.
As AB and CD react, their concentration decreases. This causes the reaction rate to decrease as well.
As AD and BC form, the reverse reaction begins to occur slowly.
As AD and BC’s concentration increases, the reverse reaction speeds up.
Eventually, both the forward and reverse reaction occur at the same rate =
DYNAMIC EQUILIBRIUM
Dynamic equilibrium is like a boarder crossing.

5. 4 Conditions of Dynamic Equilibrium*
Can be achieved in all reversible reactions when the rates of the forward and reverse reaction become equal
Represented by rather than by 
All observable properties appear constant (color, pH, etc.)
Can only be achieved in a closed system (no exchange of matter and must have a constant temperature)
Equilibrium can be approached from either direction. This means that the equilibrium concentrations will be the same regardless if you started with all reactants, all products, or a mixture of the two
Systems at equilibrium may have the following qualities:
More reactants than products (the reaction favors reactants)
More products than reactants (the reaction favors products)
Equal quantities of both reactants and products
When systems of equilibrium proceed to such a large extent in one direction, the reverse is not considered. This type of reaction is known as quantitative. You can use this type of equilibrium in stoich. Calculations.

6. Two types of Equilibrium:
Static Equilibrium;
opposing forces are balanced and transfer stops
Dynamic Equilibrium:
opposing forces proceed at the same rate
system displays constant macroscopic properties (temp, color, pH etc.)

7. Various types of dynamic equilibrium systems can occur:
Phase Equilibrium: a single substance existing in more than 1 phase
Example: Liquid water in a sealed container with water vapour in the space above it
Water evaporates until the concentration of water vapour rises to a maximum and then remains constant
Solubility Equilibrium: a saturated solution
Rate of dissolving = rate of recrystallization
Phase equilibrium
(eg. ice/water system at 0 oC )
Solubility equilibrium
(eg. saturated solution)

8. Types of Equilibrium
Chemical Equilibrium – reactants and products in a closed system
Example: The Hydrogen-Iodine Equilibrium System
Chemical reaction equilibrium
(balance btwn forward & reverse rxns.)
The rate of reaction of the reactants decreases as the number of reactant molecules decrease. The rate at which the product turns back to reactants increases as the number of product molecules increases. These two rates become equal at some point, after which the quantity of each will not change.

9. Approaching Equilibrium From Opposite Directions page 172

10. Describing the Position of Equilibrium
Percent Yield- the yield of product measured at equilibrium compared with the maximum possible yield of product.
% yield = product eq’m x 100 %
product max
The equilibrium concentration is determined experimentally, the maximum concentration is determined with stoichiometry

11. Describing the Position of Equilibrium
Percent Yield – Example
If 2.50 mol of hydrogen gas reacts with 3.0 mol of iodine gas in a 1.00L vessel, what is the percent yield if 3.90 mol of hydrogen iodide is present at equilibrium

12. Describing the Position of Equilibrium
Using an Equilibrium Constant, (Kc)
This relationship only works if all concentrations are at equilibrium at a constant temperature in a closed system
Think “products over reactants”
If the Kc > 1, the equilibrium favors products
If the Kc < 1, the equilibrium favors reactants
In this equation, A and B are reactants and C and D are products. All lowercase letters are coefficients
In this equation Kc is the equilibrium constant: therefore, units are not necessary. The equation also has the following qualities:
[C] and [D] are the equilibrium molar concentration of the products (mol/L)
[A] and [B] are the equilibrium molar concentration of the reactants (mol/L)
Exponents are the coefficients of the balanced equation
By looking at the magnitude of Keq characteristics of the reaction can be predicted:
Keq > 1 products favored
Keq < 1 reactants favored
Keq = 1 neither favored
(equilibrium yield = 50%)
Note: Keq ≠ 0

13. In this equation, A and B are reactants and C and D are products
In this equation, A and B are reactants and C and D are products. All lowercase letters are coefficients
In this equation Kc is the equilibrium constant: therefore, units are not necessary. The equation also has the following qualities:
[C] and [D] are the equilibrium molar concentration of the products (mol/L)
[A] and [B] are the equilibrium molar concentration of the reactants (mol/L)
Exponents are the coefficients of the balanced equation
Kc is the equilibrium constant for a specific temperature. Different temperatures have different equilibrium constants
All substances are assumed to be present, or else the concentration would be zero. The Kc value may be very large or very small, but it can never be equal to zero.
The value of Kc depends on temperature, meaning a temperature change will change the value of Kc
The value of Kc can be used to estimate whether reactants or products are favored at equilibrium.

14. Describing the Position of Equilibrium
Using an Equilibrium Constant, (Kc)
Example #1: Write the equilibrium law expression for the reaction of nitrogen monoxide gas with oxygen gas to form nitrogen dioxide gas.

15. Describing the Position of Equilibrium
Using an Equilibrium Constant, (Kc)
Note: The Kc value describes the extent of the forward reaction.
Kc reverse = = The reciprocal value
Kc forward
Example #2: The value of Kc for the formation of HI(g) from H2(g) and I2(g) is 40, at a given temperature. What is the value of Kc for the decomposition of HI(g) at the same temperature.

16. Describing the Position of Equilibrium
Using an Equilibrium Constant, (Kc)
Note: For heterogeneous equilibrium systems, DO NOT include liquids and solids in the expression. (They are assumed to have fixed concentrations)
Example #3: Write the equilibrium law expression for the decomposition of solid ammonium chloride to gaseous ammonia and gaseous hydrogen chloride
Example #4: Write the equilibrium law expression for the reaction of zinc in copper(II) chloride solution.
The solid is omitted from the expression
The solids, as well as spectator ions (Cl-) are omitted from the expression

17. General Rules for the Equilibrium Expression
Solids are omitted from the equilibrium expression because their concentrations are unchanging as a result of density (they cannot be significantly compressed).
Liquids are omitted from the equilibrium expression if there is only one (as in aqueous solutions). If there is more than one liquid present, the liquid can have its concentration changed by dilution. Liquid water is not included, since the dissolved substances do not change the concentration of the water.
If there is more than one liquid, include all the liquids in the equilibrium expression. Liquids must be miscible for their proportions to change

18. All other states are included
A complete description of equilibrium state includes the temperature, composition and concentration of all entities present. The concentration for gases is often expressed as pressure

19. Describing the Position of Equilibrium

20. Calculations in Equilibrium Systems
Using the equilibrium law expression to determine whether a system is at equilibrium:
Substitute in the given concentrations to the equilibrium expression. If the value is the equilibrium constant, the system is at equilibrium
If the value is larger, this means there are more products that reactants. To reach equilibrium, the reaction must proceed to the left (towards the reactants)
If the value is smaller, this means there are more reactants than products. To reach equilibrium, the reaction must proceed to the right (towards the products)

21. Calculations in Equilibrium Systems
Example #1: In the following system:
N2(g) + 3H2(g) ↔ 2NH3(g)
0.249 mol N2(g), 3.21 X mol H2(g) and 6.42 X 10-4 mol NH3(g) are combined in a 1.00 L vessel at 375oC, Kc = 1.2
Is the system at equilibrium?

22. Calculations in Equilibrium Systems
Example #2: Find the equilibrium concentration of the ions that are formed when solid silver chloride is dissolved in water. The equilibrium constant for this reaction is Kc = 5.4 X 10-4.
AgCl(s)  Ag+(aq) + Cl-(aq)

23. ICE Charts and Equilibrium Calculations
STEPS:
Always write out the equilibrium reaction and equilibrium law expression if not given.
Draw an ICE Chart (Initial, Change in and Equilibrium concentrations) (I + C = E)
Substitute values where appropriate
Solve for x
Solve for equilibrium concentrations

28. ICE Charts and Equilibrium Calculations
Example #1: Consider the following equilibrium at 100 oC:
N2O4(g) ↔ 2 NO2(g)
2.0 mol of N2O4(g) was introduced into an empty 2.0 L bulb. After equilibrium was established, only 1.6 mol of N2O4(g) remained. What is the value of Kc?

29. ICE Charts and Equilibrium Calculations
Example #2
A 10 L bulb is filled with 4.0 mol of SO2(g), 2.2 mol of O2(g) and 5.6 mol of SO3(g). The gases then reach equilibrium according to the following equation:
2SO2(g) + O2(g) ↔ 2SO3(g)
At equilibrium, the bulb was found to contain 2.6 mol of SO2(g). Calculate Kc for this reaction.

30. Your Task

31. ICE Charts and Equilibrium Calculations
 Example #4: Using the Approximation Rule 
Calculate the concentration of gases produced when mol/L COCl2(g) decomposes into carbon monoxide and chlorine gas. The Kc for this reaction is 2.2 X

32. Le Chatelier’s Principle
If a stress is applied to a dynamic equilibrium system, the system changes to relieve the stress.
Equilibrium shifts can result from:
changes in concentration
(pressure – gas only);
temperature change;
Nature of reactants (the chemical properties)
Catalyst
If the equilibrium shifts such that more products are formed and more reactants are used up, the equilibrium is said to “shift right” It is shifts such that fewer products are formed and fewer reactants are used up, the equilibrium is said to “shift left.”
- It is not necessary to memorize all the affects of Kc due to various stresses. Rather an understanding of Le Chatelier’s principle allows you to predict these effects. Simply picture the equilibrium fighting back whenever you do something to it.
Changes in the temperature, will change the Kc value. No other changes will affect this value.

33. Le Chatelier’s Principle
Effect of Changes in Concentration
If a system at equilibrium is disturbed by the addition of a reactant (or the removal of a product), Le Chatelier’s principle predicts that the equilibrium will shift right.
2N2O(g) + 3O2(g) NO2(g)
If the disturbance is the removal of a reactant (or the addition of a product), Le Chatelier’s principle predicts that the equilibrium will shift left.
Since concentrations of solids are constants and do not appear in expressions for K, removing or adding some solid does not cause shifts.
Concentration Change
adding reactant or removing product will shift the reaction toward the product side (right)
adding product will shift toward the reactant side (left)

34. Concentration Change
Addition of reactant, HF(g) – will shift equilibrium to the right
More products will be produced, and a new equilibrium is established
Removal of product, HCl(g), will shift the equilibrium to the right
More products will be produced, and a new equilibrium is established
If you see spike in either a reactant or product, which causes a gradual change in the other entities – a substance has been added or removed.

35. Le Chatelier’s Principle
Effect of Changes in Temperature
In endothermic reaction, heat acts like a reactant. Increasing the temperature shifts the reaction right. Decreasing the temperature, shifts the reaction left
Heat + 2N2O(g) + 3O2(g) NO2(g)
In exothermic reactions, heat acts like a product. Increasing the temperature shifts the reaction left. Decreasing the temperature, shifts the reaction right.
4 NO2(g) N2O(g) + 3O2(g) + Heat
The equilibrium constant, Kc is temperature dependent
Temperature Change
think of heat as if it were a product or reactant
(heat speeds an endothermic rxn. and slows an exothermic rxn.)
Remember K gets bigger if there are more products being created (i.e. shifts to the right)

36. Reaction Type
Role of heat
Effect of T 
Effect of T 
Endothermic
Reactants + heat  products
K  K
Exothermic
Reactants  products + heat
K  K

37. Temperature Change
The temperature decreases, at the time indicated by the dotted line.
This will cause the equilibrium to shift to the right, creating more products, until a new equilibrium is established.
+ energy
If you see a gradual change in the reactants with an opposite change in the products, you have a temperature change going on

38. Pressure Change Pressure affects only gases.
When volume decreases, the pressure increases.
When volume increases, the pressure decreases.
Le Chatelier’s principle states that when you change the pressure of a closed system at equilibrium, the reaction will shift to produce more or fewer molecules of gas to oppose the shift.
Example: If you decrease the pressure, the reaction will counteract the shift and increase the pressure because the opposing reaction will create more molecules. The reaction will shift to the side where there are more molecules of gas.
Pressure Change
all entities must be gases
count the number of moles on each side of the equation
increase pressure will move equilibrium toward side with least number of moles & vv

39. Le Chatelier’s Principle
Effect of Changes in the Volume of the Container
If volume is decreased, pressure increases (Boyle’s Law – Chem 20)
So the reaction will shift in the direction which contains the fewest moles of gas
2N2O(g) + 3O2(g) NO2(g)
If volume is increased, pressure decreases (Boyle’s Law – Chem 20)
So the reaction will shift in the direction which contains the most moles of gas
2N2O(g) + 3O2(g) NO2(g)
If both sides of the equation have the same number of moles of gas, the change in volume of the container has no effect on the equilibrium.
Pressure
4 moles are fewer than 5
Pressure
5 moles are more than 5
When volume increases, the pressure decreases

40. Volume Change
The volume of the container decreased, at the time indicated by the dotted line. This will cause a pressure increase.
This will cause the equilibrium to shift to the right, the side with fewer moles of gas, creating more products, until a new equilibrium is established.
If you see a spike in all of the entities = P increase, V decrease
If you see a drop in all of the entities = P decrease, V increase

41. Le Chatelier’s Principle
Effect of the Addition of a Catalyst
Catalysts speed up the rate at which equilibrium is obtained, but have no effect on the magnitude of K. They increase both the forward and backward rate of reaction.
Catalyst
decreases the time to reach equilibrium but does NOT affect the final position of the equilibrium

42. Haber-Bosch Process
Is an important process for adding nitrates to fertilizers and was used in the manufacture of explosive during the Second World War.

44. Identify the nature of the changes imposed on the following equilibrium system at the four times indicated by coordinates A, B, C and D
At A, the concentration (or pressure) of every chemical in the system is decreased by increasing the container volume. Then the equilibrium shifts to the left (the side with more moles of gas)
At B, the temperature is increased. Then the equilibrium shifts to left.
At C, C2H6(g) is added to the system. Then the equilibrium shifts to the left.
At D, no shift in equilibrium position is apparent; the change imposed must be addition of a catalyst, or of a substance that is not involved in the equilibrium reaction.

45. Effect (increase, decrease, or no change)
Practice
Change
Direction of Shift
(, , or no change)
Effect on quantity
Effect (increase, decrease, or no change)
a) Decrease in volume Kc
b) Raise temperature
Amount of CO(g)
c) Addition of I2O5(s)
d) Addition of CO2(g)
Amount of I2O5(s)
e) Removal of I2(g)
Amount of CO2(g)

46. Do You Remember? pH and pOH

47. pH Measurements & Calculations
pH = - log [H3O+ (aq)]
pH + pOH = 14.0

48. Significant Digits and pH
Significant Digits: The number of digits FOLLOWING the decimal point in the pH value is equal to the number of significant digits in the hydronium ion concentration.

49. practice

50. Arrhenius Acids & Bases
An Arrhenius:
Acid reacts with water to form hydronium ions (H3O+)
Base reacts with water to form hydroxide ions (OH-)

51. Concentration vs. Strength
Concentration refers to the number of moles dissolved per liter of solution
Dilute vs. Concentrated
Strength is the extent to which an acid ionizes

52. Acid Strength as an Equilibrium Position
Do you remember the difference between strong and weak acids?
Strong acids – ionize completely (quantitatively), even though this could be written with a double arrow, it is simpler to use a single arrow to show the reaction is >99.9%
Strong Acids: HClO4(aq), HI(aq), HBr(aq) , HCl(aq) , HSO4(aq), HNO3(aq)
Weak acids – ionize (react with water) only partially (<50%)

53. Acid Ionization
To calculate the equilibrium constant you must compare the concentration of the products with the concentration of the reactants

54. % Ionization
In a 0.10 mol/L solution of acetic acid, only 1.3% of the CH3COOH molecules have reacted at equilibrium to form hydronium ions. Calculate the hydronium ion amount concentration.

55. % Ionization
The pH of 0.10 mol/L methanoic acid solution is Calculate the percent reaction for ionization of methanoic acid.

56. Acid/Base Titrations
Standard solution – [known]
Equivalence point – amount of base added = amount of acid in solution
End point – indicator color change at pH change

57. The Water Ionization Constant, Kw
Even highly purified water has a very slight conductivity. This is due to the ionization of some water molecules into hydronium ions and hydroxide ions.
The water ionization equilibrium relationship is so important, it gets its own special symbol and name: Ionization Constant for Water, Kw

58. The Water Ionization Constant, Kw
Since the mathematical relationship is simple, we can easily use Kw to calculate either the hydronium or hydroxide ion concentration, if the other concentration is know.
The presence of substances other than water decreases the certainty of the Kw value to two significant digits;
1.0 x

59. Kw Calculations
Example: A 0.15 mol/L solution of hydrochloric acid at 25°C is found to have a hydronium ion concentration of 0.15 mol/L. Calculate the amount concentration of the hydroxide ions.

60. Kw Calculations
Example #2: Calculate the amount concentration of the hydronium ion in a 0.25 mol/L solution of barium hydroxide.
Ba2+ = 0.25 mol/L x 2 mol = mol/L
1 mol

61. Kw Calculations
Example #3: Determine the hydronium ion and hydroxide ion amount concentration in 500 mL of an aqueous solution for home soap-making containing 2.6 g of dissolved sodium hydroxide.

62. Your Task

63. Bronsted-Lowry Acid-Base Concept
Focuses on the role of the chemical species in a reaction rather than on the acidic or basic properties of their aqueous solutions.
Bronsted Lowry Definition for an Acid: proton donor
Bronsted Lowry Definition for an Base: proton acceptor
Thought of acid and base behavior in terms of proton transfers.
An acid could be any entity that loses a proton to another entity in a solution
A base could be any entity that receives the proton in a reaction

64. Add info from page 184

65. Bronsted-Lowry Acid-Base Concept
Bronsted-Lowry Reaction Equation: is an equation written to show an acid-base reaction involving the transfer of a proton from one entity (an acid) to another (a base)
Bronsted-Lowry Neutralization: is a competition for protons that results in a proton transfer from the strongest acid present to the strongest base present.
The Bronsted-Lowry concept does away with defining a substance as being an acid or base. Only an entity that is involved in a proton transfer in a reaction can be defined as an acid or base – and only for a particular reaction
Remember this is just a theoretical definition, not a theory, because it does not explain why a proton is donated or accepted, and cannot predict which reaction will occur for a given entity in any new situation.

66. Bronsted-Lowry Acid-Base Concept
Protons may be gained in a reaction with one entity, but lost in a reaction with another entity.
The empirical term, amphoteric, refers to a chemical substance with the ability to react as either an acid or base.
The theoretical term, amphiprotic, describes an entity (ion or molecule) having the ability to either accept or donate a proton.
Example: When bicarbonate ions are in aqueous solution, some react with the water molecules by acting as an acid, and some react by acting as a base. Kc values given for these reactions, show that one predominates. The number of ions acting as a base is over 2000x more than the number reacting as an acid. = BASIC Solution
Amphiprotic substances can act as acids in some reactions and as bases in others. They can either accept of donate protons

67. Polyprotic acid is an acid with more than one proton to donate.
Polybasic Base is a base that can accept more than one proton.

68. Bronsted-Lowry Acid Base Concept
The Amphoteric Nature of Baking Soda
It can partially neutralize a strong acid, but also a strong base

69. Conjugate Acids and Bases
In an acid-base reaction, there will always be two acids and two bases.
The original acid on the left and the acid on the right created by adding a proton.
The original base on the left and the base on the right created by removing a proton
A pair of substances with formulas that differ only by a proton is called a conjugate acid-base pair
This analogy shows why acetic acid is a weak acid.
The proton is more strongly attracted to the acetic acid molecule than it is to the water.

70. Acids/Bases What about strong acids:
When HCl reacts with water, the water wins the competition against the Cl- for the proton. This is why at equilibrium, essentially all of the HCl molecules have lost protons to water.
NOTE: The stronger the base, the more it attracts a proton (proton acceptor). The stronger the acid, the less it attracts its own proton (proton donor)
>99%

71. Conjugate Acids and Bases
RULE: The stronger the base, the more it attracts a proton (proton acceptor). The stronger the acid, the less it attracts its own proton (proton donor)
What does this mean about their conjugate pair??
The stronger an acid, the weaker is its conjugate base.
If you are good at donating a proton, this means the conjugate base is not good at competing for it (weak attraction for protons)
The stronger a base, the weaker is its conjugate acid.
If you are good at accepting a proton, this means the conjugate acid is not good at giving it up (strong attraction for protons).

73. Conjugate Acid-Base Pairs
Write formulae for the conjugate bases of the following acids;
HCO3- (aq) HCl (aq) H2SO4 (aq)
NH4+ (aq) H2O (l) H2CO3 (aq)
HOOCCOOH (aq) H2PO4- (aq)

74. Conjugate Acid-Base Pairs
Write formulae for the conjugate acids of the following bases;
Br - (aq) NO2- (aq) PO4 3- (aq)
HSO4- (aq) HO2- (aq) HS- (aq)
HCO3- (aq) CO3 2- (aq)

75. Diagramming B/L Acid/Base Rxns

76. Amphiprotic Substances
An amphiprotic substance will act as a(n):
Acid if it is a better proton donor than the substance with which it is reacting
Base if it is a better proton acceptor than the substance with which it is reacting.

77. Amphiprotic Substances
Note: You have seen amphiprotic substances show up on both sides of the Acids-Bases Table. When identifying substances as possible acids or bases it is important for you to remember to check BOTH sides of the table.

78. Amphiprotic Substances
Examine the Acid-Bases Table. List 10 substances that are amphiprotic.
• hydrogen sulfate ion • hydrogen phosphate ion
• hydrogen oxalate ion • dihydrogen borate ion
• hydrogen sulfide ion • dihydrogen phosphate ion
• hydrogen sulfite ion • hydrogen borate ion
• hydrogen carbonate ion • water

79. Acid-Base Reactions What is happening?
Collisions are constantly occurring. Each time, a proton is transferred to a stronger proton attractor.
Theoretically, it could transfer several times (each time to a stronger proton attractor.) But once, it is transferred to the strongest base present, the proton will remain there as nothing outcompetes it.
Likewise, once the strongest acid has given up its proton, its conjugate base cannot gain one back (as it is the weakest proton attractor in the whole system)
So what does this mean?
Proton transfer occurs between the strongest acid and strongest base. All other transfers are negligible so are ignored.

80. Predicting Acid-Base Reactions

81. Predicting Acid-Base Reactions
1) List all entities present initially (ions, atoms, molecules, H2O(l)) as they exist in aqueous solution.
H3O+(aq) is the SA that can exist. If a stronger acid is dissolved, it reacts instantly and completely with water to form H3O+(aq). So all strong acids are written as H3O+(aq)
OH-(aq) is the SB that can exist. If a stronger base is dissolved, it reacts instantly and completely with water to form OH-(aq). The only example of this: soluble ionic oxides, write the cation and the oxide ion is written as OH-(aq)
No entity can react as a base if it is weaker than water.
For this reason, the conjugate bases of the strong acids are not considered bases in aqueous solutions

82. Predicting Acid-Base Reactions
1) List all entities present initially (ions, atoms, molecules, H2O(l)) as they exist in aqueous solution.
Example: What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution is neutralized by vinegar?
List entities present:

83. Predicting Acid-Base Reactions
2) Identify and list all possible aqueous acids and bases, using the Bronsted-Lowry definitions.
Use the entity lists of the Relative Strengths of Acids and Bases (in your data booklet). Amphiprotic entities are labeled fro both possibilities. Conjugate bases on SA’s are not included. Metal ions are treated as spectators.
Example: What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution is neutralized by vinegar?
List entities present:

84. Predicting Acid-Base Reactions
3) Identify the strongest acid and the strongest base present, using the table of Relative Strengths of Acids and Bases.
Use the order of the entities in the Relative Strengths of Acids and Bases table to identify the SA (the highest one on the table) and the SB (the lowest one on the table)
Example: What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution is neutralized by vinegar?
List entities present:

85. Predicting Acid-Base Reactions
4) Write an equation showing a transfer of one proton from the SA to the SB, and predict the conjugate base and the conjugate acid to be the products.
Example: What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution is neutralized by vinegar?
List entities present:

86. Predicting Acid-Base Reactions
5) Predict the approximate position of equilibrium, using the following Learning Tip
Example: What will be the predominant reaction
if spilled drain cleaner (sodium hydroxide) solution is
neutralized by vinegar?

87. Table Building
If a reaction is >50%, it favours products and the SA is above the SB.
If a reaction is <50%, it favours reactants and the SA is below the SB.

88. Table Building

89. Acid
Base

90. The Relative Acid-Base Strength Table
Acid-base strength tables are built upon the principle of comparing acids and bases under identical conditions.
Use the Acid-Base Strength Data Table to arrange the following acids according to strength in a Table with the following headings.
Acid Name Formula % Reaction Eq. Constant (Ka)

91. Creating an Acid Strength Table
• ammonium ion • hydrosulfuric acid
• ethanoic acid • hydrogen sulfate ion
• benzoic acid • hydrogen carbonate ion
• hydroiodic acid • hydrogen borate ion
• Explain the trends you notice comparing the percent reaction to the equilibrium constants.

92. Trends on Acid Base Table
Weaker acids - % reaction smaller & Ka smaller because decreasing ability to donate protons
Example: reactants favored over products so denominator gets larger

93. The Acid Ionization Constant, Ka
One important way that chemists communicate the strength of any weak acid is by using the equilibrium constant expression for the ionization of weak acids. (Ka)
Look at your Relative Strengths of Acids and Bases Table:
All of the strong acids have a Ka value listed as very large; remember they ionize quantitatively, so the actual acid species present is H3O+(aq)
All other acids are weak and vary greatly in the extent of reaction at equilibrium.
We use the equilibrium law to write the formula for Ka
Ka Values have only two sig digs because they are somewhat inaccurate because we calculate them based on assumptions.
We omit liquid water because we assume its value is essentially constant for dilute solutions

95. Ka Calculations Two common Ka calculations:
Calculating Ka from empirical data
Using Ka to predict hydronium ion concentration from an initial concentration of weak acid.

98. Ka Calculations
Example #1: The pH of a 1.00 mol/L solution of acetic acid is carefully measured to be at SATP. What is the value of Ka for acetic acid?

101. Ka Calculations
Example #2: A student measures the pH of a 0.25 mol/L solution of carbonic acid to be Calculate the Ka for carbonic acid from this evidence.

102. Ka Calculations
Example #3: The pH of a mol/L solution of sulfurous acid is measured to be Calculate the Ka for sulfurous acid from this evidence.

105. Ka Calculations
Example #4: Predict the hydronium ion concentration and pH for a mol/L aqueous solution of methanoic acid.

108. Ka Calculations
Example #5: Predict the hydronium ion concentration and pH for a mol/L aqueous solution of hydrocyanic acid.

111. Bases page 193

112. Weak bases page 194

113. Base Strength & Ionization Constant, Kb
Ionic hydroxides, such as NaOH(aq) or Ca(OH)2(aq), are assumed to dissociate completely upon dissolving.
Finding the hydroxide ion concentration does not involve any reaction with water
Example: Find the hydroxide ion concentration of a mol/L solution of barium hydroxide

114. Base Strength & Ionization Constant, Kb
We can communicate the strength of weak bases, with the equilibrium constant for its reaction with water.
The equilibrium constant is called the base ionization constant, Kb
Two common Kb calculations:
Calculating Kb from empirical data Using Kb to predict the concentration of hydroxide ions when the initial concentration is known

115. Kb Calculations
Example #1: A student measures the pH of a mol/L solution of aqueous ammonia and finds it to Calculate the Kb for ammonia

116. The Relationship between KA and kb
Page 195

117. Example
Using the values for the weak acid NH4+, determine the Kb for its conjugate base
Page 195

118. Your Task

119. Calculating OH- from Kb
First problem, the data booklet has Ka values not Kb values.
What do we do?
Kw = KaKb So Kb = Kw/Ka remember Kw = 1.00 x 10-14
Example #1: Solid sodium benzoate forms a basic solution. Determine the Kb for the weak base present.

120. Calculating OH- from Kb
Example #2: Find the hydroxide ion amount concentration, pOH, pH and the percent reaction (ionization) of a 1.20 mol/L solution of baking soda.
Baking soda = NaHCO3(s)  Na+(aq) + HCO3-(aq)
For HCO3-(aq), the conjugate acid is H2CO3(aq) whose Ka is = 4.5 x 10-7

121. Calculating OH- from Kb
Example #2: Find the hydroxide ion amount concentration, pOH, pH and the percent reaction (ionization) of a 1.20 mol/L solution of baking soda.

122. Some Things to Consider About Acids and bases
Oxy acids etc. on page 38

123. Amphoteric Species
If an entity can react as either a Bronsted-Lowry acid or base, how do you know which will be the predominant reaction?
Find the Ka value, calculate the Kb value, which ever is larger wins!
Example: Which reaction predominates when NaHSO3(s) is dissolved in water to produce HSO3-(aq) solution? Will the solution be acidic or basic?
The Ka value far exceeds the Kb value, so an aqueous solution of this substance will be acidic because the hydrogen sulfite ion will react predominately as a Bronsted-Lowry acid.

124. Titration Terminology Review
Page 201

125. A graph showing the continuous change of pH during an acid-base titration, which continues until the titrant is in great excess, is called a pH curve
Endpoint refers to the point in a titration analysis where the addition of titrant is stopped. The endpoint is defined (empirically) by the observed colour change of an indicator.
Equivalence Point refers to the point in any chemical reaction where chemically equivalent amounts of the reactants have combined. This point is determined by stoichiometry.

127. Equivalence points
It is important to note, that the equivalence point pH is 7 ONLY for strong acid-strong base reactions.
For every other acid-base reaction, the equivalence point solution will contain ions or molecules that are not spectators – so titration curves must be done empirically to determine the equivalence point .
Indicators are substances (usually weak monoprotic acids) that change color over a given pH range
Indicators can be used to identify when a neutralization reaction is complete.

128. Interpreting pH Curves
Buffering: is the property of some solutions to resist any significant change in pH when an acid or base is added.
Buffering region (nearly level portions of the graph)
Why is it buffering?
Initially, the solution is mainly all water and OH- ions.
Any additional acid added (H3O+) immediately reacts with OH- to become water – which does not change the pH significantly.
This “leveling effect” finally fails near the equivalence point, when the OH- is almost completely consumed.
Once excess acid has been added, the solution consists of water and H3O+ ions, so the pH has dropped to the acid range.
Then any additional acid that is added, simple increase the H3O+ concentration slightly, but does not change the pH much.

129. Titration Analysis Selecting proper indicators
Alizarin yellow is not a suitable indicator because it will change colour long before the equivalence point of this strong acid-strong base reaction, which theoretically has a pH of exactly 7.
Orange IV is also unsuitable; its colour change would occur too late.
The pH at the middle of the colour change range for bromothymol blue is 6.8, which very closely matches the equivalence point pH; so bromothymol blue should give accurate results.

130. Acid Base Indicators
Any acid base indicator is really two entities for which we use the same name: a Bronsted-Lowry conjugate acid-base pair.
At lease one of the entities is visibly colored, so you can tell simply by looking when it forms or is consumed. Examples include:
Phenolphthalein: conjugate acid is colorless, conjugate base is bright pink
Bromothymol Blue: conjugate acid is yellow, conjugate base is blue, and when they are in equal quantities – (appear green to the human eye)
Litmus Paper – red (HIn) to blue (In-)
We will use the designation HIn for the conjugate acid and In- for the conjugate base as their actual formulas can be very complex.
Summary: An indicator is a conjugate weak acid-weak base pair formed when an indicator dye dissolves in water.
Do not influence the reaction

131. Notes page 207

132. Selecting an Indicator from titration curve data
To select an indicator from titration curve data use the following steps:
Decide where the equivalence point is
From the indicator list in the data tables, select an indicator that changes color at a pH close to the actual equivalence point.

133. Indicator Example
A solution that is colorless in phenolphthalein, yellow in methyl orange and yellow in cholorphenol red could have a pH of what?
Answer 5.0
Make a chart and narrow down

134. Indicator Example
A solution with a pH of 3.5 would have the following colors in cresol red, methyl orange and bromocresol green respectively
Yellow, orange and yellow

135. Indicator Challenge Question
When a 0.10 mol/L HF(aq) is titrated with a 0.10mol/L KOH(aq) the best indicator to use would be what?

136. Carbonate ion is a diprotic base
Polyprotic Entities
Polyprotic acids – can lose more than one proton
Polyprotic bases – can gain more than one proton
If more than one proton transfer occurs in a titration, chemists believe the process occurs as a series of single-proton transfer reactions.
On a graph, this means there will be more than one equivalence point
First proton transfer = 100%
Second proton transfer = 100%
Carbonate ion is a diprotic base

137. Polyprotic Entities
A pH curve for the addition of NaOH to a sample of H3PO4(aq) displays only two rapid changes in pH even though H3PO4(aq) is triprotic.
This is because only two of the transfers are quantitative. The third reaction never goes to completion, but instead establishes an equilibrium.
General Rule: Only quantitative reactions produce detectable equivalence points in an acid-base titration.

138. Example
Sketch the titration curve of a polyprotic base Na2CO3 titrated with a strong acid, hydrochloric acid. Use 25 mL of 0.20 mol/L Na2CO3 titrated with 0.20 mol/L hydrochloric acid for your data

139. Example
Sketch the titration curve of the titration of a polyprotic acid, H3PO4, titrated with a strong base, NaOH. In terms of experimental data, use 25mL of 0.12mol/L H3PO4 titrated with 0.12 mol/L NaOH.

140. Page 206 explanation

141. pH Curves

142. General Rule
Strong Acid to Weak Base: pH at equivalence point is always lower than 7
Strong Base to Weak Acid: pH at equivalence point is always higher than 7

143. Titration graphs page 34

147. pH Curve Shape
SA-SB: water is the only acid or base present = neutral solution
SA-WB: a weak acid (NH4+)is present along with water, at the equivalence point, so the solution is acidic (pH < 7)
WA-SB: a weak base (CH3COO-) is present along with water, at the equivalence point, so the solution is basic (pH > 7)
WA-WB: do not have detectable equivalence points, because the reactions are usually not quantitative.

148. Buffers
A buffer is a relatively large amount of any weak acid and its conjugate base, in the same solution. In equilibrium, they maintain a relatively constant pH when small amounts of acid or base are added.
Example: The addition of a small amount of base produces more acetate ions. The very small change in the acid-base conjugate pair ration and the complete consumption of the OH- explains why the pH change is very slight
The addition of a small amount of acid produces more acetic acid. The very small change in the acid-base conjugate pair ratio and the complete consumption of the H3O+ explains why the pH change is very slight
A buffer can be described as a solution of a weak acid or base and its conjugate pair in equilibrium that has the ability to maintain a relatively constant pH when small amounts of acid or base are added.

149. Graph from page 217

150. A weak Base-Strong Acid Titration curve page 218

151. Equilibriums in Acidic Buffer Solutions
An acidic buffer solution consists of a weak acid and its conjugate base, which are both in reasonably high concentrations. To write the equilibrium equation for an acidic buffer, simply write the equation for the ionization of the weak acid in water.

152. For example, consider a buffer solution that is prepared that contains 1.0mol/L CH3COOH and 1.0 mol/L NaCH3Coo. This buffer has the following equilibrium equation:

153. Equilibriums in Basic Buffer Solutions
A basic buffer solution consists of a weak base and the salt of its conjugate acid. To write the equation for the equilibrium existing in a basic buffer solution, write the equation for the ionization of the weak base in water.

154. Example
A buffer solution is prepared containing 1.0 mol/L NH3 and 1.0 mol/L NH4Cl. The following equation describes the equilibrium present:

155. Composition and types of buffer solutions
There are two types of buffer solutions: acidic buffers and basic buffers
In any buffer solution, the concentrations of both components should be moderately high.
Neither a strong acid nor a strong base can be used to prepare a buffer solution
Type
Composition
pH Range of Operation
Acidic Buffer
A weak acid and the salt of its conjugate base
pH of 7 or below
Basic Buffer
A weak base and the salt of its conjugate acid
pH of 7 or above

156. Some examples of acidic buffers

157. Buffers
Put in buffer examples page 32 in review book

158. Adding an acid to an acidic buffer solution
A small amount of HCl is added to a buffer solution containing 1.0 mol/L acetic acid and 1.0 mol/L and a 1.0 mol/L sodium acetate. The equilibrium equation for the buffer solution has the following equilibrium equation.

159. The following sequence of events takes place:
The addition of HCl immediately increases the hydronium ion concentration
In order to counteract the increase the hydronium ion concentration, the equilibrium shifts to the left
As the equilibrium shifts to the left, CH3COO- reacts with H3O+
The hydronium ion concentration decreases close to- but not quite as low as- its original concentration
In the overall process the final hydronium ion concentration is slightly higher (and the pH is slightly lower) than it was before the acid was added

160. Graph page 214 SNAP

161. Example
Describe the effect of adding a strong acid to a solution of carbonic acid and its conjugate base, hydrogen carbonate ion.

162. Basic Buffers

163. Adding a base to a basic buffer solution
A small amount of NaOH solution is added to a buffer solution consisting of 1.0 mol/L NH3 and 1.0 mol/L NH4Cl.
The buffer solution has the following equilibrium equation:

164. The following sequence of events takes place:
The addition of NaOH (strong base) immediately increases the concentration of hydroxide ions
In order to counteract the increase in the hydroxide ion concentration the equilibrium shifts to the left
As the equilibrium shifts to the left, NH4- reacts with OH-
The hydroxide ion concentration decreases close to- but not quite as low as- its original concentration
In the overall process, the final hydroxide ion concentration is slightly higher (and the pH slightly higher) than it was before the base was added

165. Graph from page 216

166. Buffer Example: Blood Plasma
Blood plasma has a remarkable buffering ability, as shown by the following table.
This is very useful, as a change of more than 0.4 pH units, can be lethal. If the blood were not buffered, the acid absorbed from a glass of orange juice would likely be fatal.
Optimum pH for human blood is 7.35
If the pH of blood goes below 7.20 or above 7.50 serious conditions can result

167. Some Uses of buffer solutions
Buffer solutions are widely used today. Some uses include:
Stabilizing pH in hot tubs and swimming pools
Maintaining optimum pH for production of pharmaceuticals and other chemicals
Controlling the pH in wines and foods
Providing solutions with accurate pHs for calibrating pH meters

168. Buffering Capacity
The limit of the ability of a buffer to maintain a pH level.
When one of the entities of the conjugate acid-base pair reacts with an added reagent and is completely consumed, the buffering fails and the pH changes dramatically.
All of the CH3COO-(aq) is used up, H3O+ additions will now cause the pH to drastically decrease
All of the CH3COOH(aq) is used up, OH- additions will now cause the pH to drastically increase