1. Equilibrium and acids and bases
Chapter 14 and 15 in text

2. Chemical Equilibrium
Concentrations of all reactants and products are constant with time.
That does not mean [reactant] = [product]
Rather the rate of decomposition of [react] and formation of [prod] are constant.

3. Equilibrium is a Dynamic Situation
Demonstration B
6 volunteers 3 on each side of the room. As one person moves form side A to side B a person from side B must move to side A. There is no net change in people because there is an equal amount of people leaving and entering each side. A

4. When a car from island A moves to island B there is a stress placed on the equilibrium and the reaction (the members of the groups) must shift either left or right to relive the stress.
If a stress is placed on Island A then the equilibrium will shift to the right to compensate.

5. What Just Happened?
 When a chemical reaction is at equilibrium, any disturbance of the system, such as a change in temperature, or addition or removal of one of the reaction components, will "shift" the composition of the reaction to a new equilibrium state.

6. A Chemical Example of a Shift in Equilibrium
Co(H2O) Cl CoCl H20
Pink Blue
Now lets apply this concept to a chemical reaction.
Question: what color will the solution be at equilibrium?

7. Demonstration of Equilibrium

8. Question: Which direction would the equilibrium shift if HCl was added to the reaction?
Co(H2O) Cl CoCl H20
Pink Blue
HCl

9. Answer
The addition of HCl would cause an increase in Cl- and thus the reaction would react by shifting the reaction to the right (blue side) in order to “consume” the excess Cl- and return to a state of equilibrium.
Co(H2O) Cl CoCl H20
Pink Blue
Hydrogen ions will disassociate form the Cl- ions creating an increase in Cl- which is a product. In order to maintain equilibrium the excess products will shift the equilibrium in the direction that will consume them by creating more reactants (right shift).

10. The Law of Mass Action [A]j [B]k K = equilibrium constant
For any reversible reaction:
jA + kB lC + mD
The law of mass action is represented by the following equilibrium expression. (Prod /React)
K = [C]l [D]m
[A]j [B]k
K = equilibrium constant
Solids and liquids are not written in equilibrium expressions
Mathematically: if we divide the reactants by the products and raise each species to the power of their coefficient we can find the equilibrium constant K for that reaction. K = the value we get when equilibrium concentrations of the reaction are substituted into an equilibrium expression.

11. K > 1 Favors products
K<1 Favors reactants

12. Graph of Equilibrium

13. Question Write the equilibrium expression for the following equations:
PCl5 (g) PCl3 (g) + Cl2 (g)
Cl2O7(g) + 8H2(g) HCl(g) + 7H20(g)
NOTE: Make sure the RXN is BALANCED!!!

14. Answer
K = [PCl3] [Cl2]
[PCl5]
K = [H2O]7 [HCl]2
[Cl2O7] [H2]8

15. Calculating Equilibrium
Example:
Calculate the equilibrium constant K, for the following reaction at 25°C,
H2 (g) + I2 (g) HI (g)
if the equilibrium concentrations are
[H2] = 0.106M [I2] = 0.022M [HI] = 1.29M

16. Note the units cancel out! That’s one less thing to remember!
Answer
Equilibrium Expression:
K = [HI]2
[H2] [I2]
Equilibrium Constant:
K = (1.29) = x 102
(0.106) (0.022)
No units because they cancel out
Note the units cancel out! That’s one less thing to remember!

17. Homework
Pg ,8,9,14,16
Equilibrium wks

18. How do we predict which direction equilibrium will shift to?
Le Chatelier’s Principle
If we stress a reaction out the reaction will shift (either towards R or P) to reduce the stress.
Stress = a change in: Concentration
Pressure
Volume
Temperature

19. Concentration Ask yourself:
Is what is being added a solid or liq? (No effect)
a gas or aqueous reactant or a product? (effect!)
If it’s a reactant the reaction will shift toward the products to consume the extra reactant.
If it’s a product the reaction will shift toward the reactants to consume the extra product.

20. Example
N2 g + 3H2 g NH3 g
Add NH3
Remove N2
Add H2

21. Changes in V or P P changes in L or aq have little effect
P changes of gases have huge effects (PV=nRT) (PV=PV)
Concentration is effected by pressure.

22. An increase in Pressure (due to a decrease in V) will shift to decreases the total number of moles of gas.
If the number of moles on R and P side are the same then a change in P will have no effect on equilibrium.

23. Example Lets do Pg 626 example 14.12 in your text
↓P shift to side of reaction with greatest number of moles of gas
↑P shift to side of reaction with less moles of gas

24. Temperature
ONLY a change in TEMPERATURE can change the value of K (equilibrium constant).
You must look at the H of the reaction to see how T will effect it
Temperature is our stress so the reaction will move in the direction that removes the stress.

25. ↑ temperature the rxn moves in the endothermic direction (+ΔH)
ENDO  Right ( K ↑ )
↓ temperature the rxn moves in the exothermic direction (-ΔH)
EXO  Left ( K ↓ )
**K only depends on temperature, catalysts have NO EFFECT on K**
K>>1 favors products
K<< 1 favors reactants

26. Think about heat like a R or P
Endothermic +ΔH
Heat + A  B H = 400 kJ
Favors ↑ in T
therefore K ↑ when it is heated and K ↓ when it is cooled.
Exothermic -ΔH
A  Heat + B H = kJ
Favors ↓ in T therefore K ↑ when it is cooled and K ↓ when it is heated

27. Example
Examples on pg in text

28. Homework
Pg 636 pg 49,51,53-55,62

29. Nature of Acids and Bases
Acids: Sour Taste
If a solution has a high
[H+] = acidic
Base: Bitter Taste / Slippery feel
[OH-] = base

30. About the scale P = quantity So … pH = quantity of H ion
pOH = quantity of OH ion
pH + pOH = 14

31. Concentrations [H+] = [OH-] = Neutral [H+] > [OH-] = Acid
[H+] < [OH-] = Base

32. Arrhenius Concept
Focuses on what ions were formed when acids and bases dissolved in water.
Acids dissociate in water give hydrogen ions (H+ or H3O+ hydronium ion)
Bases dissociate in water give hydroxide ions (OH- hydroxide ion ) .

33. Arrhenius acid - Any substance that ionizes when it dissolves in water to give the H+ ion. 
e.g.   
Arrhenius base - Any substance that ionizes when it dissolves in water to give the OH- ion. 
e.g.    

34. The theory can only classify substances when they are dissolved in water since the definitions are based upon the dissociation of compounds in water.
It does not explain why some compounds containing hydrogen such as HCl dissolve in water to give acidic solutions and why others such as CH4 do not.
The theory can only classify substances as bases if they contain the OH- ion and cannot explain why some compounds that don't contain the OH- such as Na2CO3 have base-like characteristics.

35. Bronsted Lowery Acid Base Concept
Acid: substance that can donate a proton (+)
Base: substance that accepts a proton (+) (aka they have a lone pair of e- to accept a proton)
Unlike Arrhenius concept this is applicable in both aqueous and non-aqueous states.

36. Equilibrium
Reaction produces reactants and products at the same rate, but not necessarily in the same amounts
Bathroom theory

37. Example NH3+ (aq) + H20 (l)  NH4 (aq) + OH- (aq)
Equilibrium will favor the formation of the weaker acid and the weaker base.
In this rxn the [NH4] and [OH- ] will be low because they are the stronger acid and base.

38. In the above reaction, the H+ from HCl is donated to H2O which accepts the H+ to form H3O+, leaving a Cl- ion. 

39. Conjugate Acid and Base Pairs
The part of the acid remaining when an acid donates a H+ ion is called the conjugate base. 
The acid formed when a base accepts a H+ ion is called the conjugate acid. 

40. For the generic acid HA:
Formed when a proton Is transferred to the base
Everything that is left after a proton to the base.
NOTE
Strong acids have weak conjugate bases.
Strong bases have weak conjugate acids.

41. Question If H20 is an acid what would its conjugate base be?
What is the conjugate acid of HPO42- ?
What is the conjugate base of HS-

42. Answer H20 take away a proton OH- HPO42- add an H+ H2PO4- HS- S2-
( we added a proton and that needs to be reflected in the molecular charge)
HS S2-

43. Amphoteric The ability of a substance to act as an acid or a base.
Ex: H2PO4- and H2O can act as both acids and bases.

44. Back to Old Faithful
In this equation the stronger base will win the competition for H+.
If H2O is a stronger base than A-, then it will have a greater affinity for the protons and the equilibrium will lie to the right favoring the formation of H3O+.
If A- is stronger then equilibrium will fall to the left and acid in the form HA will form.

45. Strong Acids Conj. Bases
HCl
HBr HI
HNO3
HClO4
HClO3
H2SO4
Cl-
Br - I-
NO3-
ClO4-
ClO3-
HSO4-
You must memorize all of these

46. Common Strong Bases Formula Name NaOH sodium hydroxide LiOH
lithium hydroxide
KOH
Mg(OH)2
potassium hydroxide
Magnesium Hydroxide
Ca(OH)2
Ba(OH)2
Sr(OH)2
Calcium hydroxide
Barium Hydroxide
Strontium Hydroxide

47. Example Identify the CA and CB for each reaction HNO3 + H2O ↔
NH3 + H2O ↔

48. Acid-dissociation equilibrium constant (Ka)
The relative strength of an acid is described as an acid-dissociation equilibrium constant.
The acid-dissociation equilibrium constant is the mathematical product of the equilibrium concentrations of the products of this reaction divided by the equilibrium concentration of the original acid

49. Think Products over reactants

50. Question
Write an ionization equation for the following and then write the acid dissociation constant for both. (all occur in water)
Hydrochloric acid
Acetic acid HC2H3O2

51. Answer HCl ↔ H+ + Cl- Ka = [H+] [Cl-] / [HCl] HC2H3O2 ↔ H+ + C2H3O2-
Ka = [H+] [C2H3O2-] / [HC2H3O2 ]

53. Temperature increases as Kw increases

54. Question
Calculate the H+ in aqueous household ammonia if the OH- concentration is M. Then tell me if the solution is an acid or a base.
This solution is in water so we can use Kw

55. Answer Kw = [H] [OH] = 1.0 x 10 -14 [H] = Kw /[OH] = 1.0 x 10 -14
[H] = Kw /[0.0025M] = 1.0 x
[H] = 4.0 x M
0.0025M OH > 4.0 x M H solution is a base

56. Question
Calculate the values of [H+] [OH-] in a neutral solution at 25ºC
If we know Kw = [H][OH] and [H] = [OH] in a neutral solution
Kw = 1.0 E-14
then 1.0 E-14 = [x][x]
1.0 E-14 = X2 (QUADRATIC TIME!!!!)
X = 1.0E-7 M [H] and [OH]

57. Example Calculate the concentrations using Kw= [H+] [OH-] =10-14
For the following and state if it is an acid, base, or neutral
[OH-] = 10-5 M
[H+] = 10.0 M

58. Answer Kw = 10-14= [H+] [10-5] = 1.0 E-9 M [H+] < [OH-] = Base
Kw = 10-14=[10] [OH-] = 1.0E-15 M
[H+] >[OH-] = Acid

59. pH and pOH pH = -log [H+] pOH = -log [OH-] [H] = 10-pH [OH] = 10-pOH
Tell us the concentration of H and OH in solution.
pH = -log [H+] pOH = -log [OH-]
[H] = 10-pH [OH] = 10-pOH
pH + pOH =14
(really strong acids may have a negative pH)

60. Find pH
pH = -log [H+]
Calculate the pH of a neutral solution that has a [H+] of 1.0 x 10-7 M
pH = - log 1.0 x 10-7 M
pH = 7.00

61. pOH pOH = -log [OH-] or pH + pOH = 14
So we can use the pH (7) from the last problem to find the pOH of that solution.
14-7 = 7 or we could have been given the [OH-]

62. Given pH find [H]
pH = -log [H]
pH = 4 find [H]
4 = -log X
10 -4 = X

63. Strong Acids
Strong acids dissociate almost completely in water and therefore have relatively large Ka values.
Equilibrium lies far to the right
HA dissociates almost completely
Yield weak conjugate bases
Strong electrolytes (100% conductivity)
Ka Large > 1 

64. Weak Acids
Weak acids dissociate only slightly in water and therefore have relatively small Ka values.
Equilibrium lies far to the left (HA does not dissociate)
Yields strong conjugate base
Ka small < 1

65. Solving Strong Acid Equations (and Bases too)
1. Strong acids dissociate 100% and are the main source of H+ in a solution.
2. Therefore [H+] equals the original concentration of the acid
ex: 0.2 M HCl = [H] = 0.2 M [Cl] = 0.2 M

66. Calculating pH for a strong acid example
What is the pH of a 0.04 M solution of HClO4?
HClO4 = strong acid equilibrium lies to the right and completely dissociates.
[H+] = [ClO4-]= 0.04 M
pH = -log [0.04] = 1.40

67. Calculating Ka Write the ionization equation for the reaction
Write the equilibrium expression for the reaction (Ka)
Calculate the [H+] using a given pH
ICE box ([initial], [change], [equilibrium] to determine the values for concentration that you will substitute into your equilibrium equation.
Insert concentration into equilibrium expression and solve for Ka

68. Solving for Ka example
A student prepared a 0.1M solution of formic acid (HCHO2) and measured its pH to be Calculate Ka.
How will you attack this problem?

69. Answer Write the ionization rxn. HCHO2 ↔H+ + CHO2- Write Ka equation
Ka = [H+] [CHO2-]
[HCHO2]
Solve for [H] using pH given
pH = -log [H+] =
2.38 = -log [H+]
[H+] = = = 4.2 X 10-3

70. Not done yet ICE Box HCHO2 ↔ H+ + CHO2- [Initial] 0 .10 0 0
[Change] x x x10-3
[Equilibrium] x10-3 M 4.2 x10-3 M x10-3M

71. Still not yet Plug values into Ka equation HCHO2 ↔ H+ + CHO2-
[Equilibrium] x x x10-3
Ka = [H+][CHO2] = [4.2 x10-3] [4.2 x10-3]
[HCHO2] [ x10-3 ]
Solve for Ka = 1.8 x 10-4

72. Lets do another one
A student calculates the pH of HOCl to be 3.5 find Ka. The initial concentration of the acid is 0.01M
Write equation for reaction
Write Ka equation
Use pH to solve for [H+]
ICE Box
Plug values into Ka equation

73. Using Ka and [acid] to calculate pH
Write the dissociation equation for the reaction.
Write the equation for Ka
ICE Box but now we know our initial concentration of our acid so we can use it to solve for the [H+]
Use found [H+] to calculate pH

74. Weak acid example
Calculate the pH of a 0.1 M aqueous solution of HOCl (Ka= 3.4x10-8 weak acid).
HOCl has a higher Ka value than water and will dissociate to produce the most H+ ions, so we use it to write our equation for Ka

75. Write the rxn: HOCl H+ + OCl-
Write the equation for: Ka = 3.4x10-8 = [H+][OCl-]/HOCL
Plug into ICE Box using X for unknown concentrations:
HOCl H OCl- I
C X X X
E X X X
Substitute E values into Ka equation
3.4x10-8 = (x) (x)
0.1-X
X = the amount of HOCl that dissociates
We can ignore this x since it is so small since stronger acid

76. Plug into Ka and solve for X = [H+]:
3.4x10-8 = X2
0.1
X = 5.9 x 10-5 M
Use pH = -log [H+] to solve for pH:
pH = -log 5.9 x 10-5
pH = 4.23

77. Percent Dissociation (aka % ionization)
% dissociation = [ H+]final X 100
[ H+]initial
This equation allows us to identify for the exact concentration of [ H+] that must dissociate for the equation to reach equilibrium.
The stronger the acid the greater the ionization.

78. Example Calculate the % dissociation for 1.0 M HC2H3O2 Ka = 1.8 X 10-5
HC2H3O H+ + C2H3O2 I
C X X X
E X X X

79. Ka = 1.8 X 10-5 = X2/ 1.0
[X] = 4.2 x 10-3
% diss = [H+ ]/ HC2H3O2 * 100
= 4.2 x 10-3/ 1.0
= %

80. Bases
When strong bases are dissolved in aqueous solutions they dissociate 100% in OH-, so we can treat strong base equations like we treat strong acid equations.
Strong Base = Large Kb = high pOH

81. All hydroxides of group 1A and 2A are strong bases.
Exception Be(OH)2
Strong bases will have large Kb values.

82. Weak Bases B (aq) + H2O BH+ (aq) + OH-(aq)
Base Acid Conj. Acid Conj Base
React with water to form conjugate acid of the base and OH- ion.
Small Kb values (low pOH)

83. **** Kb is small and tells us Base x is a weak base
Example
Calculate the pH of 0.05 M solution of Base X, Kb = 1.7 x 10-9
**** Kb is small and tells us Base x is a weak base

84. Answer B (aq) + H2O BH+ (aq) + OH-(aq) Base Acid Conj. Acid Conj Base
( H2O is and acid so we can ignore it since we are given kb) I
C -x x x
E x x x
Plug equilibrium values into Kb = [BH+][OH-]
[ B ]

85. 1.7 x 10-9 = x2 0.05 (Ignore the X) [x] = [OH-] = 9.2 x 10-6
pOH = - log (9.2 x 10-6) = 5.04
pH + pOH = 14
pH = 14
pH = 8.96

86. Lewis Acids Lewis Acid: a substance that accepts an electron pair
Lewis Base: a substance that donates and electron pair.

87. Electron pair acceptor
Keeping it straight
Model
Definition of Acid
Definition of Base
Arrehnius
H+ producer
In water
OH-
Producer
Bronsted- Lowery H+
donor
acceptor
Lewis
Electron pair acceptor
Electron pair donor

88. Lewis Acid Base Example
Brønsted-Lowry acid-base reaction = donation and acceptance of a proton
Lewis base (OH -) = hydroxide ion donates a pair of electrons for covalent bond formation, Lewis acid (H+ ) = accepts the pair of electrons.
Lewis Acid
Lewis Base
Lewis complex

89. Lewis Example For each rxn identify the Lewis acid and base.
H+ + H2O ↔ H3O+

90. Answer
The proton (H+) is the Lewis acid and the water (H2O) is the Lewis base.

91. Note
Note:
Every Bronsted-Lowery base is a Lewis base because they all have a lone pair of e- to accept the protons.
BUT not every Lewis base is a Bronsted-Lowery base because not all LB can accept protons.

92. Chemicals which have no hydrogen to donate (aka the Bronsted-Lowry scheme) can still be acids according to the lewis scheme.
example, BF3 . If we determine Lewis structure of BF3 , we find that B is octet deficient and can accept a lone pair. Thus it can act as a Lewis acid. Thus, when reacting with ammonia, the reaction would look like:

94. Titration Demo https://www.youtube.com/watch?v=3P8FPsbseqM
Titration Curves

95. Label the Diagram

96. Selecting an Indicator

98. Which indicator would you choose to use in the lab?

99. Sample problem
30 mL of 0.10M NaOH neutralized 25.0mL of hydrochloric acid. Determine the concentration of the acid
Write the balanced chemical equation for the reaction          
Calculate moles NaOH (known volume and concentration)
From the balanced chemical equation find the mole ratio  
Find moles HCl    
Calculate concentration of HCl: M = n ÷ V      

100.       NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
Extract the relevant information from
the question:       NaOH      V = 30mL , M = 0.10M   HCl      V = 25.0mL, M = ?
Calculate moles NaOH
NaOH      V = 30mL , M = 0.10M     
3 x 10-3 moles
From the balanced chemical equation find the mole ratio      
Find moles HCl       NaOH: HCl is 1:1       So [NaOH] = [HCl] = 3 x 10-3 moles at the equivalence point
Calculate concentration of HCl: M = n ÷ V       n = 3 x 10-3 mol,       V = 25.0 x 10-3L       0.12 M HCl