Chemistry Review
Round 4 (final countdown)
Buffers, Ksp, Thermo, and Electro
1
Chapter 15
Applying equilibrium
2
The Common Ion Effect
When the salt with the anion of a weak
acid is added to that acid,
 It reverses the dissociation of the acid.
 Lowers the percent dissociation of the
acid.
 The same principle applies to salts with
the cation of a weak base..
 The calculations are the same as last
chapter.

3
Buffered solutions
A solution that resists a change in pH.
 Either a weak acid and its salt or a weak
base and its salt.
 We can make a buffer of any pH by
varying the concentrations of these
solutions.
 Same calculations as before.
 Calculate the pH of a solution that is .50
M HAc and .25 M NaAc (Ka = 1.8 x 10-5)

4
Adding a strong acid or base
Do the stoichiometry first. (BAAM)
 A strong base will grab protons from
the weak acid reducing [HA]0
 A strong acid will add its proton to the
anion of the salt reducing [A-]0
 Then do the equilibrium problem.
 What is the pH of 1.0 L of the previous
solution when 0.010 mol of solid NaOH
is added?

5
General equation
Ka = [H+] [A-]
[HA]
 so [H+] = Ka [HA]
[A-]
 The [H+] depends on the ratio [HA]/[A]
 taking the negative log of both sides
 pH = -log(Ka [HA]/[A-])
 pH = -log(Ka)-log([HA]/[A-])
 pH = pKa + log([A-]/[HA])

6
This is called the HendersonHasselbach equation
pH = pKa + log([A-]/[HA])
 pH = pKa + log(base/acid)
 Calculate the pH of the following
mixtures
 0.75 M lactic acid (HC3H5O3) and 0.25
M sodium lactate (Ka = 1.4 x 10-4)
 0.25 M NH3 and 0.40 M NH4Cl
 (Kb = 1.8 x 10-5)

7
Buffer capacity
The pH of a buffered solution is
determined by the ratio [A-]/[HA].
 As long as this doesn’t change much the
pH won’t change much.
 The more concentrated these two are
the more H+ and OH- the solution will
be able to absorb.
 Larger concentrations bigger buffer
capacity.

8
Buffer Capacity
Calculate the change in pH that occurs
when 0.010 mol of HCl(g) is added to
1.0L of each of the following:
 5.00 M HAc and 5.00 M NaAc
 0.050 M HAc and 0.050 M NaAc
 Ka= 1.8x10-5

9
Buffer capacity
The best buffers have a ratio
[A-]/[HA] = 1
 This is most resistant to change
 True when [A-] = [HA]
 Make pH = pKa (since log1=0)

10
Titrations
Millimole (mmol) = 1/1000 mol
 Molarity = mmol/mL = mol/L
 Makes calculations easier because we
will rarely add Liters of solution.
 Adding a solution of known
concentration until the substance being
tested is consumed.
 This is called the equivalence point.
 Graph of pH vs. mL is a titration curve.

11
Strong acid with Strong Base
Do the stoichiometry.
 There is no equilibrium .
 They both dissociate completely.
 The titration of 50.0 mL of 0.200 M HNO3
with 0.100 M NaOH
 Analyze the pH

12
Weak acid with Strong base
There is an equilibrium.
 Do stoichiometry.
 Then do equilibrium.
 Titrate 50.0 mL of 0.10 M HF
(Ka = 7.2 x 10-4) with 0.10 M NaOH

13
Titration Curves
14
Strong acid with strong Base
 Equivalence at pH 7

pH
7
15
mL of Base added
Weak acid with strong Base
 Equivalence at pH >7

pH
>7
16
mL of Base added
Strong base with strong acid
 Equivalence at pH 7

pH
7
17
mL of Base added
Weak base with strong acid
 Equivalence at pH <7

pH
<7
18
mL of Base added
Summary
Strong acid and base just stoichiometry.
 Determine Ka, use for 0 mL base
 Weak acid before equivalence point
–Stoichiometry first
–Then Henderson-Hasselbach
 Weak acid at equivalence point Kb
 Weak base after equivalence - leftover
strong base.

19
Summary
Determine Ka, use for 0 mL acid.
 Weak base before equivalence point.
–Stoichiometry first
–Then Henderson-Hasselbach
 Weak base at equivalence point Ka.
 Weak base after equivalence - leftover
strong acid.

20
Solubility Equilibria
Will it all dissolve, and if not, how
much?
21
All dissolving is an equilibrium.
 If there is not much solid it will all
dissolve.
 As more solid is added the solution will
become saturated.
 Solid
dissolved
 The solid will precipitate as fast as it
dissolves .
 Equilibrium

22
General equation
M+ stands for the cation (usually metal).
 Nm- stands for the anion (a nonmetal).

But the concentration of a solid doesn’t
change.
 Ksp = [M+]a[Nm-]b
 Called the solubility product for each
compound.

23
Watch out
Solubility is not the same as solubility
product.
 Solubility product is an equilibrium
constant.
 it doesn’t change except with
temperature.
 Solubility is an equilibrium position for
how much can dissolve.
 A common ion can change this.

24
Calculating Ksp
The solubility of iron(II) oxalate FeC2O4
is 65.9 mg/L
 The solubility of Li2CO3 is 5.48 g/L

25
Calculating Solubility
The solubility is determined by
equilibrium.
 Its an equilibrium problem.
 Calculate the solubility of SrSO4, with a
Ksp of 3.2 x 10-7 in M and g/L.
 Calculate the solubility of Ag2CrO4,
with a Ksp of 9.0 x 10-12 in M and g/L.

26
Relative solubilities
Ksp will only allow us to compare the
solubility of solids the at fall apart into
the same number of ions.
 The bigger the Ksp of those the more
soluble.
 If they fall apart into different number
of pieces you have to do the math.

27
Common Ion Effect
If we try to dissolve the solid in a
solution with either the cation or anion
already present less will dissolve.
 Calculate the solubility of SrSO4, with a
Ksp of 3.2 x 10-7 in M and g/L in a
solution of 0.010 M Na2SO4.
 Calculate the solubility of SrSO4, with a
Ksp of 3.2 x 10-7 in M and g/L in a
solution of 0.010 M SrNO3.

28
Precipitation
Ion Product, Q =[M+]a[Nm-]b
 If Q>Ksp a precipitate forms.
 If Q<Ksp No precipitate.
 If Q = Ksp equilibrium.
 A solution of 750.0 mL of 4.00 x 10-3M
Ce(NO3)3 is added to 300.0 mL of 2.00 x
10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x
10-10M)precipitate and if so, what is the
concentration of the ions?

29
Selective Precipitations
Used to separate mixtures of metal ions
in solutions.
 Add anions that will only precipitate
certain metals at a time.
 Used to purify mixtures.


30
Often use H2S because in acidic
solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4
will precipitate.
Selective Precipitation
In Basic adding OH-solution S-2 will
increase so more soluble sulfides will
precipitate.
 Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3,
Al(OH)3

31
B
32
D
33
A
34
D
35
E
36
37
38
39
Chapter 6
Energy
Thermodynamics
40
Energy is...
The ability to do work.
 Conserved.
 made of heat and work.
 a state function.
 independent of the path, or how you get
from point A to B.
 Work is a force acting over a distance.
 Heat is energy transferred between
objects because of temperature difference.

41
The universe
is divided into two halves.
 the system and the surroundings.
 The system is the part you are
concerned with.
 The surroundings are the rest.
 Exothermic reactions release energy to
the surroundings.
 Endo thermic reactions absorb energy
from the surroundings.

42
Potential energy
CH 4 + 2O 2  CO 2 + 2H 2 O + Heat
43
CH 4 + 2O 2
Heat
CO 2 + 2 H 2 O
N 2 + O 2 + heat  2NO
Potential energy
2NO
44
Heat
N2 + O2
Direction
Every energy measurement has three
parts.
1. A unit ( Joules of calories).
2. A number how many.
3. and a sign to tell direction.
 negative - exothermic
 positive- endothermic

45
Surroundings
System
Energy
DE <0
46
Surroundings
System
Energy
DE >0
47
Same rules for heat and work
Heat given off is negative.
 Heat absorbed is positive.
 Work done by system on surroundings
is positive.
 Work done on system by surroundings
is negative.
 Thermodynamics- The study of energy
and the changes it undergoes.

48
First Law of Thermodynamics
The energy of the universe is constant.
 Law of conservation of energy.
 q = heat
 w = work
 DE = q + w
 Take the systems point of view to
decide signs.

49
What is work?
Work is a force acting over a distance.
 w= F x Dd
 P = F/ area
 d = V/area
 w= (P x area) x D (V/area)= PDV
 Work can be calculated by multiplying
pressure by the change in volume at
constant pressure.
 units of liter - atm L-atm

50
Work needs a sign
If the volume of a gas increases, the
system has done work on the
surroundings.
 work is negative
 w = - PDV
 Expanding work is negative.
 Contracting, surroundings do work on
the system w is positive.
 1 L atm = 101.3 J

51
Examples
What amount of work is done when 15
L of gas is expanded to 25 L at 2.4 atm
pressure?
 If 2.36 J of heat are absorbed by the gas
above. what is the change in energy?
 How much heat would it take to change
the gas without changing the internal
energy of the gas?

52
Enthalpy
abbreviated H
 H = E + PV (that’s the definition)
 at constant pressure.
 DH = DE + PDV


the heat at constant pressure qp can be
calculated from
53

DE = qp + w = qp - PDV

qp = DE + P DV = DH
Calorimetry
Measuring heat.
 Use a calorimeter.
 Two kinds
 Constant pressure calorimeter (called a
coffee cup calorimeter)
 heat capacity for a material, C is
calculated
 C= heat absorbed/ DT = DH/ DT
 specific heat capacity = C/mass

54
Calorimetry
molar heat capacity = C/moles
 heat = specific heat x m x DT
 heat = molar heat x moles x DT
 Make the units work and you’ve done
the problem right.
 A coffee cup calorimeter measures DH.
 An insulated cup, full of water.
 The specific heat of water is 1 cal/gºC
 Heat of reaction= DH = sh x mass x DT

55
Examples
The specific heat of graphite is 0.71
J/gºC. Calculate the energy needed to
raise the temperature of 75 kg of
graphite from 294 K to 348 K.
 A 46.2 g sample of copper is heated to
95.4ºC and then placed in a calorimeter
containing 75.0 g of water at 19.6ºC. The
final temperature of both the water and
the copper is 21.8ºC. What is the specific
heat of copper?

56
Properties
intensive properties not related to the
amount of substance.
 density, specific heat, temperature.
 Extensive property - does depend on
the amount of stuff.
 Heat capacity, mass, heat from a
reaction.

57
Hess’s Law
Enthalpy is a state function.
 It is independent of the path.
 We can add equations to to come up
with the desired final product, and add
the DH
 Two rules
 If the reaction is reversed the sign of DH
is changed
 If the reaction is multiplied, so is DH

58
H (kJ)
O2 NO2
-112 kJ
180 kJ
N2 2O2
59
NO2
68 kJ
Standard Enthalpy
The enthalpy change for a reaction at
standard conditions (25ºC, 1 atm , 1 M
solutions)
 Symbol DHº
 When using Hess’s Law, work by
adding the equations up to make it look
like the answer.
 The other parts will cancel out.

60
Example
Given
5
C 2 H 2 (g) + O 2 (g)  2CO 2 (g) + H 2 O( l)
2
DHº= -1300. kJ
C(s) + O 2 (g)  CO 2 (g)
DHº= -394 kJ
1
H 2 (g) + O 2 (g)  H 2 O(l)
2
DHº= -286 kJ

calculate DHº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
61
Example
Given
O 2 (g) + H 2 (g)  2OH(g) DHº= +77.9kJ
O 2 (g)  2O(g) DHº= +495 kJ
H 2 (g)  2H(g) DHº= +435.9kJ
Calculate DHº for this reaction
O(g) + H(g)  OH(g)
62
Standard Enthalpies of Formation
Hess’s Law is much more useful if you
know lots of reactions.
 Made a table of standard heats of
formation. The amount of heat needed
to for 1 mole of a compound from its
elements in their standard states.
 Standard states are 1 atm, 1M and 25ºC
 For an element it is 0
 There is a table in Appendix 4 (pg A22)

63
Standard Enthalpies of Formation
Need to be able to write the equations.
 What is the equation for the formation
of NO2 ?
 ½N2 (g) + O2 (g)  NO2 (g)
 Have to make one mole to meet the
definition.
 Write the equation for the formation of
methanol CH3OH.

64
Since we can manipulate the
equations
We can use heats of formation to figure
out the heat of reaction.
 Lets do it with this equation.
 C2H5OH +3O2(g)  2CO2 + 3H2O
 which leads us to this rule.

65
Since we can manipulate the
equations
We can use heats of formation to figure
out the heat of reaction.
 Lets do it with this equation.
 C2H5OH +3O2(g)  2CO2 + 3H2O
 which leads us to this rule.

( DH of products) - ( DH of reactants) = DH o
66
Chapter 16
Spontaneity, entropy and free
energy
67
Spontaneous
A reaction that will occur without
outside intervention.
 We can’t determine how fast.
 We need both thermodynamics and
kinetics to describe a reaction
completely.
 Thermodynamics compares initial and
final states.
 Kinetics describes pathway between.

68
Thermodynamics
1st Law- the energy of the universe is
constant.
 Keeps track of thermodynamics doesn’t
correctly predict spontaneity.
 Entropy (S) is disorder or randomness
 2nd Law the entropy of the universe
increases.

69
Entropy
Defined in terms of probability.
 Substances take the arrangement that is
most likely.
 The most likely is the most random.
 Calculate the number of arrangements
for a system.

70
2 possible
arrangements
 50 % chance of
finding the left
empty

71
4
possible
arrangements
 25% chance of
finding the left
empty
 50 % chance of
them being
evenly
dispersed
72
4
possible
arrangements
 8% chance of
finding the left
empty
 50 % chance of
them being
evenly
dispersed
73
Gases

Gases completely fill their chamber
because there are many more ways to
do that than to leave half empty.
Ssolid <Sliquid <<Sgas
there are many more ways for the
molecules to be arranged as a liquid
than a solid.
 Gases have a huge number of positions
possible.

74
Gibb's Free Energy
G=H-TS
 Never used this way.
 DG=DH-TDS at constant temperature
 Divide by -T
 -DG/T = -DH/T-DS
 -DG/T = DSsurr + DS
 -DG/T = DSuniv
 If DG is negative at constant T and P,
the Process is spontaneous.

75
Let’s Check
For the reaction H2O(s)  H2O(l)
 DSº = 22.1 J/K mol DHº =6030 J/mol
 Calculate DG at 10ºC and -10ºC
 Look at the equation DG=DH-TDS
 Spontaneity can be predicted from the
sign of DH and DS.

76
Spontaneous?
DS
DH
+
-
At all Temperatures
+
At high temperatures,
“entropy driven”
-
At low temperatures,
“enthalpy driven”
+
Not at any temperature,
Reverse is spontaneous
+
77
DG=DH-TDS
Third Law of Thermo
The entropy of a pure crystal at 0 K is 0.
 Gives us a starting point.
 All others must be>0.
 Standard Entropies Sº ( at 298 K and 1
atm) of substances are listed.
 Products - reactants to find DSº (a state
function).
 More complex molecules higher Sº.

78
Free Energy in Reactions
DGº = standard free energy change.
 Free energy change that will occur if
reactants in their standard state turn to
products in their standard state.
 Can’t be measured directly, can be
calculated from other measurements.
 DGº=DHº-TDSº
 Use Hess’s Law with known reactions.

79
Free Energy in Reactions
There are tables of DGºf .
 Products-reactants because it is a state
function.
 The standard free energy of formation
for any element in its standard state is
0.
 Remember- Spontaneity tells us
nothing about rate.

80
Free energy and Pressure
DG = DGº +RTln(Q) where Q is the
reaction quotients (P of the products /P
of the reactants).
 CO(g) + 2H2(g)  CH3OH(l)
 Would the reaction be spontaneous at
25ºC with the H2 pressure of 5.0 atm and
the CO pressure of 3.0 atm?
 DGºf CH3OH(l) = -166 kJ
 DGºf CO(g) = -137 kJ
DGºf H2(g) = 0 kJ

81
How far?
DG tells us spontaneity at current
conditions. When will it stop?
 It will go to the lowest possible free
energy which may be an equilibrium.
 At equilibrium DG = 0, Q = K
 DGº = -RTlnK

82
DGº
=0
<0
>0
83
K
=1
>0
<0
Temperature dependence of K
DGº= -RTlnK = DHº - TDSº
 ln(K) = DHº/R(1/T)+ DSº/R
 A straight line of lnK vs 1/T

84
Free energy And Work
Free energy is that energy free to do
work.
 The maximum amount of work
possible at a given temperature and
pressure.
 Never really achieved because some of
the free energy is changed to heat
during a change, so it can’t be used to
do work.

85
C
86
87
E
D
88
E
89
D
90
91
92
Electrochemistry
Applications of Redox
93
Review
Oxidation reduction reactions involve a
transfer of electrons.
 OIL- RIG
 Oxidation Involves Gain
 Reduction Involves Loss
 LEO-GER
 Lose Electrons Oxidation
 Gain Electrons Reduction

94
Applications


Moving electrons is electric current.
8H++MnO4-+ 5Fe+2 +5e-  Mn+2 + 5Fe+3 +4H2O
Helps to break the reactions into half
reactions.
+
+2
 8H +MnO4 +5e  Mn +4H2O
 5(Fe+2  Fe+3 + e- )
 In the same mixture it happens without
doing useful work, but if separate

95
Connected this way the reaction starts
 Stops immediately because charge builds
up.

H+
MnO4-
96
Fe+2
Galvanic Cell
Salt
Bridge
allows
current
to flow
H+
MnO4-
97
Fe+2
Electricity travels in a complete circuit
 Instead of a salt bridge

H+
MnO4-
98
Fe+2
Porous
Disk
H+
MnO4-
99
Fe+2
e-
e-
e-
e-
Anode
e-
oxidation
100
Cathode
e-
reduction
Cell Potential


The push or pull (“driving force”) is
called the cell potential Ecell
 Also called the electromotive force
(emf)
 Unit is the volt(V)
 = 1 joule of work/coulomb of charge
 Measured with a voltmeter

101
Species being reduced pushes the electron.
Species being oxidized pulls the electron.
0.76
H2 in
Anode
Zn+2
SO4-2
1 M ZnSO4
102
Cathode
H+
Cl1 M HCl
Standard Hydrogen Electrode

This is the reference
all other oxidations
are compared to
 Eº = 0

103
º indicates standard H+
states of 25ºC,
1 Clatm, 1 M solutions.
1 M HCl
H2 in
Cell Potential
Zn(s) + Cu+2 (aq)  Zn+2(aq) + Cu(s)
 The total cell potential is the sum of the
potential at each electrode.


Eº cell = EºZn Zn+2 + Eº Cu+2 Cu
We can look up reduction potentials in
a table.
 One of the reactions must be reversed,
so change it sign.

104
Cell Potential
Determine the cell potential for a
galvanic cell based on the redox
reaction.
 Cu(s) + Fe+3(aq)  Cu+2(aq) + Fe+2(aq)

Eº = 0.77 V
 Cu+2(aq)+2e- Cu(s)
Eº = 0.34 V
+2
 Cu(s) Cu (aq)+2e
Eº = -0.34 V
 2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V

105
Fe+3(aq) + e- Fe+2(aq)
Line Notation


solidAqueousAqueoussolid
Anode on the leftCathode on the right
Single line different phases.
 Double line porous disk or salt bridge.
 If all the substances on one side are
aqueous, a platinum electrode is
indicated.
 For the last reaction
 Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)

106
Galvanic Cell


1)
2)
3)
4)
107
The reaction always runs
spontaneously in the direction that
produced a positive cell potential.
Four things for a complete description.
Cell Potential
Direction of flow
Designation of anode and cathode
Nature of all the componentselectrodes and ions
Potential, Work and DG
DGº = -nFE º
 if E º < 0, then DGº > 0 spontaneous
 if E º > 0, then DGº < 0 nonspontaneous
 In fact, reverse is spontaneous.

Calculate DGº for the following reaction:
 Cu+2(aq)+ Fe(s) Cu(s)+ Fe+2(aq)

108

Fe+2(aq) + e-Fe(s)

Cu+2(aq)+2e- Cu(s)
Eº = 0.44 V
Eº = 0.34 V
Cell Potential and
Concentration
Qualitatively - Can predict direction of
change in E from LeChâtelier.
 2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)

Predict if Ecell will be greater or less than
Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 M
 if [Al+3] = 1.0 M and [Mn+2] = 1.5M
 if [Al+3] = 1.5 M and [Mn+2] = 1.5 M
109

The Nernst Equation
DG = DGº +RTln(Q)
 -nFE = -nFEº + RTln(Q)
 E = Eº - RTln(Q)
nF
 2Al(s) + 3Mn+2(aq)  2Al+3(aq) +
3Mn(s)
Eº = 0.48 V

110

Always have to figure out n by balancing.

If concentration can gives voltage, then
from voltage we can tell concentration.
The Nernst Equation

As reactions proceed concentrations of
products increase and reactants decrease.
Reach equilibrium where Q = K and
Ecell = 0
 0 = Eº - RTln(K)
nF
 Eº = RTln(K)
nF
 nFEº
= ln(K)
RT
111

Corrosion
Rusting - spontaneous oxidation.
 Most structural metals have reduction
potentials that are less positive than O2
.
+2
 Fe  Fe +2e
Eº= 0.44 V
 O2 + 2H2O + 4e- 4OHEº= 0.40 V
 Fe+2 + O2 + H2O Fe2 O3 + H+
 Reaction happens in two places.

112
Salt speeds up process by increasing
conductivity
Water
Rust
e-
Iron Dissolves- Fe  Fe+2
113
Preventing Corrosion
Coating to keep out air and water.
 Galvanizing - Putting on a zinc coat
 Has a lower reduction potential, so it is
more. easily oxidized.
 Alloying with metals that form oxide
coats.
 Cathodic Protection - Attaching large
pieces of an active metal like
magnesium that get oxidized instead.

114
Electrolysis
Running a galvanic cell backwards.
 Put a voltage bigger than the potential
and reverse the direction of the redox
reaction.
 Used for electroplating.

115
1.10
e-
e-
Zn
116
1.0 M
+2
Zn
Anode
1.0 M
Cu+2
Cathode
Cu
e-
e-
A battery
>1.10V
Zn
117
1.0 M
+2
Zn
Cathode
1.0 M
Cu+2
Anode
Cu
Calculating plating
Have to count charge.
 Measure current I (in amperes)
 1 amp = 1 coulomb of charge per
second
q=Ixt
 q/nF = moles of metal
 Mass of plated metal
 How long must 5.00 amp current be
applied to produce 15.5 g of Ag from
118 Ag+

Other uses
Electroysis of water.
 Seperating mixtures of ions.
 More positive reduction potential
means the reaction proceeds forward.
 We want the reverse.
 Most negative reduction potential is
easiest to plate out of solution.

119
B
120
A
121
122
123
124