Transcript Lecture 12 Notes

Announcements

Homework –
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Chapter 4
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8, 11, 13, 17, 19, 22
Chapter 6
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6, 9, 14, 15
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Exam
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Thursday
4-8
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Meaning of Confidence interval?
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Is an interval around the experimental
mean that most likely contains the true
mean (m).
Homework
4.11
x  0.148
s  0.034
(2.015)(0.034 )
90% confidence: m  0.148 
6
90% confidence: m  0.148  0.028
95% confidence: m  0.148  0.056
Question 4-13.
4-13. A trainee in a medical lab will be released
to work on her own when her results agree
with those of an experienced worker at the
95% confidence interval. Results for a
blood urea nitrogen analysis are shown ….
a)
What does abbreviation dL refer to?
dL = deciliter = 0.1 L = 100 mL
b) Should the trainee work alone?
Comparison of Means with
Student’s t
Is there a significant
difference?
First you must ask,
is there a significant difference in their
standard deviations?
YES
NO
tcalculated 
x1  x2
s pooled
f-test
n1n2
n1  n2
tcalculated 
x1  x2
2
1
2
2
s
s

n1 n2
4-13. dL = deciliter = 0.1 L = 100 mL
x  14.57 mg / dL
s  0.53 mg / dL
n6
x  13.95 mg / dL
s  0.42 mg / dL
n5
2
Fcalc
 0.053 
  1.59
 
 0.042 
Find spooled and t
Ftable = 6.26
No difference
s pooled
0.532 (5)  0.422 (4)

 0.484
652
tcalc 
14.57  13.95
0.484
(6)(5)
 2.12
65
ttable = 2.262
No significant difference between two workers …
Therefore trainee should be “Released”
Homework
4-17. If you measure a quantity four times
and the standard deviation is 1.0 % of the
average, can you be 90 % confident that
the true value is within 1.2% of the
measured average
(2.353)(1.0%)
m x
 x  1.18%
4
Yes
Homework
4-19. Hydrocarbons in the cab of an
automobile … Do the results differ at 95%
CL? 99% CL?
x  31.4  30.0mg / m3
x  52.9  29.8mg / m3
n  32
n  32
2
 30.0 
Fcalc  
  1.013
 29.8 
Find spooled and t
Ftable ~ 1.84
No Difference
Homework
s pooled
30.02 (31)  29.82 (31)

 29.9
32  32  2
tcalc 
52.9  31.4
29.9
(32)(32)
 2.88
32  32
The table gives t for 60 degrees of freedom, which is close to 62.
ttable = 1.671 and 2.000 at the 90 and 95% CL, respectively.
The difference IS significant at both confidence levels.
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4-22. Q-test, Is 216 rejectable?
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192, 216, 202, 195, 204
216  204
Q
 0.50
216  192
Qtable = 0.64
Retain the “outlier” 216
Chapter 6
Chemical
Equilibrium
Chemical Equilibrium
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
Equilibrium Constant
Equilibrium and Thermodynamics
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Enthalpy
Entropy
Free Energy
Le Chatelier’s Principle
Solubility product (Ksp)
Common Ion Effect
Separation by precipitation
Complex formation
Example
The equilibrium constant for the reaction
-14
+
K
=
1.0
x
10
H2O H + OH
w
NH3 + H2O
NH4+ + OH-
KNH3 = 1.8 x 10-5
Find the Equilibrium constant for the following
reaction
NH4+ NH3 + H+
K3 = ?
Equilibrium and
Thermodynamics
A brief review …
Equilibrium and
Thermodynamics
enthalpy => H
enthalpy change => DH
exothermic vs. endothermic
entropy => S
free energy
Gibbs free energy => G
Gibbs free energy change => DG
Equilibrium and
Thermodynamics
DGo = DHo - TDSo
DGo = -RT ln (K)
K = e-(DGo/RT)
Equilibrium and
Thermodynamics

The case of HCl
HCl
H+ + Cl-
K=?
DHo = -74.83 x 103 J/mol
DS0 = -130.4 kJ/mol
DGo = DHo - TDSo
DGo = (-74.83 kJ/mol) – (298.15 K) (-130.4 kJ/mol)
DGo = -35.97 kJ/mol
Equilibrium and
Thermodynamics

The case of HCl
HCl
H+ + Cl-
K=?
DGo = (-74.83 kJ/mol) – (298.15 K) (-130.4 kJ/mol)
DGo = -35.97 kJ/mol
K e
35.97 x103 J / mol

J
[8.314472
]( 298.15 K )
molK
 2.00x10
6
Predicting the direction in which
an equilibrium will initially move
LeChatelier’s Principle and
Reaction Quotient
Le Chatelier's Principle
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
If a stress, such as a change in
concentration, pressure, temperature, etc.,
is applied to a system at equilibrium, the
equilibrium will shift in such a way as to
lessen the effect of the stress.
Stresses –
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Adding or removing reactants or products
Changing system equilibrium temperature
Changing pressure (depends on how the change
is accomplished
Consider
6CO2 (g) + 6 H2O(g)
C6H12O6(s) + 6O2(g)
Equilibrium moves Right
Predict in which direction the equilibrium moves
as a result of the following stress:
Increasing [CO2]
Consider
6CO2 (g) + 6 H2O(g)
C6H12O6(s) + 6O2(g)
Equilibrium moves Left
Predict in which direction the equilibrium moves
as a result of the following stress:
Increasing [O2]
Consider
6CO2 (g) + 6 H2O(g)
C6H12O6(s) + 6O2(g)
Equilibrium moves Left
Predict in which direction the equilibrium moves
as a result of the following stress:
Decreasing [H2O]
Consider
6CO2 (g) + 6 H2O(g)
C6H12O6(s) + 6O2(g)
NO CHANGE
Predict in which direction the equilibrium moves
as a result of the following stress:
Removing C6H12O6(s)
[02 ]6
K1 
6
6
[CO2 ] [ H 2O]
K does not depend on
concentration of solid C6H12O6
Consider
6CO2 (g) + 6 H2O(g)
C6H12O6(s) + 6O2(g)
Equilibrium moves Right
Predict in which direction the equilibrium moves
as a result of the following stress:
Compressing the system
System shifts towards the
direction which occupies the
smallest volume. Fewest
moles of gas.
Consider
6CO2 (g) + 6 H2O(g)
C6H12O6(s) + 6O2(g)
Equilibrium moves Right
DH = + 2816 kJ
Predict in which direction the equilibrium moves
as a result of the following stress:
Increasing system temperature
System is endothermic …
heat must go into the system
(think of it as a reactant)
Consider this
CoCl2 (g)
Co (g) + Cl2(g)
K=2.19 x 10-10
When [COCl2] is 3.5 x 10-3 M, [CO] is 1.1 x 10-5 M, and
[Cl2] is 3.25 x 10-6M is the system at equilibrium?
Q= Reaction quotient
[Co][Cl2 ] [1.1x105 ][3.25x106 ]
8
Q

 1.02x10

3
CoCl2 
3.5 x10


Compare Q and K
Q = 1.02 x 10-8
K = 2.19 x 10-10
System is not at equilibrium, if it were the
ratio would be 2.19x10-10
When
Q>K TOO MUCH PRODUCT TO BE AT EQUILIRBIUM
Equilibrium moves to the left
Q<K TOO MUCH REACTANT TO BE AT EQUILIRBIUM
Equilibrium moves to the Right
Q=K System is at Equilibrium
Solubility Product
Introduction to Ksp
Solubility Product
solubility-product
the product of the solubilities
solubility-product constant => Ksp
constant that is equal to the solubilities
of the ions produced when a substance
dissolves
Solubility Product
In General:
AxBy
<=>
xA+y + yB-x
[A+y]x [B-x]y
K = -----------[AxBy]
[AxBy] K = Ksp = [A+y]x [B-x]y
Solubility Product
For silver sulfate
Ag2SO4 (s) <=>
2 Ag+(aq) + SO4-2(aq)
Ksp = [Ag+]2[SO4-2]
Solubility of a Precipitate
in Pure Water
EXAMPLE: How many grams of AgCl (fw
= 143.32) can be dissolved in 100. mL
of water at 25oC?
AgCl <=> Ag+ + ClKsp = [Ag+][Cl-] = 1.82 X 10-10
(Appen. F)
let x = molar solubility = [Ag+] = [Cl-]
EXAMPLE: How many grams of AgCl (fw
= 143.32) can be dissolved in 100. mL of
water at 25oC?
AgCl(s)
Initial
Change
Equilibrium
Ag+ (aq) + Cl- (aq)
Some
-
-
-x
-x
+x
+x
+x
+x
(x)(x) = Ksp = [Ag+][Cl-] = 1.82 X 10-10
x = 1.35 X 10-5M
EXAMPLE: How many grams of AgCl (fw
= 143.32) can be dissolved in 100. mL of
water at 25oC?
x = 1.35 X 10-5M

How many grams is that in 100 ml?
# grams = (M.W.) (Volume) (Molarity)
= 143.32 g mol-1 (.100 L) (1.35 x 10-5 mol L-1)
= 1.93X10-4 g = 0.193 mg
The Common Ion Effect
The Common Ion Effect
common ion effect
 a salt will be less soluble if one of its
constituent ions is already present in
the solution
The Common Ion Effect
EXAMPLE: Calculate the molar solubility
of Ag2CO3 in a solution that is 0.0200 M
in Na2CO3.
Ag2CO3 <=> 2 Ag+ + CO3-2
Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12
EXAMPLE: Calculate the molar solubility
of Ag2CO3 in a solution that is 0.0200 M in
Na2CO3.
Ag2CO3 <=> 2 Ag+ + CO3-2
Initial
Solid
-
0.0200M
Change
-x
Solid
+2x
+2x
+x
0.0200+x
Equilibrium
Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12
Ksp=(2x)2(0.0200M + x) = 8.1 X 10-12
4x2(0.0200M + x) = 8.1 X 10-12
EXAMPLE: Calculate the molar solubility
of Ag2CO3 in a solution that is 0.0200 M in
Na2CO3.
4x2(0.0200M + x) = 8.1 X 10-12
no exact solution to a 3rd order equation,
need to make some approximation
first, assume the X is very small compared to
0.0200 M
4X2(0.0200M) = 8.1 X 10-12
4X2(0.0200M) = 8.1 X 10-12
X= 1.0 X 10-5
EXAMPLE: Calculate the molar solubility
of Ag2CO3 in a solution that is 0.0200 M
in Na2CO3.
X = 1.0 X 10-5 M
(1.3 X 10-4 M in pure water)
Second check assumption
[CO3-2] = 0.0200 M + X
~ 0.0200 M
0.0200 M + 0.00001M ~ 0.0200M
Assumption is ok!
Separation by
Precipitation
Separation by
Precipitation
Complete separation can mean a lot …
we should define complete.
Complete means that the concentration of
the less soluble material has decreased
to 1 X 10-6M or lower before the more
soluble material begins to precipitate
Separation by
Precipitation
EXAMPLE:
Can Fe+3 and Mg+2 be separated
quantitatively as hydroxides from a solution that is
0.10 M in each cation? If the separation is possible,
what range of OH- concentrations is permissible.
Two competing reactions
Fe(OH)3(s)
Fe3+ + 3OH-
Mg(OH)2(s)
Mg2+ + 2OH-
EXAMPLE: Separate Iron and
Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
Ksp = [Mg+2][OH-]2 = 7.1 X 10-12
Assume quantitative separation requires
that the concentration of the less
soluble material to have decreased to <
1 X 10-6M before the more soluble
material begins to precipitate.
EXAMPLE: Separate Iron and
Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
Ksp = [Mg+2][OH-]2 = 7.1 X 10-12
Assume [Fe+3] = 1.0 X 10-6M
What will be the [OH-] required to reduce the
[Fe+3] to [Fe+3] = 1.0 X 10-6M ?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
EXAMPLE: Separate Iron and
Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
(1.0 X 10-6M)*[OH-]3 = 2 X 10-39
[OH  ]3  2 1033
[OH  ]  1.31011
Add OH-
Mg2+
Mg2+
Fe3+
Fe3+
Fe3+
3+
Fe
2+
Mg2+
Mg
Mg2+
Mg2+
Fe3+
3+
Fe
3+
Mg2+
2+
Fe
Mg
Fe3+
Fe3+
Fe3+
Mg2+
2+
Mg2+ Mg
Fe3+ Fe3+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+ Mg
2+
Mg2+
Mg2+
Fe3+
@ equilibrium
What is the [OH-] ^when
this happens
Is this [OH-] (that is in
solution) great enough
to start precipitating
Mg2+?
Fe(OH)3(s)
EXAMPLE: Separate Iron and
Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
(1.0 X 10-6M)*[OH-]3 = 2 X 10-39
[OH  ]3  2 1033
[OH  ]  1.31011
EXAMPLE: Separate Iron and
Magnesium?
What [OH-] is required to begin the
precipitation of Mg(OH)2?
[Mg+2] = 0.10 M
Ksp = (0.10 M)[OH-]2 = 7.1 X 10-12
[OH-] = 8.4 X 10-6M
EXAMPLE: Separate Iron and
Magnesium?
@ equilibrium
[OH-] to ‘completely’ remove Fe3+
^ -11
= 1.3 X 10 M
[OH-] to start removing Mg2+
= 8.4 X 10-6M
“All” of the Iron will be precipitated b/f any
of the magnesium starts to precipitate!!