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Announcements Homework –
Chapter 4 8, 11, 13, 17, 19, 22
Chapter 6 6, 9, 14, 15
Exam Next Thursday
Comparison of Means with Student’s t
A t test is used to compare one set of measurements with another to decide whether or not they are “The Same” Compare measured result with a
“true” value Comparing two experimental means
Comparison of Means with Student’s t
Experimental Experimental MeanMean Bromothymol Bromothymol BlueBlue
Are the end-pointsAre the end-points different?different?
Experimental MeanExperimental Mean ErthyrosineErthyrosine
First you must ask, First you must ask, is there a significant difference in is there a significant difference in their _____________________?their _____________________?
21
2121
nn
nn
s
xxt
pooledcalculated
NONO YESYES
2
22
1
21
21
ns
ns
xxtcalculated
Comparison of Standard deviations between data
The F-test may be used to provide insights into: Whether there is a difference in the
precision of two methods. (may warrant a new calculation to
compare means! ) Is method A more precise than
method B?
F-test (comparison of std. dev.)
2Indicator
2Indicator
2
1
pooled
pooledcalculated s
sF
We always put the larger standard deviation in the numerator, so that Fcalc>1.If Fcalculated > Ftable then the difference is significant at the 95% CL.
Throwing out “Bad data”
For an analysis of alcohol content in wine
Dr. Skeels finds the following: 12.51,12.56, 12.47, 12.67, and 12.48%
Q-test for Bad Data
Compare to Qcritical
Qcalc > Qcritical can reject
range
gapQcalculated
12.47 12.48 12.51 12.56 12.67
GapGap
RangeRange
20.0
11.0calculatedQ 55.0 64.0tableQ
20.0
11.0calculatedQ 55.0
64.0tableQ
Recommended treatment of Outliers (When good Data goes BAD)
Reexamine carefully all data relating to the outlying result to see if a gross error could have affected its value.
If possible, estimate the precision that can be reasonably expected.
Repeat the analysis if sufficient sample and time are available.
If more data cannot be secured, apply the Q test to the existing set to see if doubtful result should be retained or rejected based on statistical grounds
If the Qtest indicates retention, consider reporting the median instead of the mean.
The minute paper
Please answer each question in 1 or 2 sentences
1) What was the most useful or meaningful thing you learned during this session?
2) What question(s) remain uppermost in your mind as we end this session?
Chapter 6
Chemical Equilibrium
Chemical Equilibrium Equilibrium Constant Equilibrium and Thermodynamics
Enthalpy Entropy Free Energy Le Chatelier’s Principle
Solubility product (Ksp) Common Ion Effect Separation by precipitation Complex formation
Equilibrium Constant
aA + bB + . . . <=> cC + dD + .
. .
Evaluating an Equilibrium Constant
1. The concentrations of solutes should be expressed as moles per liter.
2. The concentrations of gases should be expressed in atmospheres.
3. The concentrations of pure solids, pure liquids, and solvents are omitted because they are unity.
ManipulatingEquilibrium Constants
HA H+ + A-
ManipulatingEquilibrium Constants
H+ + A- HA
Adding Equilibrium Expressions
Consider an acid that donates its proton to another atom.
HA H+ + A-
H+ + C CH+
K1
K2
HA + C C H+ + A- K3 = ?
ManipulatingEquilibrium Constants
HA + C C H+ + A- K3 = ?
Example
The equilibrium constant for the reaction
H2O H+ + OH-
Kw = 1.0 x 10-14
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 => Henthalpy change => H
exothermic vs. endothermicentropy => Sfree energy
Gibbs free energy => GGibbs free energy change => G
Equilibrium and Thermodynamics
Go = Ho - TSo
Go = -RT ln (K)
K = e-(Go/RT)
Equilibrium and Thermodynamics
The case of HCl
HCl H+ + Cl-
Ho = -74.83 x 103 J/molS0 = -130.4 kJ/mol
Go = Ho - TSo
K=?K=?
Equilibrium and Thermodynamics
The case of HCl
HCl H+ + Cl-
Go = (-74.83 kJ/mol) – (298.15 K) (-130.4 kJ/mol)Go = -35.97 kJ/mol
K=?K=?
Predicting the direction in which an equilibrium will initially move
LeChatelier’s Principle and Reaction Quotient
Le Chatelier's Principle
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 – 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)
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)
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)
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)
Predict in which direction the equilibrium moves as a result of the following stress:
Removing C6H12O6(s)
Consider
6CO2 (g) + 6 H2O(g) C6H12O6(s) + 6O2(g)
Predict in which direction the equilibrium moves as a result of the following stress:
Compressing the system
Consider
6CO2 (g) + 6 H2O(g) C6H12O6(s) + 6O2(g)
Predict in which direction the equilibrium moves as a result of the following stress:
Increasing system temperature
H = + 2816 kJ
Consider this
CoCl2 (g) Co (g) + Cl2(g)
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
K=2.19 x 10-10
2
2 ]][[
CoCl
ClCoQ
Compare Q and K
Q = ____________K = 2.19 x 10-10
System is ______________, if it were the ratio would be 2.19x10-10
When
Q>K
Q<K
Q=K
Solubility Product
Introduction to Ksp
Solubility Product
solubility-productthe 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 sulfateAg2SO4 (s) <=> 2 Ag+
(aq) + SO4-
2(aq)
Solubility of a Precipitatein Pure WaterEXAMPLE: How many grams of
AgCl (fw = 143.32) can be dissolved in 100. mL of water at 25oC?AgCl <=> Ag+ + Cl-
Ksp = 1.82 X 10-10 (Appen. F)
EXAMPLE: How many grams of AgCl (fw = 143.32) can be dissolved in 100. mL of water at 25oC?
AgCl(s) Ag+ (aq) + Cl- (aq)
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?
How many grams is that in 100 ml?
x = 1.35 X 10-5M
The Common Ion Effect
