Embed Size (px)
Citation preview
Introduction to Chem II
• Instructors
• Course Objectives
• Course Topics
• Laboratory Exercises
• Course Website
• Today’s Agenda
• Syllabus
Course Objectives
• Review some familiar topics
• Investigate some of these topics at a more in-depth level
• Model sound pedagogy
• Obtain hand-on practice with Venier Data Collection
• Show some effective demonstrations
Course Topics
• Stoichiometry
• Calorimetry
• Equilibrium
• Solubility
• Acid-base chemistry
• Redox chemistry
• Thermochemistry
5 Lab Exercises
• A calorimetry experiment using a temperature probe
• Solubility using a Ca ion selective electrode
• Equilibrium constant using a Colorimeter
• Acid-base titration
• Ag Ion Indicator electrode
Course website
http://alpha.chem.umb.edu/chemistry/bpschemII/
Syllabus
Lab experiments
Course notes
Homework solutions
Today’s Agenda
• Take a 2 hr exam• Paperwork, surveys• Lunch• Lecture; g/mol, Classification of reactions,
Stoichiometry, LR, Energetics of Reactions • Lab Lecture; Calorimetry• Lab Experiment 1• Early start on HW
Investigating Stoichiometry using Calorimetry
Experiment 1
Heat of Reaction - H
• At constant pressure – most lab experiments
• aA + bB → products H/mol A
• . H = q (heat produced or absorbed)
Calorimetry
• Method of measuring the heat of reaction
• Calorimeter-coffee cup
• q = cmT– c is the specific heat [J/(g ºC)] of solution – m = mass of solution – .T is change in temperature
• .T is directly proportional to the heat of reaction
The experiment
• Mix reactants in different molar ratios
• Predict the stoichiometry of the reaction from the ratio that gives the maximum temperature increase
Example of the Experiment
• 1 to 1, A + B → products• Mixing molar ratios• Constant total volume - cmT
mmol A mmol B LR A cons fraction ratio T
5.0 20.0 A 5.0 0.25 1 to 4 10
7.5 17.5 A 7.5 0.43 3 to 7 15
10.0 15.0 A 10.0 0.67 2 to 3 20
12.5 12.5 A 12.5 1.00 1 to 1 25
15.0 10.0 B 10.0 1.50 3 to 2 20
17.5 7.5 B 7.5 2.33 7 to 3 15
20.0 5.0 B 5.0 4.00 4 to 1 10
0
510
15
2025
30
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00
Mole Ratio
Tem
p C
han
ge
(C)
1:1 Stoichiometry (mol ratio A/B)
Example 2 • 2 to 1, 2A + B → products
mmol A mmol B LR A cons fraction ratio T
5.0 20.0 A 5.0 0.25 1 to 4 10
7.5 17.5 A 7.5 0.43 3 to 7 15
10.0 15.0 A 10.0 0.67 2 to 3 20
12.5 12.5 A 12.5 1.00 1 to 1 25
15.0 10.0 B 15.0 1.50 3 to 2 30
17.5 7.5 B 15.0 2.33 7 to 3 30
20.0 5.0 B 10.0 4.00 4 to 1 20
16.7 8.4 B 16.7 1.99 2 to 1 32
0
10
20
30
40
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00
Mole Ratio
Tem
p C
han
ge
(C)
2:1 Stoichiometry (mol ratio A/B)
Example 3 • 3 to 1, 3A + B → products
mmol A mmol B LR A cons fraction ratio T
5.0 20.0 A 5.0 0.25 1 to 4 10
7.5 17.5 A 7.5 0.43 3 to 7 15
10.0 15.0 A 10.0 0.67 2 to 3 20
12.5 12.5 A 12.5 1.00 1 to 1 25
15.0 10.0 A 15.0 1.50 3 to 2 30
17.5 7.5 A 17.5 2.33 7 to 3 35
20.0 5.0 B 15.0 4.00 4 to 1 30
18.8 6.3 B 18.8 3.00 3 to 1 36
3:1 Stoichiometry (mol ratio A/B)
0
10
20
30
40
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00
Mole Ratio
Tem
p C
han
ge
(C)
Determining the Hm
• . H = cmT = (4.4 J/gC)*(50 g)*(36)
= 7920 J
• mol A reacted = 18.8 mmol A
• .Hm = H/(mol A reacted)
= (7920)/(.0188 mol) = 421276 J/mol
= 421 kJ/mol
Products
• Thiosulfate is a classic reducing agent
• 2S2O32- ↔ S4O6
2- + 2e-
• Cl- is the product of the reduction of OCl-
• Write a balanced redox equation– Step 1: determine half reactions.– Step 2 Make the reduction half reaction and oxidation
half reaction have the same number of electrons by multiply reactions by common denominator
– Step 3: Add reactions
• OCl- + H2O + 2 e- ↔ Cl- + 2OH-
• 2S2O32- ↔ S4O6
2- + 2e-
• ________________________
• OCl- + H2O + 2S2O32- → Cl- + 2OH- +
S4O62-
Solubility of CaSO4
Experiment 2
Goals
• Determine the solubility of CaSO4 in three different solution– Saturated CaSO4 in H2O
– Saturated CaSO4 in 0.10 M KNO3
– Saturated CaSO4 in 0.10 M Na2SO4
• Compare and rationalize the results
Major concepts
• Solubility Product Constants and saturated solution
• LeChatlier’s principle and the common ion effect
• Effect of ionic strength and ion activities on Ksp
• Ion Selective Electrodes
Ksp of CaSO4
• CaSO4(s) ↔ Ca2+ + SO42-
• Ksp(CaSO4) = [Ca2+][SO42-] = 2.4∙10-5
Saturated solution in water
• Add several grams of CaSO4 to 1 L of water
• Shake and mix for weeks
• Allow CaSO4 that did not dissociate to settle to bottom
• Ksp(CaSO4) = [Ca2+][SO42-] = 2.4∙10-5
= x2
[Ca2+] = 5.0∙10-3 M
Saturated solution in 0.10 M Na2SO4
• Add several grams of CaSO4 to 1 L of 0.10 M Na2SO4
• Common Ion effect
• Ksp(CaSO4) = [Ca2+][SO42-] = 2.4∙10-5
= x(x+0.10)
Assume x <<< 0.10 x = 2.4∙10-4 M
[Ca2+] = 2.4∙10-4 M
Saturated solution in 0.10 M KNO3
• Activities
• Ksp(CaSO4) = ACa2+ASO42- = [Ca2+]Ca2+
[SO42-]SO4
2- = 2.4∙10-5
• Activity coefficient () is dependent on the ionic strength of the solution, and the size and charge of the ion. It is a number between 0 and 1. At very low ionic strength, approaches 1
Ionic strength• A measure of the concentration of ions in
solution
= ½ ∑ cizi2
Sat. solution in 0.10 M KNO3
= ½ ([K+](+1)2 + [NO3-](-1)2 + [Ca2+](2+)2
+ [SO42-](-2)2) = 0.12 M
Ca2+@=0.12 =
Take home message
• The common ion effect decreases the solubility by over an order of magnitude
• At high ionic strengths, solubility increases slightly ( by a factor of 1.5 -5).
Ion Selective Electrode
• A probe that consists of two reference electrodes connected electrically through a specific type of salt bridge through the solution being measured.
• The salt bridge is a membrane that specifically binds the ion of interest
• A junction potential develops at this membrane that is proportional to the concentration of the ion of interest
voltmeterCathode
Reference electrode
Ag/AgCl, sat. KCl
Anode
Reference electrode
Ag/AgCl, sat. KCl
Ion selective
membrane
solution
Engineer this whole set-up in one probe
pH meter
Ca2+ selective electrode
Response of Ca2+ Selective Electrode
• Ecell = constant + 29.58 logACa2+
0
50
100
150
200
-6 -5 -4 -3 -2 -1 0
log A(Ca2+)
Ecell (
mV
)
