1. Introduction to Chem II Instructors Course Objectives Course Topics Laboratory Exercises Course Website Today’s Agenda Syllabus

2. 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

3. Course Topics Stoichiometry Calorimetry Equilibrium Solubility Acid-base chemistry Redox chemistry Thermochemistry

4. 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

5. Course website http://alpha.chem.umb.edu/chemistry/bpschemII/ Syllabus Lab experiments Course notes Homework solutions

6. 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

7. Investigating Stoichiometry using Calorimetry Experiment 1

8. Heat of Reaction -  H At constant pressure – most lab experiments aA + bB → products  H/mol A.  H = q (heat produced or absorbed)

9. Calorimetry Method of measuring the heat of reaction Calorimeter-coffee cup q = cm  T –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

10. The experiment Mix reactants in different molar ratios Predict the stoichiometry of the reaction from the ratio that gives the maximum temperature increase

11. Example of the Experiment 1 to 1,A + B → products Mixing molar ratios Constant total volume - cm  T mmol Ammol BLRA consfractionratio TT 5.020.0A5.00.251 to 410 7.517.5A7.50.433 to 715 10.015.0A10.00.672 to 320 12.5 A 1.001 to 125 15.010.0B 1.503 to 220 17.57.5B 2.337 to 315 20.05.0B 4.004 to 110

12. 1:1 Stoichiometry (mol ratio A/B)

13. Example 2 2 to 1,2A + B → products mmol Ammol BLRA consfractionratio TT 5.020.0A5.00.251 to 410 7.517.5A7.50.433 to 715 10.015.0A10.00.672 to 320 12.5 A 1.001 to 125 15.010.0B15.01.503 to 230 17.57.5B15.02.337 to 330 20.05.0B10.04.004 to 120 16.78.4B16.71.992 to 132

14. 2:1 Stoichiometry (mol ratio A/B)

15. Example 3 3 to 1,3A + B → products mmol Ammol BLRA consfractionratio TT 5.020.0A5.00.251 to 410 7.517.5A7.50.433 to 715 10.015.0A10.00.672 to 320 12.5 A 1.001 to 125 15.010.0A15.01.503 to 230 17.57.5A17.52.337 to 335 20.05.0B15.04.004 to 130 18.86.3B18.83.003 to 136

16. 3:1 Stoichiometry (mol ratio A/B)

17. Determining the  H m.  H = cm  T = (4.4 J/gC)*(50 g)*(36) = 7920 J mol A reacted = 18.8 mmol A.  H m =  H/(mol A reacted) = (7920)/(.0188 mol) = 421276 J/mol = 421 kJ/mol

18. Products Thiosulfate is a classic reducing agent 2S 2 O 3 2- ↔ S 4 O 6 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

19. OCl - + H 2 O + 2 e- ↔Cl - + 2OH - 2S 2 O 3 2- ↔ S 4 O 6 2- + 2e- ________________________ OCl - + H 2 O + 2S 2 O 3 2- → Cl - + 2OH - + S 4 O 6 2-

20. Solubility of CaSO 4 Experiment 2

21. Goals Determine the solubility of CaSO 4 in three different solution –Saturated CaSO 4 in H 2 O –Saturated CaSO 4 in 0.10 M KNO 3 –Saturated CaSO 4 in 0.10 M Na 2 SO 4 Compare and rationalize the results

22. 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

23. Ksp of CaSO 4 CaSO 4(s) ↔ Ca 2+ + SO 4 2- Ksp(CaSO 4 ) = [Ca 2+ ][SO 4 2- ] = 2.4∙10 -5

24. Saturated solution in water Add several grams of CaSO 4 to 1 L of water Shake and mix for weeks Allow CaSO 4 that did not dissociate to settle to bottom Ksp(CaSO 4 ) = [Ca 2+ ][SO 4 2- ] = 2.4∙10 -5 = x 2 [Ca 2+ ] = 5.0∙10 -3 M

25. Saturated solution in 0.10 M Na 2 SO 4 Add several grams of CaSO 4 to 1 L of 0.10 M Na 2 SO 4 Common Ion effect Ksp(CaSO 4 ) = [Ca 2+ ][SO 4 2- ] = 2.4∙10 -5 = x(x+0.10) Assume x <<< 0.10x = 2.4∙10 -4 M [Ca 2+ ] = 2.4∙10 -4 M

26. Saturated solution in 0.10 M KNO 3 Activities Ksp(CaSO 4 ) = A Ca2+ A SO42- = [Ca 2+ ]  Ca2+ [SO 4 2- ]  SO 4 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

27. Ionic strength A measure of the concentration of ions in solution  = ½ ∑ c i z i 2 Sat. solution in 0.10 M KNO 3  = ½ ([K + ](+1) 2 + [NO 3 - ](-1) 2 + [Ca 2+ ](2+) 2 + [SO 4 2- ](-2) 2 ) = 0.12 M  Ca2+@  =0.12 =

28. 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).

29. 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

30. voltmeter Cathode 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 Ca 2+ selective electrode

31. Response of Ca 2+ Selective Electrode Ecell = constant + 29.58 log A Ca2+