1. Announcements Homework – Chapter 4 8, 11, 13, 17, 19, 22 Chapter 6 6, 9, 14, 15 Exam Next Thursday

2. 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

3. Comparison of Means with Student’s t Experimental Mean Bromothymol Blue Bromothymol Blue Are the end-points different? different? Experimental Mean Erthyrosine Erthyrosine First you must ask, is there a significant difference in their _____________________? NO YES

4. 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?

5. F-test (comparison of std. dev.) We always put the larger standard deviation in the numerator, so that F calc >1. If F calculated > F table then the difference is significant at the 95% CL.

7. Throwing out “Bad data”

8. For an analysis of alcohol content in wine Dr. Skeels finds the following: 12.51, 12.56, 12.47, 12.67, and 12.48%

9. Q-test for Bad Data Compare to Q critical Q calc > Q critical can reject

10. 12.47 12.48 12.51 12.56 12.67 Gap Range

12. 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 Q test indicates retention, consider reporting the median instead of the mean.

13. 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?

14. Chapter 6 Chemical Equilibrium

15. Equilibrium Constant Equilibrium and Thermodynamics Enthalpy Entropy Free Energy Le Chatelier’s Principle Solubility product (K sp ) Common Ion Effect Separation by precipitation Complex formation

16. Equilibrium Constant aA + bB +... cC + dD +...

17. 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.

18. Manipulating Equilibrium Constants HA H + + A -

19. Manipulating Equilibrium Constants H + + A - HA

20. Adding Equilibrium Expressions Consider an acid that donates its proton to another atom. HA H + + A - H + + C CH + K1K1 K2K2 HA + C  C H + + A - K 3 = ?

21. Manipulating Equilibrium Constants HA + C  C H + + A - K 3 = ?

22. Example The equilibrium constant for the reaction H 2 O H + + OH - K w = 1.0 x 10 -14 NH 3 + H 2 O NH 4 + + OH - K NH3 = 1.8 x 10- 5 Find the Equilibrium constant for the following reaction NH 4 + NH 3 + H + K 3 = ?

23. Equilibrium and Thermodynamics A brief review …

24. Equilibrium and Thermodynamics enthalpy => H enthalpy change =>  H exothermic vs. endothermic entropy => S free energy Gibbs free energy => G Gibbs free energy change =>  G

25. Equilibrium and Thermodynamics  G o =  H o - T  S o  G o = - RT ln (K) K = e -(  Go/RT)

26. Equilibrium and Thermodynamics The case of HCl HCl H + + Cl -  H o = -74.83 x 10 3 J/mol  S 0 = -130.4 kJ/mol  G o =  H o - T  S o K=?

27. Equilibrium and Thermodynamics The case of HCl HCl H + + Cl -  G o = ( -74.83 kJ/mol) – (298.15 K) (-130.4 kJ/mol)  G o = -35.97 kJ/mol K=?

28. Predicting the direction in which an equilibrium will initially move LeChatelier’s Principle and Reaction Quotient

29. 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

30. Consider 6CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (s) + 6O 2 (g) Predict in which direction the equilibrium moves as a result of the following stress: Increasing [CO 2 ]

31. Consider 6CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (s) + 6O 2 (g) Predict in which direction the equilibrium moves as a result of the following stress: Increasing [O 2 ]

32. Consider 6CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (s) + 6O 2 (g) Predict in which direction the equilibrium moves as a result of the following stress: Decreasing [H 2 O]

33. Consider 6CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (s) + 6O 2 (g) Predict in which direction the equilibrium moves as a result of the following stress: Removing C 6 H 12 O 6 (s)

34. Consider 6CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (s) + 6O 2 (g) Predict in which direction the equilibrium moves as a result of the following stress: Compressing the system

35. Consider 6CO 2 (g) + 6 H 2 O(g) C 6 H 12 O 6 (s) + 6O 2 (g) Predict in which direction the equilibrium moves as a result of the following stress: Increasing system temperature  H = + 2816 kJ

36. Consider this CoCl 2 (g) Co (g) + Cl 2(g) When [COCl 2 ] is 3.5 x 10 -3 M, [CO] is 1.1 x 10 -5 M, and [Cl 2 ] is 3.25 x 10 -6 M is the system at equilibrium? Q= Reaction quotient K=2.19 x 10 -10

37. 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

38. Solubility Product Introduction to K sp

39. Solubility Product solubility-product the product of the solubilities solubility-product constant => K sp constant that is equal to the solubilities of the ions produced when a substance dissolves

40. Solubility Product In General: A x B y xA +y + yB -x [A +y ] x [B -x ] y K = ------------ [A x B y ] [A x B y ] K = K sp = [A +y ] x [B -x ] y

41. Solubility Product For silver sulfate Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 -2 (aq)

42. 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 25 o C? AgCl Ag + + Cl - K sp = 1.82 X 10 -10 (Appen. F)

43. EXAMPLE: How many grams of AgCl (fw = 143.32) can be dissolved in 100. mL of water at 25 o C? AgCl(s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] = 1.82 X 10 -10 x = 1.35 X 10 -5 M

44. EXAMPLE: How many grams of AgCl (fw = 143.32) can be dissolved in 100. mL of water at 25 o C? How many grams is that in 100 ml? x = 1.35 X 10 -5 M

45. The Common Ion Effect