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

2. 4-8 Meaning of Confidence interval? Is an interval around the experimental mean that most likely contains the true mean (  ).

3. Homework 4.11

4. 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?

5. Comparison of Means with Student’s t Is there a significant difference? difference? First you must ask, is there a significant difference in their standard deviations? NO YES f-test

6. 4-13. dL = deciliter = 0.1 L = 100 mL F table = 6.26No difference Find s pooled and t

7. t table = 2.262 No significant difference between two workers … Therefore trainee should be “Released”

8. 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 Yes

9. Homework 4-19. Hydrocarbons in the cab of an automobile … Do the results differ at 95% CL? 99% CL? F table ~ 1.84 No Difference Find s pooled and t

10. Homework The table gives t for 60 degrees of freedom, which is close to 62. t table = 1.671 and 2.000 at the 90 and 95% CL, respectively. The difference IS significant at both confidence levels.

11. 4-22. Q-test, Is 216 rejectable? 192, 216, 202, 195, 204 Q table = 0.64 Retain the “outlier” 216

12. Chapter 6 Chemical Equilibrium

13. 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

14. 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 = ?

15. Equilibrium and Thermodynamics A brief review …

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

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

18. 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  G o = ( -74.83 kJ/mol) – (298.15 K) (-130.4 kJ/mol)  G o = -35.97 kJ/mol K=?

19. 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=?

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

21. 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

22. 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 ] Equilibrium moves Right

23. 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 ] Equilibrium moves Left

24. 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] Equilibrium moves Left

25. 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) NO CHANGE solid K does not depend on concentration of solid C 6 H 12 O 6

26. 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 System shifts towards the direction which occupies the smallest volume. Fewest moles of gas. Equilibrium moves Right

27. 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 System is endothermic … heat must go into the system (think of it as a reactant) Equilibrium moves Right  H = + 2816 kJ

28. 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

29. 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 System is at Equilibrium Q=K System is at Equilibrium

30. Solubility Product Introduction to K sp

31. 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

32. 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

33. Solubility Product For silver sulfate Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 -2 (aq) K sp = [Ag + ] 2 [SO 4 -2 ]

34. 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 = [Ag + ][Cl - ] = 1.82 X 10 -10 (Appen. F) let x = molar solubility = [Ag + ] = [Cl - ]

35. 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) InitialSome-- Change-x+x Equilibrium-x+x (x)(x) = K sp = [Ag + ][Cl - ] = 1.82 X 10 -10 x = 1.35 X 10 -5 M

36. 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? # 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 x = 1.35 X 10 -5 M

37. The Common Ion Effect

38. common ion effect a salt will be less soluble if one of its constituent ions is already present in the solution

39. The Common Ion Effect EXAMPLE: Calculate the molar solubility of Ag 2 CO 3 in a solution that is 0.0200 M in Na 2 CO 3. Ag 2 CO 3 2 Ag + + CO 3 -2 K sp = [Ag + ] 2 [CO 3 -2 ] = 8.1 X 10 -12

40. EXAMPLE: Calculate the molar solubility of Ag 2 CO 3 in a solution that is 0.0200 M in Na 2 CO 3. Ag 2 CO 3 2 Ag + + CO 3 -2 InitialSolid-0.0200M Change-x+2x+x EquilibriumSolid+2x0.0200+x K sp = (2x) 2 (0.0200M + x) = 8.1 X 10 -12 K sp = [Ag + ] 2 [CO 3 -2 ] = 8.1 X 10 -12 4x 2 (0.0200M + x) = 8.1 X 10 -12

41. EXAMPLE: Calculate the molar solubility of Ag 2 CO 3 in a solution that is 0.0200 M in Na 2 CO 3. 4x 2 (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 4X 2 (0.0200M) = 8.1 X 10 -12 X= 1.0 X 10 -5 M

42. EXAMPLE: Calculate the molar solubility of Ag 2 CO 3 in a solution that is 0.0200 M in Na 2 CO 3. X = 1.0 X 10 -5 M (1.3 X 10 -4 M in pure water) Second check assumption [CO 3 -2 ] = 0.0200 M + X ~ 0.0200 M 0.0200 M + 0.00001M ~ 0.0200M Assumption is ok!

43. Separation by Precipitation

44. 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 -6 M or lower before the more soluble material begins to precipitate

45. 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) Fe 3+ + 3OH - Mg(OH) 2 (s) Mg 2+ + 2OH -

46. EXAMPLE: Separate Iron and Magnesium? K sp = [Fe +3 ][OH - ] 3 = 2 X 10 -39 K sp = [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 -6 M before the more soluble material begins to precipitate.

47. EXAMPLE: Separate Iron and Magnesium? K sp = [Fe +3 ][OH - ] 3 = 2 X 10 -39 K sp = [Mg +2 ][OH - ] 2 = 7.1 X 10 -12 Assume [Fe +3 ] = 1.0 X 10 -6 M What will be the [OH - ] required to reduce the [Fe +3 ] to [Fe +3 ] = 1.0 X 10 -6 M ? K sp = [Fe +3 ][OH - ] 3 = 2 X 10 -39

48. EXAMPLE: Separate Iron and Magnesium? K sp = [Fe +3 ][OH - ] 3 = 2 X 10 -39 (1.0 X 10 -6 M)*[OH - ] 3 = 2 X 10 -39

49. Fe 3+ Mg 2+ Fe 3+ Mg 2+ Fe 3+ Mg 2+ Add OH -

50. Fe 3+ Mg 2+ Fe(OH) 3 (s) What is the [OH - ] when this happens ^ @ equilibrium Is this [OH - ] (that is in solution) great enough to start precipitating Mg 2+?

51. EXAMPLE: Separate Iron and Magnesium? K sp = [Fe +3 ][OH - ] 3 = 2 X 10 -39 (1.0 X 10 -6 M)*[OH - ] 3 = 2 X 10 -39

52. EXAMPLE: Separate Iron and Magnesium? What [OH - ] is required to begin the precipitation of Mg(OH) 2 ? [Mg +2 ] = 0.10 M K sp = (0.10 M)[OH - ] 2 = 7.1 X 10 -12 [OH - ] = 8.4 X 10 -6 M

53. EXAMPLE: Separate Iron and Magnesium? [OH - ] to ‘completely’ remove Fe 3+ = 1.3 X 10 -11 M [OH - ] to start removing Mg 2+ = 8.4 X 10 -6 M “All” of the Iron will be precipitated b/f any of the magnesium starts to precipitate!! ^ @ equilibrium