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Homework – Chapter 4 8, 11, 13, 17, 19, 22 Chapter 6 6, 9, 14, 15. Exam Thursday. Announcements. 4-8. Meaning of Confidence interval? Is an interval around the experimental mean that most likely contains the true mean ( m ). Homework. 4.11. Question 4-13.

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

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

  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 …. • 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? First you must ask, is there a significant difference in their standard deviations? f-test YES NO

  6. 4-13. dL = deciliter = 0.1 L = 100 mL Ftable = 6.26 No difference Find spooled and t

  7. ttable = 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? Ftable ~ 1.84 No Difference Find spooled and t

  10. Homework The table gives t for 60 degrees of freedom, which is close to 62. ttable = 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 Qtable = 0.64 Retain the “outlier” 216

  12. Chapter 6 Chemical Equilibrium

  13. 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

  14. NH3 + H2O NH4+ + OH- Find the Equilibrium constant for the following reaction NH4+ NH3 + H+ Example The equilibrium constant for the reaction H2O H+ + OH- Kw = 1.0 x 10-14 KNH3 = 1.8 x 10-5 K3 = ?

  15. Equilibrium and Thermodynamics A brief review …

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

  17. Equilibrium and Thermodynamics DGo =DHo - TDSo DGo= -RT ln (K) K = e-(DGo/RT)

  18. Equilibrium and Thermodynamics • The case of HCl HCl H+ + Cl- K=? DHo = -74.83 x 103 J/mol DS0 = -130.4 kJ/mol DGo =DHo - TDSo DGo = (-74.83 kJ/mol) – (298.15 K) (-130.4 kJ/mol) DGo = -35.97 kJ/mol

  19. Equilibrium and Thermodynamics • The case of HCl HCl H+ + Cl- K=? DGo = (-74.83 kJ/mol) – (298.15 K) (-130.4 kJ/mol) DGo = -35.97 kJ/mol

  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. Equilibrium moves Right 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]

  23. Equilibrium moves Left 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]

  24. Equilibrium moves Left 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]

  25. Consider 6CO2 (g) + 6 H2O(g) C6H12O6(s) + 6O2(g) NO CHANGE Predict in which direction the equilibrium moves as a result of the following stress: Removing C6H12O6(s) K does not depend on concentration of solid C6H12O6

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

  27. Equilibrium moves Right Consider 6CO2 (g) + 6 H2O(g) C6H12O6(s) + 6O2(g) DH = + 2816 kJ 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)

  28. 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

  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 Q=K System is at Equilibrium

  30. Solubility Product Introduction to Ksp

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

  32. 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

  33. Solubility Product For silver sulfate Ag2SO4 (s)<=>2 Ag+(aq) + SO4-2(aq) Ksp = [Ag+]2[SO4-2]

  34. Solubility of a Precipitatein Pure Water EXAMPLE: How many grams of AgCl (fw = 143.32) can be dissolved in 100. mL of water at 25oC? AgCl <=> Ag+ + Cl- Ksp = [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 25oC? AgCl(s) Ag+ (aq) + Cl- (aq) (x)(x) = Ksp = [Ag+][Cl-] = 1.82 X 10-10 x = 1.35 X 10-5M

  36. 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? # 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 10-5M

  37. The Common Ion Effect

  38. The Common Ion Effect 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 Ag2CO3 in a solution that is 0.0200 M in Na2CO3. Ag2CO3<=>2 Ag+ + CO3-2 Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12

  40. EXAMPLE: Calculate the molar solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3. Ag2CO3<=>2 Ag+ + CO3-2 Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12 Ksp=(2x)2(0.0200M + x) = 8.1 X 10-12 4x2(0.0200M + x) = 8.1 X 10-12

  41. EXAMPLE: Calculate the molar solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3. 4x2(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 4X2(0.0200M) = 8.1 X 10-12 4X2(0.0200M) = 8.1 X 10-12 X= 1.0 X 10-5 M

  42. EXAMPLE: Calculate the molar solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3. X = 1.0 X 10-5 M (1.3 X 10-4 M in pure water) Second check assumption [CO3-2] = 0.0200 M + X~0.0200 M 0.0200 M + 0.00001M ~ 0.0200M Assumption is ok!

  43. Separation by Precipitation

  44. Separation by Precipitation Completeseparation can mean a lot … we should define complete. Complete means that the concentration of the less soluble material has decreased to 1 X 10-6M or lower before the more soluble material begins to precipitate

  45. Fe(OH)3(s) Fe3+ + 3OH- Mg(OH)2(s) Mg2+ + 2OH- 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

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

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

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

  49. Add OH- Mg2+ Mg2+ Fe3+ Fe3+ Fe3+ Fe3+ Mg2+ Mg2+ Mg2+ Mg2+ Fe3+ Fe3+ Mg2+ Fe3+ Mg2+ Fe3+ Fe3+ Fe3+ Mg2+ Mg2+ Mg2+ Fe3+ Fe3+

  50. Mg2+ Mg2+ Mg2+ Mg2+ Mg2+ @ equilibrium Mg2+ Fe3+ What is the [OH-] when this happens ^ Mg2+ Mg2+ Is this [OH-] (that is in solution) great enough to start precipitating Mg2+? Mg2+ Mg2+ Mg2+ Fe(OH)3(s)

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