1. Chapter 6 Problems • 6.19, 6.21, 6.24 • 6-29, 6-31, 6-39, 6.41 • 6-42, 6-48,

2. 6-19. • A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+? What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate?

3. 6-19. • A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+? What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate? Ca2+ at equilibrium CaSO4 Ca2+ + SO42- Ksp= 2.4 x 10-5 = [0.000500][SO42-] CaSO4 Ca2+ + SO42- Ksp= 2.4 x 10-5 = [Ca2+][SO42-] [SO42-] = 0.048 M Sep. is NOT feasible Q>K Ag2SO4 2Ag+ + SO42- Ksp = 1.5 x 10-5 Q = [Ag+]2[SO42-] Q = [0.03]2[0.000500] Q = 4.3 x 10-5

4. 6-19. • A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+? What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate? Ag2SO4 2Ag+ + SO42- Ksp = 1.5 x 10-5 K/[Ag2+]2 = [SO42-] 1.5 x 10-5/[0.0300]2 = [SO42-] 1.67 x 10-2= [SO42-] Find Ca2+ CaSO4 Ca2+ + SO42- Ksp= 2.4 x 10-5 = [Ca2+][1.67 x 10-2] [Ca2+] = 0.0014 M About 2.8 % remains in solution

5. 6.21 • If a solution containing 0.10 M Cl-, Br-, I- and Cr2O42- is treated with Ag+, in what order will the anions precipitate? AgCl  Ag+ + Cl- Ksp = 1.8 x 10-10=[Ag][Cl] AgBr  Ag+ + Br- Ksp = 5.0 x 10-13=[Ag][Br] AgI  Ag+ + I- Ksp = 8.3 x 10-17=[Ag][I] Ag2CrO4 2Ag+ + CrO4-Ksp = 1.2 x 10-12=[Ag]2[Cl] AgCl  Ag+ + Cl- Ksp = 1.8 x 10-10=[Ag][0.1] AgBr  Ag+ + Br- Ksp = 5.0 x 10-13=[Ag][0.1] AgI  Ag+ + I- Ksp = 8.3 x 10-17=[Ag][0.1] Ag2CrO4 2Ag+ + CrO4-Ksp = 1.2 x 10-12=[Ag]2[0.1] 1.8 x 10-9=[Ag] 5.0 x 10-12=[Ag] 8.3 x 10-16=[Ag] 3.5 x 10-6=[Ag] SOLVE for Ag+ required at equilibrium

6. 6-24. SnCl2(aq) • The cumulative formation constant for SnCl2(aq) in 1.0 M NaNO3 is b2=12. Find the concentration of SnCl2 for a solution in which the concentration of Sn2+ and Cl- are both somehow fixed at 0.20 M. Sn2+ (aq) + 2Cl-(aq) SnCl2 (aq) b2=12 b2=12

8. Complex Formation complex ions (also called coordination ions) Lewis Acids and Bases acid => electron pair acceptor (metal) base => electron pair donor (ligand)

9. Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions

10. Overall constants are designated with b This one is b2 Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions

11. Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions

12. Acids and Bases & Equilibrium Section 6-7

13. Strong Bronsted-Lowry Acid • A strong Bronsted-Lowry Acid is one that donates all of its acidic protons to water molecules in aqueous solution. (Water is base – electron donor or the proton acceptor). • HCl as example

14. Strong Bronsted-Lowry Base • Accepts protons from water molecules to form an amount of hydroxide ion, OH-, equivalent to the amount of base added. • Example: NH2- (the amide ion)

16. Question • Can you think of a salt that when dissolved in water is not an acid nor a base? • Can you think of a salt that when dissolved in water IS an acid or base?

17. Weak Bronsted-Lowry acid • One that DOES not donate all of its acidic protons to water molecules in aqueous solution. • Example? • Use of double arrows! Said to reach equilibrium.

18. Weak Bronsted-Lowry base • Does NOT accept an amount of protons equivalent to the amount of base added, so the hydroxide ion in a weak base solution is not equivalent to the concentration of base added. • example: NH3

19. Common Classes of Weak Acids and Bases Weak Acids • carboxylic acids • ammonium ions Weak Bases • amines • carboxylate anion

20. Equilibrium and Water Question: Calculate the Concentration of H+ and OH- in Pure water at 250C.

21. EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C. H2O H+ + OH- Kw = [H+][OH-] = ? KW=(X)(X) = ?

23. EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C. H2O H+ + OH- Kw = [H+][OH-] = 1.01 X 10-14 KW=(X)(X) = 1.01 X 10-14 (X) = 1.00 X 10-7

24. Example • What is the concentration of OH- in a solution of water that is 1.0 x 10-3 M in [H+] (@ 25 oC)? “From now on, assume the temperature to be 25oC unless otherwise stated.” Kw = [H+][OH-] 1 x 10-14 = [1 x 10-3][OH-] 1 x 10-11 = [OH-]

25. pH ~ -3 -----> ~ +16 pH + pOH = - log Kw = pKw = 14.00

26. Is there such a thing as Pure Water? • In most labs the answer is NO • Why? • A century ago, Kohlrausch and his students found it required to 42 consecutive distillations to reduce the conductivity to a limiting value. CO2 + H2O HCO3- + H+

27. HA + H2O(l) H3O+ + A- Weak Acids and Bases Ka HA H+ + A- Ka’s ARE THE SAME

28. Weak Acids and Bases Kb B + H2O BH+ + OH-

29. Relation Between Ka and Kb

30. Kb NH3 + H2O NH4+ + OH- Ka NH4+ + H2O NH3 + H3O+ H2O + H2O OH- + H3O+ Relation between Ka and Kb • Consider Ammonia and its conjugate acid.

31. Example The Ka for acetic acid is 1.75 x 10-5. Find Kb for its conjugate base. Kw = Ka x Kb

32. Example • Calculate the hydroxide ion concentration in a 0.0100 M sodium hypochlorite solution. OCl- + H2O  HOCl + OH- The acid dissociation constant = 3.0 x 10-8

33. 1st Insurance Problem Challenge on page 120

34. Chapter 8 Activity

35. Write out the equilibrium constant for the following expression Fe3+ + SCN-D Fe(SCN)2+ Q: What happens to K when we add, say KNO3 ? A: Nothing should happenbased on our K, our K is independent of K+ & NO3-

36. K decreases when an inert salt is added!!! Why?

37. 8-1 Effect of Ionic Strength on Solubility of Salts • Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s)D Hg22+ + 2IO3- Ksp=1.3x10-18 some - - -x +x +2x some-x +x +2x I C E A seemingly strange effect is observed when a salt such as KNO3 is added. As more KNO3 is added to the solution, more solid dissolves until [Hg22+] increases to 1.0 x 10-6 M. Why?

38. Increased solubility • Why? • Complex Ion? • No • Hg22+ and IO3- do not form complexes with K+ or NO3-. • How else?

39. The Explanation • Consider Hg22+ and theIO3- Electrostatic attraction - 2+

40. The Explanation • Consider Hg22+ and theIO3- Electrostatic attraction - 2+ Hg2(IO3)2(s) The Precipitate!!

41. The Explanation • Consider Hg22+ and theIO3- NO3- K+ NO3- K+ NO3- NO3- Electrostatic attraction K+ - K+ 2+ NO3- NO3- NO3- NO3- NO3- K+ K+ NO3- Add KNO3

42. The Explanation • Consider Hg22+ and theIO3- NO3- K+ K+ NO3- NO3- NO3- K+ - K+ 2+ NO3- NO3- NO3- NO3- K+ K+ NO3- NO3- Hg22+ and IO3- can’t get CLOSE ENOUGH to form Crystal lattice Or at least it is a lot “Harder” to form crystal lattice

44. The potassium hydrogen tartrate example

45. Alright, what do we mean by Ionic strength? • Ionic strength is dependent on the number of ions in solution and their charge. Ionic strength (m) = ½ (c1z12+ c2z22 + …) Or Ionic strength (m) = ½ S cizi2

46. Examples • Calculate the ionic strength of (a) 0.1 M solution of KNO3and (b) a 0.1 M solution of Na2SO4 (c) a mixture containing 0.1 M KNO3 and 0.1 M Na2SO4. (m) = ½ (c1z12+ c2z22 + …)

47. Alright, that’s great but how does it affect the equilibrium constant? • Activity = Ac = [C]gc • AND

48. Relationship between activity and ionic strength Debye-Huckel Equation m = ionic strength of solution g = activity coefficient Z = Charge on the species x a = effective diameter of ion (nm) 2 comments • What happens to g when m approaches zero? • Most singly charged ions have an effective radius of about 0.3 nm Anyway … we generally don’t need to calculate g – can get it from a table

49. Activity coefficients are related to the hydrated radius of atoms in molecules

50. Relationship between m and g