Chapter 6 Problems

1. Chapter 6 Problems • 6.6, 6.9, 6.15, 6.16, • 6.19, 6.21, 6.24

2. Chapter 6 Chemical Equilibrium

3. Chemical Equilibrium • Equilibrium Constant • Solubility product (Ksp) • Common Ion Effect • Separation by precipitation • Complex formation

4. Solubility Product Introduction to Ksp

5. 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

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

7. 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)

8. 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

9. 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

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

11. The Common Ion Effect Ag2CO32 Ag+ + CO3-2 What is the effect on solubility of Ag2CO3 if I add CO3-2?

12. The Common Ion Effect Add common ion effect • a salt will be less soluble if one of its constituent ions is already present in the solution

13. The Common Ion Effect EXAMPLE: Calculate the molar solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3. Ag2CO32 Ag+ + CO3-2 Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12

14. 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

15. 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

16. 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!

17. Separation by Precipitation

18. 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

19. 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.

20. 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+

21. 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)

22. 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

23. 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.

24. 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-] @ equilibrium required to reduce the [Fe+3] to [Fe+3] = 1.0 X 10-6M ? Ksp = [Fe+3][OH-]3 = 2 X 10-39

25. 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

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

27. EXAMPLE: Separate Iron and Magnesium? @ equilibrium [OH-] to ‘completely’ remove Fe3+ = 1.3 X 10-11 M ^ [OH-] to start removing Mg2+ = 8.4 X 10-6M “All” of the Iron will be precipitated b/f any of the magnesium starts to precipitate!!

28. Complex Ion Formation

29. Complex Formation • Consider Lead Iodide PbI2 (s) Pb2+ + 2I- Ksp= 7.9 x 10-9 What should happen if I- is added to a solution? Should the solubility go up or down?

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

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

32. 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

34. Protic Acids and Bases Section 6-7

35. 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?

36. Protic Acids and Bases - Salts • Consider Ammonium chloride • Can ‘generally be thought of as the product of an acid-base reaction. NH4+Cl- (s) NH4+ + Cl- From general chemistry – single positive and single negative charges are STRONG ELECTROLYTES – they dissolve completely into ions in dilute aqueous solution

37. Protic Acids and Bases Conjugate Acids and Bases in the B-L concept CH3COOH + H2O  CH3COO- + H3O+ acid + base <=> conjugate base + conjugate acid conjugate base => what remains after a B-L acid donates its proton conjugate acid => what is formed when a B-L base accepts a proton

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

40. 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

41. Example • Concentration of OH- if [H+] is 1.0 x 10-3 M @ 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-]

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

43. 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+

44. 6-9 Strengths of Acids and Bases

45. 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

46. 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)

48. 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.

49. 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

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