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 M Ca2+ and 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 M Ca2+ and 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 = [Ca2+][SO42-]
CaSO4  Ca2+ + SO42- Ksp= 2.4 x 10-5 = [ ][SO42-]
[SO42-] = M
Sep. is NOT feasible
Ag2SO4  2Ag+ + SO42- Ksp = 1.5 x 10-5
Q>K
Q = [Ag+]2[SO42-]
Q = [0.03]2[ ]
Q = 4.3 x 10-5

4. 6-19.
A solution contains M Ca2+ and 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+] = 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.
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.
SnCl2(aq)
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
Pb2+ + I PbI+
PbI+ + I PbI2
K2 = 1.4 x 101
PbI2 + I PbI3-
K3 =5.9
PbI3+ I PbI42-
K4 = 3.6

10. Effects of Complex Ion Formation on Solubility
Consider the addition of I- to a solution of Pb+2 ions
Pb2+ + I- <=> PbI+
PbI+ + I- <=> PbI2
K2 = 1.4 x 101
Pb2+ + 2I- <=> PbI2
K’ =?
Overall constants are designated with b
This one is b2

11. Effects of Complex Ion Formation on Solubility
Consider the addition of I- to a solution of Pb+2 ions
Pb2+ + I PbI+
PbI+ + I PbI2
K2 = 1.4 x 101
PbI2 + I PbI3-
K3 =5.9
PbI3+ I PbI42-
K4 = 3.6

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
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-
Initial
liquid -
Change -x +x
Equilibrium
Liquid-x
Kw = [H+][OH-] = ?
KW=(X)(X) = ?

23. EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C.
H2O H OH-
Initial
liquid -
Change -x +x
Equilibrium
Liquid-x
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 = [1 x 10-3][OH-]
1 x = [OH-]

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

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

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

29. Relation Between Ka and Kb

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

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

32. Example
Calculate the hydroxide ion concentration in a 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
Fe SCN- D Fe(SCN)2+
Q: What happens to K when we add, say KNO3 ?
A: Nothing should happen based 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 Hg IO Ksp=1.3x10-18 I C E
some
-x +x +2x
some-x +x +2x
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? How else? No
Hg22+ and IO3- do not form complexes with K+ or NO3-.
How else?

39. The Explanation Consider Hg22+ and the IO3- - 2+
Electrostatic attraction - 2+

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

41. The Explanation Consider Hg22+ and the IO3- - 2+ Add KNO3 NO3- K+ NO3-
Electrostatic attraction K+ - K+ 2+
NO3-
NO3-
NO3-
NO3-
NO3- K+ K+
NO3-
Add KNO3

42. The Explanation Consider Hg22+ and the IO3- - 2+
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 (m) = ½ (c1z12+ c2z22 + …)
Calculate the ionic strength of (a) 0.1 M solution of KNO3 and (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

53. Back to our original problem
Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.
Hg2(IO3)2(s) D Hg IO Ksp=1.3x10-18
At low ionic strengths g -> 1

54. Back to our original problem
Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.
Hg2(IO3)2(s) D Hg IO Ksp=1.3x10-18
In 0.1 M KNO3 - how much Hg22+ will be dissolved?

55. Back to our original problem
Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.
Hg2(IO3)2(s) D Hg IO Ksp=1.3x10-18