1. Ch 15 Part III: Solubility, Ksp, and Precipitation

2. Topics
Day 1: Solubility Equilibria and the Solubility Product Day 2: Relative Solubilities; Common Ions Day 3: Precipitation

3. Solubility Equilibria
In introductory chemistry classes, any “slightly soluble” or “sparingly soluble” substance is classified as “insoluble”.
How is “soluble vs. insoluble” determined?
If 3.0 g (or more) of salt dissolve in 100 g of water, it is considered soluble.
What if 2.9 g of salt dissolve?
According to the criteria, it would be considered insoluble.
In reality, very small amounts of these so called “insoluble salts” do dissolve to establish equilibrium.

4. When equilibrium is reached, no more solid dissolves and the solution is said to be saturated.
Ksp respresents the maximum concentration of ions that can be in solution. Any more and the reverse reaction dominates, and the solid precipitates out of the solution.

5. In a saturated solution, [products] is independent of volume
In a saturated solution, [products] is independent of volume. If some of the solution the solution evaporates, precipitation occurs until Ksp is reestablished. This means [products] is the same as before.

6. CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
Example
CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
Ksp = [Ca2+] [F-]2
Remember, pure solids are not included in the equilibrium expression.

7. Solubility vs. Solubility Product
Solubility Product (Ksp):
Equilibrium constant; no units
One value for a given solid at a given temperature

8. Solubility vs. Solubility Product
Molar Solubility (x):
The amount of salt that dissolves to reach equilibrium
Units are mol/L (M)
Also known as “solubility”
It is an equilibrium position (left vs. right) which indicates if products or reactants are favored

9. Solubility vs. Solubility Product
The amount of exposed surface area, stirring, or grinding affects the speed of the reaction but does not change molar solubility (x)

10. Solubility vs. Solubility Product
Before you go on, make sure you understand the difference between solubility and solubility product (Ksp):
The solubility is an equilibrium position and may be affected by the presence of a common ion.
The solubility product is an equilibrium constant and has only one value for a given solid at a given temperature.

11. Calculating Ksp from Solubility
You will now work two examples where you are given molar solubility (x) and solve for Ksp.

12. Example 1
Copper(I) bromide (CuBr) has a measured solubility of 2.0x10-4 mol/L at 25C. Calculate its Ksp value. R CuBr(s) ⇌ Cu+(aq) + Br- (aq) I C E
+ 2.0x10-4
+ 2.0x10-4
2.0x10-4
2.0x10-4
Ksp = [Cu+] [Br-]
= [2.0x10-4] [2.0x10-4]
= 4.0x10-8

13. Example 2
Calculate Ksp for bismuth sulfide (Bi2S3) which has a solubility of 1.0x10-15 mol/L at 25°C. Bi2S3(s) ⇌ 2Bi3+(aq) + 3S2-(aq)
The solubility represents x, the amount of solid that dissolves. So you will have 2x for Bi+3 and 3x for S-2. 2x 3x
+ 2(1.0x10-15)
+ 3(1.0x10-15)
2.0x10-15
3.0x10-15
Ksp = [Bi3+]2 [S2-]3
= [2.0x10-15]2 [3.0x10-15]3
= 1.1x10-73

14. Calculating Solubility from Ksp
Now, you will work an example where you are given Ksp and solve for molar solubility (x).

15. Example
The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4x10-7 at 25°C. Calculate its solubility at 25°C. Cu(IO3)2(s) ⇌ Cu2+(aq) + 2IO3-(aq)
+ x
+ 2x x 2x
Ksp = [Cu2+][IO3-]2
= [x][2x]2
1.4x10-7 = 4x3
On calculator:
Math 4: cube root
2nd Answer
x3 = 3.5x10-8
x = 3.3x10-3 mol/L

16. Calculating Concentration
Finally, you will use molar solubility to determine the concentration of an ion in solution.

17. Example
The molar solubility of silver phosphate, Ag3PO4, at 25 oC is 1.55 x Determine the concentration of the ions in a saturated solution.

18. Ag3PO4 (s) ↔ 3Ag+ (aq) + PO43- (aq) 3x x [Ag+] = 3x = 3(1
Ag3PO4 (s) ↔ 3Ag+ (aq) + PO43- (aq) 3x x [Ag+] = 3x = 3(1.55 x 10-5) = 4.65 x 10-5 [PO43-] = 1.55 x 10-5 M

19. b) 1.00 L of this solution is allowed to evaporate to a final volume of 500. mL. What is the [PO43-] in the solution? Justify your answer.

20. [PO43-] = x 10-5 M
The [PO43-] in a saturated solution of Ag3PO4
is independent of the volume of the solution.
The [PO43-] before evaporation is the same as the [PO43-] after evaporation!

21. and do today’s homework

22. Relative Solubilities
A salt’s Ksp gives us information about its solubility. However, we must be careful in using Ksp to predict the relative solubilities of a group of salts.
There are two possible cases:
The salts being compared dissociate to form the same number of ions.
The salts being compared dissociate to form different numbers of ions.

23. Relative Solubilities
If the salts being compared produce the same number of ions, then the larger the Ksp the more soluble the salt
If the salts being compared produce different numbers of ions, then the Ksp values cannot be compared directly. The molar solubilities (x) of each salt must be calculated and then compared

24. Example 1
Rank the following salts in order of decreasing solubility AgI(s) Ksp = 1.5 x CuI(s) Ksp = 5.0 x CaSO4(s) Ksp = 6.1 x 10-5 CaSO4(s) > CuI(s) > AgI(s)
How many moles of ions does each salt produce?
1 mole salt → 2 moles ions

25. Rank the following salts in order of increasing solubility:
CuS(s) Ksp = 8.5 x 10-45
Ag2S(s) Ksp = 1.6 x 10-49
Bi2S3(s) Ksp = 1.1 x 10-73
Stop and calculate the molar solubility of each of these salts.
How many moles of ions does each salt produce?
2, 3, and 5 moles of ions

26. CuS(s) Ksp = 8.5 x 10-45 CuS (s) ↔ Cu2+ (aq) + S2- (aq) x x
x = 9.2 x mol/L

27. Ag2S(s) Ksp = 1.6 x 10-49 Ag2S (s) ↔ 2Ag+ (aq) + S2- (aq) 2x x
Ksp = (2x)2 (x)
1.6 x = 4x3
x3 = 4 x 10-50
x = 3.4 x mol/L
MATH 4: 3√
2ND Answer

28. Bi2S3(s) Ksp = 1.1 x 10-73 Bi2S3 (s) ↔ 2Bi3+ (aq) + 3S3- (aq) 2x 3x
1.1 x = (2x)2 (3x)3
= (4x2)(27x3)
= 108x5
x5 = x 10-75
x = 1.0 x mol/L
5 MATH
5: x√
2ND Answer

29. Now, rank the salts in order of decreasing solubility
Now, rank the salts in order of decreasing solubility. Bi2S3(s) solubility (x) = 1.0 x mol/L Ag2S(s) solubility (x) = 3.4 x mol/L CuS(s) solubility (x) = 9.2 x mol/L

30. Answer Check:
Bi2S3(s) > Ag2S(s) > CuS(s)

31. Common Ion Effect
The common ion effect is exhibited when the dissolving solution and dissolving salt have a common ion.
When a salt is dissolved in water containing a common ion, its solubility is decreased.
This is another way of saying that when a common ion is added, percent dissociation of the salt decreases.

32. Example 1
Compare the solubility of Ag2CrO4 (Ksp=9.0x10-12) dissolved in pure water to its solubility in M AgNO3 In pure water: Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42-(aq)
Initial
Change
Equilibrium
Ksp = [Ag+]2 [CrO42-]
= [2x]2[x]
+ 2x
+ x
9.0 x = 4x3 2x x
x = 1.3 x 10-4 mol/L

33. In 0.100 M AgNO3 Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42-(aq) Initial Change
Change
+2x +x
Equilibrium x x
Ksp = [Ag+]2 [CrO42-]
9.0x10-12= (0.100)2(x)
x = 9.0x10-10 mol/L
assume x ≈ 0.100
Note that Ag2CrO4 is much less soluble in the presence of Ag+ ions from AgNO3.

34. Example 2
Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in a M NaF solution. CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
Initial
0.025
Change +x
+2x
Equilibrium x x
assume
x ≈ 0.025
Ksp = [Ca2+][F-]2
4.0x10-11 = [x][0.025]2
x = 6.4x10-8 mol/L

35. pH and Solubility
The pH of a solution can greatly affect a salt’s solubility.

36. Example
Why does an increase in pH decrease the solubility of magnesium hydroxide?
Mg(OH)2 ⇌ Mg OH-

37. Mg(OH)2 ⇌ Mg OH-
Increasing [OH-] will drive the reaction left which decreases solubility of the salt because the [products] decreases while the [reactants] increases
The reverse reaction is favored and Mg(OH)2 will be reformed until equilibrium is re-established

38. Why does decreasing the pH increase the solubility of magnesium hydroxide? Mg(OH)2 ⇌ Mg2+ + 2 OH-

39. Mg(OH)2 ⇌ Mg OH-
The addition of H+ will remove OH- from solution because they will react to form water. Removal of the OH- ions will cause the forward reaction to run which will increase the solubility of Mg(OH)2.

40. Example
What is the effect of decreasing pH for salts that contain the conjugate base of a weak acid?
Ag3PO4 ⇌ 3 Ag+ + PO43-
The solubility of these salts will increase.
Decreasing pH will cause the reaction to shift right because the additional H+ will react to completion with the best available base, PO43-, to form HPO4 2-
This will decrease the [PO4-3] in solution which favors the formation of additional products until equilibrium is re-established

41. General Rule General Rule:
If the anion is an effective base (OH- or the conjugate base of a weak acid), decreasing the pH by making the solution more acidic will increase the solubility of the salt.

42. Anions That Are Effective Bases
OH- (will always react to completion with the best available acid)
F-, CN-, S2-, CO32-, C2O42-, and CrO42- are a few common examples
Salts containing these anions are much more soluble in acidic solution than in pure water

43. Some Neutral Anions
Cl-, Br-, I-, NO3-, ClO4- (these are the weak conjugate bases of strong acids)
The solubility of salts containing these anions are unaffected by changes in pH

44. Example What is the effect of decreasing pH of AgCl?
AgCl will have the same molar solubility in an acidic solution as it does in water because Cl- is the weak conjugate base of the strong acid HCl. It will not react with H+ in solution so solubility is unaffected.

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46. Precipitation
We have been studying solids dissolving in solution. Now, we will study the reverse- the formation of a solid from a solution.

47. Ion Solubility Product Quotient (Qsp)
Qsp has the same form as Ksp except that it is calculated using initial concentrations instead of equilibrium concentrations. CaF2(s) ⇌ Ca2+(aq) + 2F-(aq) Qsp = [Ca2+][F-]2

48. Ksp vs. Qsp
If Ksp < Qsp, precipitation will occur and reverse reaction will be favored until Ksp is achieved. If Ksp ˃ Qsp, no precipitate forms.

49. Predicting if Precipitation will Occur when Two Solutions are Mixed

50. Example
A solution is prepared by adding mL of 4.00 x 10-3M Ce(NO3)3 to mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 (Ksp=1.9 x 10-10) precipitate from this solution?
Determine if Ce(IO3)3 will precipitate when the solutions are mixed.
a. Calculate the initial concentrations of Ce3+ and IO3-
[Ce3+]0= 750.0mL (4.00x10-3mmol/mL)
(750.0mL mL)
= 2.857x10-3M
[IO3-]0= 300.0mL (2.00x10-2mmol/mL)
(750.0mL mL)
= 5.714x10-3M

51. b. Calculate Qsp for Ce(IO3) c. Compare Ksp and Qsp
Qsp = [Ce3+]0[IO3-]30
= [2.857x10-3][5.714x10-3]3
Qsp = 5.33 x 10-10
Ksp vs Qsp
1.9x < x10-10
Ce(IO3)3 will precipitate from the solution

52. Calculating Equilibrium Concentrations in Solution after Precipitation has Occurred

53. Example 1
Calculate the equilibrium concentrations of Pb2+ and I- ions in solution after mixing mL of M Pb(NO3)2
and mL of M NaI.
Determine if PbI2 (Ksp = 1.4 x 10-8) precipitates when the solutions are mixed.
a. Calculate the initial concentrations of Pb2+ and I-.
[Pb2+]0= 100.0mL (0.0500mmol/mL)
(100.0mL mL)
= M
[I-]0= 200.0mL (0.100mmol/mL)
(100.0mL mL)
= M

54. b. Calculate Qsp for PbI2. c. Compare Ksp and Qsp.
Qsp = [Pb2+]0[I-]0 2
= [ ][ ]2
= 7.40x10-5
Ksp vs Qsp
1.4x < x10-5
PbI2 will precipitate from the solution.

55. Pb2+ (aq) + 2I- (aq) ↔ PbI2 (s) 5 mmol 20 mmol -5 -2(5) 0 10 mmol
Complete a stoichiometry calculation (BCA) as the reaction goes to completion.
Pb2+ (aq) I- (aq) ↔ PbI2 (s)
5 mmol mmol
(5)
mmol
100 mL x .05 M
200 mL x .1 M B C A

56. Determine equilibrium concentrations.
Calculate concentration of excess reactant.
10 mmol/300 mL = M = [I-]eq
We are finished with I-

57. b. Calculate concentration of limiting reactant using Ksp
b. Calculate concentration of limiting reactant using Ksp. Because the reaction is reversible, the equilibrium concentration of Pb2+ is not 0. Ksp = [Pb2+][I-]2 1.4 x 10-8 = (x) (0.0333)2 [Pb2+] = 1.3 x 10-5 M

58. These are the equilibrium concentrations: [Pb2+] = 1
These are the equilibrium concentrations: [Pb2+] = 1.3 x 10-5 M [I-]eq = M

59. Example 2
A solution is prepared by mixing mL of 1.00x10-2 M Mg(NO3)2 and mL of 1.00x10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp=6.4x10-9).
Determine if MgF2 precipitates.
a. Calculate the initial concentrations of Mg2+ and F―
[Mg2+]0= 150.0mL (1.00x10-2mmol/mL)
(150.0mL mL)
= 3.75x10-3M
[F-]0= 250.0mL (1.00x10-1mmol/mL)
(150.0mL mL)
= 6.25x10-2M

60. MgF2 will precipitate from the solution
Calculate Qsp for MgF2
c. Compare Ksp to Qsp
Qsp = [Mg2+]0[F-]02
= [3.75x10-3][6.25x10-2]2
= 1.46x10-5
Ksp vs. Qsp
1.4x ˂ x10-5
MgF2 will precipitate from the solution

61. 2. Complete a stoichiometry (BCA) calculation
2. Complete a stoichiometry (BCA) calculation. Mg2+(aq) + 2F-(aq)  MgF2(s)
150.0mL (1.00x10-2M)
250.0mL (1.00x10-1M)
1.50 mmol
25.0 mmol B C A
-1.50 mmol
-2(1.50 mmol)
22.0 mmol

62. Determine equilibrium concentrations
a. Calculate concentration of excess reactant
[F-]excess= 22.0 mmol
400.0 mL
= 5.50x10-2M

63. b. Calculate concentration of limiting reactant using
Ksp.
Because the reaction is reversible, the equilibrium concentration of LR is not 0.
Ksp = [Mg2+][F-]2
6.4x10-9 = [x][0.055]2
[Mg2+] = 2.10x10-6 mol/L

64. These are the equilibrium concentrations:
[Mg2+] = 2.10x10-6 M [F-] = 5.50x10-2 M

65. Selective Precipitation
Use a reagent whose anion forms a precipitate with only one of the metal ions in the mixture.
Example:
Solution contains Ba2+ and Ag+ ions.
Adding NaCl will form a precipitate with Ag+ (AgCl), while still leaving Ba2+ in solution.

66. Precipitation of each salt will begin at these concentrations
Example
A solution contains 2.0x10-3 M Pb2+ and 1.0x10-4 M Cu+. If a source of I- is added gradually to this solution, will PbI2 (Ksp=1.4x10-8) or CuI (Ksp=5.3x10-12) precipitate first? What [I-] is necessary to begin the precipitation of each salt.
PbI2(s) ⇌ Pb2+(aq)+ 2I-(aq)
CuI(s) ⇌ Cu+(aq)+ I-(aq)
Ksp = [Pb2+][I-]2
1.4x10-8 = 2.0x10-3 [I-]2
[I-] = 2.6 x 10-3 M
Ksp = [Cu+][I-] 5.3x10-12 = 1.0x10-4 [I-] [I-] = 5.3x10-8 M
Precipitation of each salt will begin at these concentrations

67. CuI will precipitate first because a lower [I-] is needed to start the precipitation. For PbI2 to precipitate: For CuI to precipitate: [I-] = 2.6x10-3 M [I-] = 5.3x10-8 M

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