1. Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 8 (Chp 15,17): Chemical & Solubility Equilibrium (Keq , Kc , Kp , Ksp )
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall

2. (forward & reverse rxns occur simultaneously)
Dynamic
Equilibrium:
(forward & reverse rxns occur simultaneously)
N2O4(g) NO2(g)
double arrow

3. Dynamic Equilibrium: N2O4(g) 2 NO2(g)
Initially, forward & reverse rxns occur at different rates.
(based on collisions, one slows down; other speeds up)
At equilibrium: Rateforward = Ratereverse
N2O4(g) NO2(g)
Ratef
Rater

4. At Equilibrium N2O4(g) 2 NO2(g)
At equilibrium, concentrations (M, mol, P, etc.) of reactant and product remain constant.
N2O4(g) NO2(g)

5. At Equilibrium… N2O4(g) 2 NO2(g)
…forward and reverse rates are ________
…concentrations are ________
equal
constant
N2O4(g) NO2(g)
HW p. 660 #2

6. The Equilibrium Constant (Keq)

7. The Equilibrium Constant
Consider the reaction
aA + bB
cC + dD
At equilibrium…
Ratef = Rater
kf [A]a[B]b = kr [C]c[D]d
kf [C]c[D]d
kr [A]a[B]b =
Keq =
[C]c[D]d
[A]a[B]b
[products]
[reactants]

8. The Equilibrium Constant
Consider the reaction
aA + bB
cC + dD
The equilibrium constant expression (Keq) is
Kc =
[C]c[D]d
[A]a[B]b
K =
[products]
[reactants]
[ ] is conc. in M
K expressions do not include:
pure solids(s) or pure liquids(l)
(b/c concentrations are constant)

9. The Equilibrium Constant
In three experiments at the same temperature, equilibrium was achieved & data were collected.
Exp.
Butanoic acid
(moles)
Ethanol
Ethyl butanoate
Water
Products
Reactants
(Kc) 1
10.0
20.0 2
5.0
40.0 3
30.0
1.0
12.0 4 4
Dingle apnotes13 Task 13a 4
same ratio
(constant)
Calculate Kc for each experiment at this temperature and compare the values.

10. What Does the Value of K Mean?
[products]
[reactants]
Reactants  Products
If K > 1, the reaction is product-favored;
more product
at equilibrium.
If K < 1, the reaction is reactant-favored;
more reactant
at equilibrium.

11. The Equilibrium Constant
Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written
Kp =
(PC)c (PD)d
(PA)a (PB)b
Example:
3 Fe(s) + 4 H2O(g) ↔ Fe3O4(s) + 4 H2(g)

12. Heterogeneous Equilibria
The concentrations of solids and liquids do not appear in the K expression.
PbCl2 (s)
Pb2+ (aq) + 2 Cl−(aq)
Kc = [Pb2+] [Cl−]2
Kc = ?

13. CaCO3 (s) CO2 (g) + CaO(s) Kc = ? Kc = [CO2] Kp = ? Kp = PCO2
As long as some CaCO3 or CaO remains, the amount of CO2 above the solid be constant.
Kc = ?
Kc =
[CO2]
Kp = ?
Kp =
PCO2

14. ↔ ↔ ↔ Manipulating K N2O4 2 NO2 N2O4 2 NO2 K of reverse rxn = 1/K
Kc = = 4.0
[NO2]2
[N2O4]
N2O4
2 NO2
Manipulating K ↔
Kc = =
(4.0)
[N2O4]
[NO2]2
N2O4
2 NO2
K of reverse rxn = 1/K ↔
Kc = = (4.0)2
[NO2]4
[N2O4]2
4 NO2
2 N2O4
K of multiplied reaction = K^# (raised to power)

15. K of combined reaction = K1 x K2 …
Manipulating K
A  3 B + 2 C K1 = 2.5
2 C + 3 D  4 E K2 = 60
A + 3 D  3 B + 4 E Kovr = ?
Kovr = (2.5)(60)
K of combined reaction = K1 x K2 …
HW p. 661
#14, 16, 20

16. Equilibrium Calculations

17. Equilibrium Calculations
A closed system initially containing
0.100 M H2 and M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is M. Calculate Kc at 448C for the reaction taking place, which is
H2 (g) + I2 (g)
2 HI (g)
find limiting reactant
mol reactant completely to mol product, but…
NOW we still have some reactant left and some product formed at equilibrium.

18. RICE Tables
Reaction
Initial
Change
Equilibrium
H I2
2 HI +
0.100 M H2 and M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is M.
Calculate Kc at 448C.

19. What Do We Know? H2 I2 2 HI + [H2]in = 0.100 M [I2]in = 0.200 M
[HI]in = 0 M
Reaction
Initial
Change
Equilibrium
H I2
2 HI +
0.100 M
0.200 M
0 M
0.187 M
[HI]eq = M
0.100 M H2 and M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is M.

20. [HI] Increases by 0.187 M H2 I2 2 HI +
Initial
0.100 M
0 M
Change
+0.187
Equilibrium
0.187 M
Reaction
Initial
0.200 M
Change
Equilibrium
H I2
2 HI +
0.100 M H2 and M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is M.
Calculate Kc at 448C.

21. Stoichiometry shows [H2] and [I2] decrease by half as much
Initial
Change
–0.0935
+0.187
Equilibrium
0.187 M
Reaction
Initial
0.100 M
0.200 M
0 M
Change
Equilibrium
H I2
2 HI +
+0.187
0.187 M
0.187 M HI x 1 mol H2 = M H2
2 mol HI

22. Kc = [HI]2 [H2] [I2] = (0.187)2 (0.0065)(0.1065) = 51
We can now calculate the equilibrium concentrations of all three compounds…
Initial
0.100 M
0.200 M
0 M
Change
–0.0935
+0.187
Equilibrium M M
0.187 M
Reaction
Initial
Change
Equilibrium
H I2
2 HI +
Calculate Kc at 448C.
HW p. 662 #27,30,40,34
Kc =
[HI]2
[H2] [I2] =
(0.187)2
(0.0065)(0.1065)
= 51

23. RICE Practice given K, solve for x
HW p. 663 #43
At 2000oC the equilibrium constant for the reaction
2 NO(g) ↔ N2(g) + O2(g)
is Kc = 2.4 x If the initial concentration of NO is M , what are the equilibrium concentrations of NO , N2 , and O2 ?

24. What Do We Know? 2 NO N2 O2 + Kc = 2.4 x 103
Reaction
Initial
Change
Equilibrium
2 NO
N O2 +
0.200 M
0 M
Kc = 2.4 x 103
the initial concentration of NO is M

25. ? ? ? What Do We Know? What do we NOT know? 2 NO N2 O2 +
Reaction
Initial
0.200 M
0 M
Change
Equilibrium
2 NO
N O2 + ? ? ?
Kc = 2.4 x 103
the initial concentration of NO is M

26. Stoichiometry shows [NO] decreases by twice as much as [N2] and [O2] increases.
Reaction
Initial
Change
Equilibrium
Initial
0.200 M
0 M
Change
Equilibrium
2 NO
N O2 +
– 2x
+ x

27. Now, what was the question again?
We now have the equilibrium concentrations of all three compounds… (in terms of x)
Initial
0.200
Change
– 2x
+ x
Equilibrium
Reaction
Initial
Change
Equilibrium
2 NO
N O2 +
0.200 – 2x x
Now, what was the question again?
what are the equilibrium concentrations of NO , N2 , and O2 ?

28. √ √ Kc = [N2] [O2] [NO]2 Kc = 2.4 x 103 2.4 x 103 = (x)2 (0.200 – 2x)2
Equilibrium
0.200 – 2x x
Kc =
[N2] [O2]
[NO]2
Kc = 2.4 x 103 √
2.4 x 103 =
(x)2
(0.200 – 2x)2 √
HW p. 663 # 44 x
(0.200 – 2x)
49 =
(next slide)
9.8 – 98x = x
[N2]eq = M
[O2]eq = M
[NO]eq = M
9.8 = 99x
x =

29. RICE Practice given K, solve for x
HW p. 663 #44
For the equilibrium
Br2(g) + Cl2(g) ↔ 2 BrCl(g)
at 400 K, Kc = If 0.30 mol of Br2 and 0.30 mol of Cl2 are introduced into a 1.0 L container at 400 K, what will be the equilibrium concentrations of Br2, Cl2, and BrCl?

30. the initial concentrations of Br2 and Cl2 are 0.30 M
What Do We Know?
Reaction
Initial
Change
Equilibrium
Br Cl2
2 BrCl +
0.30 M
0 M
Kc = 7.0
[Br2]in = 0.30 M
[Cl2]in = 0.30 M
the initial concentrations of Br2 and Cl2 are 0.30 M

31. the initial concentrations of Br2 and Cl2 are 0.30 M
What Do We Know?
What do we NOT know?
Reaction
Initial
0.30 M
0 M
Change
Equilibrium
Br Cl2
2 BrCl + ? ? ?
Kc = 7.0
[Br2]in = 0.30 M
[Cl2]in = 0.30 M
the initial concentrations of Br2 and Cl2 are 0.30 M

32. Stoichiometry shows [BrCl] increases by twice as much as [Br2] and [Cl2] decrease.
Reaction
Initial
0.30 M
Change
Equilibrium
Initial
0.30 M
0 M
Change
Equilibrium
Br Cl2
2 BrCl +
– x
+ 2x

33. Now, what was the question again?
We now have the equilibrium concentrations of all three compounds… (in terms of x)
Initial
0.30 M
0 M
Change
– x
+ 2x
Equilibrium
Reaction
Initial
Change
Equilibrium
Br Cl2
2 BrCl +
0.30 – x 2x
Now, what was the question again?
what are the equilibrium concentrations of Br2, Cl2, and BrCl?

34. √ √ Kc = [BrCl]2 [Br2][Cl2] Kc = 7.0 7.0 = (2x)2 (0.30 – x)2 2x
Equilibrium
0.30 – x 2x
Kc =
[BrCl]2
[Br2][Cl2]
Kc = 7.0 √
7.0 =
(2x)2
(0.30 – x)2 √
HW p. 664 #64 2x
(0.30 – x)
2.6 =
(next slide)
0.78 – 2.6x = 2x
[Br2]eq = 0.13 M
[Cl2]eq = 0.13 M
[BrCl]eq = 0.34 M
0.78 = 4.6x
x = 0.17

35. RICE Practice given K, solve for x
HW p. 664 #64
For the equilibrium
2 IBr(g) ↔ I2(g) + Br2(g)
Kp = 8.5 x 10–3 at 150oC. If atm of IBr is placed in a 2.0 L container, what is the partial pressure of this substance after equilibrium is reached?

36. What Do We Know? 2 IBr I2 Br2 + Kp = 8.5 x 10–3
Reaction
Initial
Change
Equilibrium
2 IBr
I Br2 +
0.025 atm
0 atm
Kp = 8.5 x 10–3
the initial pressure of IBr is atm

37. ? ? ? What Do We Know? What do we NOT know? 2 IBr I2 Br2 +
Reaction
Initial
0.025 atm
0 atm
Change
Equilibrium
2 IBr
I Br2 + ? ? ?
Kp = 8.5 x 10–3
the initial pressure of IBr is atm

38. Stoichiometry shows [NO] decreases by twice as much as [N2] and [O2] increases.
Reaction
Initial
0 atm
Change
Equilibrium
Initial
0.025 atm
0 atm
Change
Equilibrium
2 IBr
I Br2 +
– 2x
+ x

39. Now, what was the question again?
We now have the equilibrium concentrations of all three compounds… (in terms of x)
Initial
0.025 atm
0 atm
Change
Equilibrium
Reaction
Initial
0 atm
Change
Equilibrium
2 IBr
I Br2 +
– 2x
+ x
0.025 – 2x x
Now, what was the question again?
What is the equilibrium partial pressure of IBr?

40. √ √ Kp = (PI2)(PBr2) (PIBr)2 8.5 x 10–3 = (x)2 (0.025 – 2x)2 x
Equilibrium
0.025 – 2x x
Kp =
(PI2)(PBr2)
(PIBr)2
Kp = 8.5 x 10–3 √
8.5 x 10–3 =
(x)2
(0.025 – 2x)2 √ x
(0.025 – 2x)
HW p. 664 #66 =
(next slide)
– 0.184x = x
= x
(PIBr)eq = atm
x =

41. RICE Practice
HW p. 664 #66
Solid NH4HS is introduced into an evacuated flask at 24oC. The following reaction takes place:
NH4HS(s) ↔ NH3(g) + H2S(g)
At equilibrium the total pressure in the container is atm.
What is Kp for this equilibrium at 24oC?

42. What Do We Know? NH4HS(s) NH3 H2S +
Solid NH4HS is introduced into an evacuated flask at 24oC.
Reaction
Initial
Change
Equilibrium
Reaction
Initial …
0 atm
Change
Equilibrium
NH4HS(s)
NH H2S +
At equilibrium,
the total pressure (PT)eq is atm.
Kp = ?
0.614 atm

43. ? ? ? ? ? What Do We Know? What do we NOT know? NH4HS(s) NH3 H2S +
Reaction
Initial …
0 atm
Change
Equilibrium
NH4HS(s)
NH H2S + ? ? ? ? ?
At equilibrium,
the total pressure (PT)eq is atm.
Kp = ?
0.614 atm

44. Stoichiometry shows [NH3] increases by the same amount as [H2S] increases.
Initial …
0 atm
Change
– x
+ x
Equilibrium
Reaction
Initial
0 atm
Change
Equilibrium
NH4HS(s)
NH H2S +
0.614 atm

45. Now, what was the question again?
We now have the equilibrium concentrations of all three compounds… (in terms of x)
Reaction
Initial
0 atm
Change
Equilibrium
Initial …
0 atm
Change
– x
+ x
Equilibrium x
NH4HS(s)
NH H2S +
0.614 atm
Now, what was the question again?
What is Kp ?

46. Kp = (PNH3)(PH2S) PT = 0.614 atm PT = PNH3 + PH2S Kp = (x)2
Equilibrium … x
Kp =
(PNH3)(PH2S)
PT = atm
PT = PNH3 + PH2S
Kp =
(x)2
= x + x
= 2x
Kp = (0.307)2
x =
Kp =
WS Equil Calc’s III

47. The Reaction Quotient (Q)
aA + bB
cC + dD
Kc =
[C]c[D]d
[A]a[B]b
(given on exam)
A Q expression gives the same ratio as the equilibrium (K) expression, but …
…may NOT be at equilibrium.
Q =
[C]c[D]d
[A]a[B]b
NOT
(given on exam)
Calculate Q by substituting INITIAL (current) concentrations into the Q expression.

48. If Q = K, system is at equilibrium (K).
[P]
[R]
= K
R P
ratef = rater Q K Q K K Q
Q = K
If Q = K, system is at equilibrium (K).

49. Q R P [R] K Q K K [P] Q = [R] R P Q
ratef > rater
R P
ratef = rater
Q =
[P]
[R] Q K Q K K Q
Q < K
If Q < K, too much reactant, system will
shift right faster to reach equilibrium (K).

50. Q Q = [P] K Q K K [P] Q = [R] R PHW
p. 662
#36, 38
ratef < rater
R P
Q =
[P]
[R]
ratef = rater Q K Q K K Q
Q > K
If Q > K, too much product, system will
shift left faster to reach equilibrium (K).

51. Le Châtelier’s Principle

52. Le Châtelier’s Principle:
“Systems at equilibrium disturbed by a change
amount (M , PP) of reactants or products
volume (V) of container
temperature (T) (changes Keq value)
…will shift ( or ) to counteract the change.”
(that affects collision frequency) like:
Because…
R P
R P
ratef > rater
ratef < rater or
…the rxn goes faster in one direction (Q  K)
until ratef = rater (Q = K again, R P ).

53. The Haber Process N2(g) + 3 H2(g)  2 NH3(g)
The conversion of nitrogen (N2) & hydrogen (H2) into ammonia (NH3) is tremendously significant in agriculture, where ammonia-based fertilizers are of utmost importance.
N2(g) H2(g)  2 NH3(g)

54. [NH3]2 [NH3]2 K = [N2][H2]3 [NH3]2 [N2][H2]3 K =
Add
reactant:
(changes Q)
N2(g) + 3 H2(g)  2 NH3(g)
Q = K
K =
[NH3]2
[N2][H2]3
Q < K
Q =
[NH3]2
[N2][H2]3
Q = K
K =
[NH3]2
[N2][H2]3
(same K)

55. N2(g) + 3 H2(g)  2 NH3(g) Remove product: (changes Q)
This apparatus pushes the equilibrium to the right by removing the product ammonia (NH3) from the system as a liquid.

56. Change Volume (moles of gas)
N2(g) + 3 H2(g)  2 NH3(g)
(changes Q)
↓V (↑PT) shifts to
↑V (↓PT) shifts to
V has no shift if
fewer mol of gas (↓ngas)
more mol of gas (↑ngas)
equal mol of gas R & P
↑PT by adding noble gas?
no shift b/c same R & P pressures
1. 2 NO(g)  N2(g) + O2(g) , ↑V will shift ____.
2. N2O4(g)  2 NO2(g) , ↓V will shift ____.
3. CaCO3(s)  CO2(g) + CaO(s) , shift  by _V.
NOT  ↓

57. [P] [R] K2 = Change Temperature [P] K1 = [R]
Changing temp. is the ONLY way to change the value of ___. K
Why?
T (add heat) shifts…
T (remove heat) shifts…
…in the _____thermic direction to ______ heat
endo
…in the _____thermic direction to ______ heat
exo
produce
absorb
(original)
K2 =
[P]
[R]
(smaller)
K2 =
[P]
[R]
(larger)
K1 =
[P]
[R]
N2(g) + 3 H2(g)  2 NH3(g) DH = –

58. CoCl42– + 6 H2O(l)  4 Cl– + Co(H2O)62+
Change Temperature
CoCl42– + 6 H2O(l)  4 Cl– + Co(H2O)62+
heat +
+ heat –
∆H = __
Add product:
Remove product:
ice
Add
heat
Remove heat

59. [P] [R] K = increase ratef and rater . Catalysts (same) R
Equilibrium occurs faster, but…
at no shift (composition [P]/[R] is same). P

60. Change in external factor Shift to restore equilibrium
Le Châtelier’s Principle
(practice)
N2(g) + 3 H2(g)  2 NH3(g) ∆H = –92
+ heat
Change in external factor
Shift to restore equilibrium
Reason
Increase pressure
(decrease volume)
Increase temp.
Increase [N2]
Increase [NH3]
Add a catalyst
↑P (↓V) shifts to side of fewer moles of gas
Right 
∆H = –
∆H = – , adding heat shifts in the endo- dir.
 Left
Adding reactant shifts right to consume it
Right 
Adding product shifts left to consume it
 Left
Catalysts change how fast, but not how far.
No Shift

61. Change in external factor Shift to restore equilibrium
Le Châtelier’s Principle
(practice)
2 H2O(g) + O2(g)  2 H2O2(l) ∆H = +108
heat +
Change in external factor
Shift to restore equilibrium
Reason
________ pressure
(_______ volume)
_________ temp.
_________ PO2
_________ PH2O
Decrease H2O2
Decrease
↓P (↑V) shifts to side of more moles of gas
 Left
Increase
∆H = +
∆H = + , remove heat shifts in the exo- dir.
 Left
Decrease
Adding reactant shifts right to consume it
Right 
Increase
Removing reactant shifts left to produce it
Decrease
 Left
(s) & (l) do not affect Q & K (usually)
No Shift

62. Le Châtelier’s Principle
(summary)
R  P
(M, PPR, PPP)
add R or P:
remove R or P:
volume: ↓V shifts to
↑V shifts to
temp.
↑T shifts
↓T shifts
catalyst:
shift away faster (consume)
shift toward faster (replace )
fewer mol of gas (↓ngas)
(Ptotal)
more mol of gas (↑ngas)
(changes K)
(H + R  P) (R  P + H)
in endo dir. to use up heat
in exo dir. to make more heat
no shift
WS Eq Pract. 2 #4

63. Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 8 (Chp 15, 17): Chemical & Solubility Equilibria (Keq , Kc , Kp , Ksp )
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall

64. Solubility Product Constant (Ksp)
Write the K expression for a saturated solution of PbI2 in water:
[Pb2+][I−]2
PbI2(s) 
Pb2+(aq) + 2 I−(aq)
K = sp
XaYb(s) 
aX+(aq) + bY−(aq)
Ksp = [X+]a [Y–]b
For “insoluble” solids, the equilibrium constant,
Ksp , is the solubility product constant, or the…
…product of M’s of dissolved ions at equilibrium.

65. Ksp is NOT the Solubility
grams of solid (s) dissolved in 1 L (g/L)
mol of solid (s) dissolved in 1 L (M)
product of conc.’s (M) of ions(aq) at equilibrium
solubility:
molar solubility:
Ksp :
XaYb  aX+ + bY–
[XaYb] [X+] [Y–]
molar solubility molar conc.’s of ions
Ksp = [X+]a[Y–]b

66. Ksp is NOT the Solubility
molar solubility:
Ksp :
grams of solid (s) dissolved in 1 L (g/L)
mol of solid (s) dissolved in 1 L (M)
product of conc.’s (M) of ions (aq) at equilibrium
4.2 g/L
0.027 M CaCrO4
0.027 M Ca2+
0.027 M CrO42–
molar mass:
g/mol
CaCrO4(s)  Ca CrO42–
[CaCrO4] [Ca2+] [CrO42–]
Ksp = [Ca2+][CrO42–]
HW p. 763 #46

67. 1 PbBr2 dissociates into…
Ksp Calculations
HW p. 763 #48a
If solubility (or molar solubility) is known, solve for Ksp .
[PbBr2] is M at 25oC .
(maximum that can dissolve) R I C E
PbBr2(s)  Pb Br–
0.010 M M M –
0 M M M
Ksp = [Pb2+][Br–]2
(all dissolved = saturated)
(any excess solid is irrelevant)
Ksp = (0.010)(0.020)2
1 PbBr2 dissociates into…
1 Pb2+ ion & 2 Br– ions
Ksp = 4.0 x 10–6

68. Ksp Calculations
HW p. 763 #50
Solubility (or molar solubility) is known, solve for Ksp .
Ksp = (0.0012)(0.0024)2
[PbI2] = M R I C E
Ksp = [Pb2+][I–]2
PbI2(s)  Pb I–
M M M –
0 M M M
Ksp = 6.9 x 10–9
solubility:
0.54 g/L
Molar solubility: M
1 mol
461 g
0.54 g x =
mol

69. Ksp Calculations If only Ksp is known, solve for x (M).
Ksp for AgCl is 1.8 x 10–10 .
Ksp = [Ag+][Cl–]
Ksp = x2 R I C E
AgCl(s)  Ag+ + Cl–
x M M
–x +x x
0 M x x
1.8 x 10–10 = x2
√1.8 x 10–10 = x
1.3 x 10–5 = x
(molar solubility)
[AgCl] = 1.3 x 10–5 M
[Ag+] = 1.3 x 10–5 M
[Cl–] = 1.3 x 10–5 M

70. Ksp Calculations If only Ksp is known, solve for x (M).
HW p. 663 #47
If only Ksp is known, solve for x (M).
Ksp for CaSO4 is 2.4 x 10–5 . R I C E
CaSO4(s)  Ca2+ + SO42–
x M M
–x x x
0 M x x
Ksp = [Ca2+][SO42–]
Ksp = x2
2.4 x 10–5 = x2
√2.4 x 10–5 = x
= x
(molar solubility)
[CaSO4] = M
[Ca2+] = M
[SO42–] = M

71. Ksp Calculations If only Ksp is known, solve for x (M).
Ksp for PbCl2 is 1.6 x 10–5 . R I C E
PbCl2(s)  Pb Cl–
x M M
–x x x
0 M x x
Ksp = [Pb2+][Cl–]2
Ksp = (x)(2x)2
Ksp = 4x3
(molar solubility)
1.6 x 10–5 = 4x3
3√4.0 x 10–6 = x
= x
[PbCl2] = M
[Pb2+] = M
[Cl–] = M

72. Ksp Calculations If only Ksp is known, solve for x (M).
Ksp for Cr(OH)3 is 1.6 x 10–30 . R I C E
Cr(OH)3(s)  Cr OH–
x M M
–x x +3x
0 M x x
Ksp = [Cr3+][OH–]3
Ksp = (x)(3x)3
Ksp = 27x4
1.6 x 10–30 = 27x4
4√5.9 x 10–32 = x
1.6 x 10–8 = x
(molar solubility)
[Cr(OH)3] = 1.6 x 10–8 M
[Cr3+] = 1.6 x 10–8 M
[OH–] = 4.8 x 10–8 M

73. Ksp Calculations (in pure H2O) (in 0.010 M KF) LaF3(s)  La3+ + 3 F–
HW p. 763 #52a
#52b
(in pure H2O)
(in M KF)
LaF3(s)  La F–
x M M
–x x +3x
0 M x x
LaF3(s)  La F–
x M
–x x +3x
0 M x x
0.010 M
≈ 0.010
b/c K <<<1
Solubility is lower in _________________
sol’n w/ common ion
Ksp = [La3+][F–]3
Ksp = (x)(3x)3
2 x 10–19 = 27x4
x = 9 x 10–6 M LaF3
Ksp = [La3+][F–]3
Ksp = (x)( x)3
2 x 10–19 = (x)(0.010)3
x = 2 x 10–13 M LaF3

74. Factors Affecting Solubility
Common-Ion Effect (more Le Châtelier)
If a common ion is added to an equilibrium solution, the equilibrium will shift left and the solubility of the salt will decrease. OR
adding common ion shifts left (less soluble)
BaSO4(s) 
Ba2+(aq) + SO42−(aq)
BaSO4 would be least soluble in which of these 1.0 M aqueous solutions?
Na2SO4 BaCl2 Al2(SO4)3 NaNO3
most soluble?
WS Ksp #1-2

75. Factors Affecting Solubility
Mg(OH)2(s) 
Mg2+(aq) + 2 OH−(aq)
Addition of HCl (acid) to this solution would cause…
Greater solubility because…
Added acid (H+) reacts with OH– thereby removing product causing a shift to the right to dissolve more solid Mg(OH)2. 75

76. Basic anions, more soluble in acidic solution.
HW p. 763 #55
Basic anions, more soluble in acidic solution.
H+ NO Effect on:
Cl– , Br–, I–,
NO3–, SO42–, ClO4–
Adding H+ would cause…
shift  , more soluble. H+
Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq)

77. Factors Affecting Solubility
Complex Ions
Metal ions can and form complex ions with lone e– pairs in the solvent.

78. AgCl(s)  Ag+(aq) + Cl−(aq)
forming complex ions…
…increases solubility
p. 765 #59
Ag(NH3)2+
NH3
AgCl(s)  Ag+(aq) + Cl−(aq)

79. Will a Precipitate Form?
XaYb(s) 
aX+(aq) + bY−(aq)
Ksp = [X+]a [Y–]b
Q = [X+]a [Y−]b
WS Ksp #4
In a solution,
If Q = Ksp, at equilibrium (saturated).
If Q < Ksp, more solid will dissolve (unsaturated) until Q = Ksp . (products too small, proceed right→)
If Q > Ksp, solid will precipitate out (saturated) until Q = Ksp . (products too big, proceed left←)
HW p. 764 #62b, 66

80. Will a Precipitate Form?
(OR…is Q > K ?)
HW p. 764 #62b
AgIO3(s)  Ag+ + IO3–
Ksp = [Ag+][IO3–]
Ksp = 3.1 x 10–8
100 mL of M AgNO3
10 mL of M NaIO3
Q = [Ag+][IO3–]
(mixing changes M and V)
M1V1 = M2V2
Q = (0.0091)(0.0014)
Q =
[Ag+] = ________
(0.010 M)(100 mL) = M2(110 mL) M
Q = 1.3 x 10–5
Q > K , so…
rxn shifts left
prec. will form
[IO3–] = ________
(0.015 M)(10 mL) = M2(110 mL) M

81. Which Will Precipitate First?
HW p. 764 #65
(AgI)
(PbI2)
AgI(s)  Ag+ + I–
PbI2(s)  Pb I–
Ksp = [Ag+][I–]
= 8.3 x 10–17
Ksp = [Pb2+][I–]2
= 7.9 x 10–9
AgI will precipitate first b/c…
less I– is
needed to reach equilibrium at a smaller Ksp.
Ksp = [Ag+][I–]
Ksp = [Pb2+][I–]2
8.3 x 10–17 = (2.0 x 10–4)(x)
7.9 x 10–9 = (1.5 x 10–3)(x)2
x = 4.2 x 1013
x = 2.3 x 10–3
[I–] = 4.2 x 10–13 M
[I–] = 2.3 x 10–3 M