1. Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 5 (Chp 15,17): Chemical & Solubility Equilibrium (Kc , Kp , Ksp , Q)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall

2. SUMMARY Equilibrium At Equilibrium… N2O4(g) 2 NO2(g)
…forward and reverse rates are ________
…concentrations are ________
equal
constant
N2O4(g) NO2(g)

3. The Equilibrium Constant
For: 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:
solids(s) or pure liquids(l)
K > 1, the reaction is product-favored;
more product
at equilibrium.
K < 1, the reaction is reactant-favored;
more reactant
at equilibrium.
[P]
[R]
[P]
[R]

4. ↔ ↔ ↔ Manipulating K K of reverse rxn = 1/K N2O4 2 NO2 N2O4 2 NO2
Kc = = 4.0
[NO2]2
[N2O4]
N2O4
2 NO2 ↔
Kc = =
(4.0)
[N2O4]
[NO2]2
N2O4
2 NO2
K of multiplied reaction = K^# (raised to power)
K of combined reactions = K1 x K2 … ↔
Kc = = (4.0)2
[NO2]4
[N2O4]2
4 NO2
2 N2O4
A  B K1 = 2.5
C  2 B K2 = 60
A + C  3 B Kovr = (2.5)(60)

5. 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.

6. What Do We Know? H2 I2 2 HI + [H2]in = 0.100 M [I2]in = 0.200 M
[HI]in = 0 M
Reaction
Initial
0.100
0.200
Change
Equilibrium
0.187
H I2
2 HI +
[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.

7. [HI] Increases by 0.187 M H2 I2 2 HI +
Initial
0.100
Change
+0.187
Equilibrium
0.187
Reaction
Initial
0.200
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.

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

9. 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
0.200
Change
–0.0935
+0.187
Equilibrium
0.0065
0.1065
0.187
Reaction
Initial
Change
Equilibrium
H I2
2 HI +
Calculate Kc at 448C.
Kc =
[HI]2
[H2] [I2] =
(0.187)2
(0.0065)(0.1065)
= 51

10. 2 NO N2 O2 + At 2000oC the equilibrium constant for the rxn
2 NO(g) ↔ N2(g) + O2(g)
is Kc = 2.4 x 103.
Reaction
Initial
Change
Equilibrium
2 NO
N O2 +
If the initial concentration of NO is M , what are the equilibrium concentrations of NO , N2 , and O2 ?

11. ? ? ? 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

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

13. 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
0.200 – 2x x
Reaction
Initial
Change
Equilibrium
2 NO
N O2 +
Now, what was the question again?
what are the equilibrium concentrations of NO , N2 , and O2 ?

14. √ √ Kc = [N2] [O2] [NO]2 2.4 x 103 = (x)2 (0.200 – 2x)2 x (0.200 – 2x)
Equilibrium
0.200 – 2x x
Kc =
[N2] [O2]
[NO]2 √
2.4 x 103 =
(x)2
(0.200 – 2x)2 √ x
(0.200 – 2x)
49 =
9.8 – 98x = x
[N2]eq = M
[O2]eq = M
[NO]eq = M
9.8 = 99x
x =

15. Q = Reaction Quotient, Q R P [P] [R] R P Q = [P] [R] Q = K K Q K K Q
ratef < rater
ratef = rater
ratef > rater
Q =
[P]
[R]
Q =
[P]
[R] Q
Q =
[P]
[R]
= K K Q K K Q
Q < K
Q > K
Q = K
too much R,
shift faster
at equilibrium
too much P,
shift  faster

16. [NH3]2 [NH3]2 K = [N2][H2]3 [NH3]2 [N2][H2]3 K =
Add
reactant:
N2(g) 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)

17. Le Châtelier’s Principle
System at equilibrium disturbed by change (affecting collisions) will shift ( or ) to counteract the change.
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

18. Change in external factor Shift to restore equilibrium
Le Châtelier’s Principle
(practice)
N2(g) + 3 H2(g)  2 NH3(g) ∆H = –92 WS
+ 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, heat as product, adding prod. shifts left
 Left
Adding reactant shifts right faster to consume
Right 
Adding product shifts left faster to consume
 Left
Catalysts inc. both rates, but not how far.
No Shift

19. Solubility Product Constant (Ksp)
XaYb(s) 
aX+(aq) + bY−(aq)
Ksp = [X+]a [Y–]b
(Always solid reactant)
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 R I C E
XaYb  aX bY–
# 0 M M
–# +a# b#
0 M a# b#
Ksp = [X+]a[Y–]b

20. 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

21. 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

22. 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?

23. Common Ion  less soluble
(in pure H2O)
(in M KF)
LaF3(s)  La F– x
–x x +3x
x x
LaF3(s)  La F– x
–x x +3x
x x
0.010
≈ 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

24. 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)

25. more soluble by forming complex ions
Adding :NH3 causes…
shift  , more soluble.
Ag(NH3)2+
NH3
AgCl(s)  Ag+(aq) + Cl−(aq)

26. Will a Precipitate Form?
(OR…is Q > K ?)
XaYb(s) 
aX+(aq) + bY−(aq)
Ksp = [X+]a [Y–]b
Q = [X+]a [Y−]b
In a solution,
If Q = Ksp, at equilibrium (saturated).
If Q < Ksp, more solid will dissolve (unsaturated) until Q = Ksp . (products too small, shift right→)
If Q > Ksp, solid will precipitate out (saturated) until Q = Ksp . (products too big, shift left←)

27. Will a Precipitate Form?
(is Q > K ?)
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

28. When Will a Precipitate Form?
p. 764 #66a
(BaSO4)
(SrSO4)
BaSO4(s)  Ba2+ + SO42–
SrSO4(s)  Sr2+ + SO42–
Ksp = [Ba2+][SO42–]
Ksp = [Sr2+][SO42–]
1.1 x 10–10 = (0.010)(x)
3.2 x 10–7 = (0.010)(x)
x = 1.1 x 10–8
x = 3.2 x 10–5
[SO42–] = 1.1 x 10–8 M
[SO42–] = 3.2 x 10–5 M
#66b
less SO42– is
needed to reach equilibrium (Ksp).
Ba2+ will precipitate first b/c…