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
SUMMARY Equilibrium
At Equilibrium…
…forward and reverse
equal
rates are ________
…concentrations
constant
are ________
The Equilibrium Constant
For: aA + bB
cC + dD
…the equilibrium constant expression (Keq) is
[products]
[C]c[D]d
K
=
Kc =
[reactants]
[A]a[B]b
K expressions do not include:
solids(s) or pure liquids(l)
K > 1, the reaction is
product-favored;
more product
[P]
at equilibrium. [R]
[ ] is conc. in M
K < 1, the reaction is
reactant-favored;
[P]
more reactant
at equilibrium. [R]
K of reverse rxn
= 1/K
2 NO2 ↔ N2O4
Manipulating K
N2O4
↔
2 NO2
[NO2]2
Kc =
= 4.0
[N2O4]
K of combined
reactions = K1 x K2 …
A B
C2B
Kc = [N2O4] = 1
[NO2]2 (4.0)
.
K of multiplied reaction
#
= K^ (raised to power)
K1 = 2.5
K2 = 60
A + C  3 B Kovr = (2.5)(60)
2 N2O4 ↔ 4 NO2
4
[NO
]
2
2
Kc =
=
(4.0)
[N2O4]2
RICE Tables
Reaction
H2
+
I2
2 HI
Initial
Change
Equilibrium
0.100 M H2 and 0.200 M I2 at 448C is allowed
to reach equilibrium. At equilibrium, the
concentration of HI is 0.187 M.
Calculate Kc at 448C.
What Do We Know?
[H2]in = 0.100 M
[I2]in = 0.200 M
Reaction
H2
Initial
0.100
+
[HI]in = 0 M
I2
2 HI
0.200
0
Change
Equilibrium
0.187
[HI]eq = 0.187 M
0.100 M H2 and 0.200 M I2 at 448C is allowed
to reach equilibrium. At equilibrium, the
concentration of HI is 0.187 M.
[HI] Increases by 0.187 M
Reaction
H2
Initial
0.100
+
I2
2 HI
0.200
0
Change
+0.187
Equilibrium
0.187
0.100 M H2 and 0.200 M I2 at 448C is allowed
to reach equilibrium. At equilibrium, the
concentration of HI is 0.187 M.
Calculate Kc at 448C.
Stoichiometry shows [H2] and [I2]
decrease by half as much
Reaction
H2
Initial
Change
+
I2
2 HI
0.100
0.200
0
–0.0935
–0.0935
+0.187
Equilibrium
0.187 M HI x 1 mol H2 = 0.0935 M H2
2 mol HI
0.187
We can now calculate the equilibrium
concentrations of all three compounds…
Reaction
H2
Initial
+
I2
2 HI
0.100
0.200
0
Change
–0.0935
–0.0935
+0.187
Equilibrium
0.0065
0.1065
0.187
Calculate Kc at 448C.
[HI]2
(0.187)2
Kc =
=
= 51
[H2] [I2]
(0.0065)(0.1065)
At 2000oC the equilibrium constant for the rxn
2 NO(g) ↔ N2(g) + O2(g)
is Kc = 2.4 x 103.
Reaction
2 NO
N2
+
O2
Initial
Change
Equilibrium
If the initial concentration of NO is 0.200 M ,
what are the equilibrium concentrations of
NO , N2 , and O2 ?
What Do We Know?
What do we NOT know?
Reaction
Initial
Change
2 NO
N2
+
0.200 M
0M
0M
?
?
?
O2
Equilibrium
Kc = 2.4 x 103
the initial concentration of NO is 0.200 M
Stoichiometry shows [NO] decreases by
twice as much as [N2] and [O2] increases.
Reaction
2 NO
N2
Initial
0.200
0
0
– 2x
+x
+x
Change
Equilibrium
+
O2
We now have the equilibrium
concentrations of all three compounds…
(in terms of x)
Reaction
2 NO
N2
Initial
0.200
0
0
– 2x
+x
+x
0.200 – 2x
x
x
Change
Equilibrium
+
O2
Now, what was the question again?
what are the equilibrium concentrations
of NO , N2 , and O2 ?
Equilibrium
0.200 – 2x
x
x
[N2] [O2]
Kc =
[NO]2
√
√
2
(x)
2.4 x 103 =
(0.200 – 2x)2
x
49 =
(0.200 – 2x)
9.8 – 98x = x
9.8 = 99x
x = 0.099
[N2]eq = 0.099 M
[O2]eq = 0.099 M
[NO]eq = 0.0020 M
Reaction Quotient, Q
R
P
rate > rate
f
r
Q
=
[P]
[R]
K
R
P
=
rate
f
r
rate
[P]
=K
Q=
[R]
Q K
P
R
rate
<
rate
f
r
Q
[P]
Q = [R]
K
Q
Q<K
too much R,
shift faster
Q=K
at equilibrium
Q>K
too much P,
shift  faster
Add
reactant:
N2(g) + 3 H2(g)  2 NH3(g)
Q=K
2
[NH
]
3
K=
[N2][H2]3
Q<K
Q=
[NH3]2
3
[N2][H2]
Q=K
2
[NH3]
K=
3
[N
]
[H
]
2
(same K)
2
Le Châtelier’s Principle
System at equilibrium disturbed by change (affecting
collisions) will shift ( or ) to counteract the change.
- add R or P: shift away faster (consume)
- remove R or P: shift toward faster (replace)
- volume:
(Ptotal)
↓V shifts to fewer mol of gas (↓ngas)
↑V shifts to more mol of gas (↑ngas)
- temp. (changes K) (H + R  P) (R  P + H)
↑T shifts in endo dir. to use up heat
↓T shifts in exo dir. to make more heat
- catalyst: no shift
Le Châtelier’s Principle (practice)
WS
N2(g) + 3 H2(g)  2 NH3(g) + heat ∆H = –92
Change in
Shift to restore
external factor
equilibrium
Increase pressure
Right 
(decrease volume)
∆H = –
Increase temp.
 Left
Increase [N2]
Increase [NH3]
Add a catalyst
Reason
↑P (↓V) shifts to side
of fewer moles of gas
Right 
–∆H, heat as product,
adding prod. shifts left
Adding reactant shifts
right faster to consume
 Left
Adding product shifts
left faster to consume
No Shift
Catalysts inc. both
rates, but not how far.
Solubility Product Constant (Ksp)
XaYb(s)  aX+(aq) + bY−(aq)
(Always solid reactant)
Ksp = [X+]a [Y–]b
solubility: grams of solid (s) dissolved in 1 L (g/L)
molar solubility: mol of solid (s) dissolved in 1 L (M)
Ksp: product of conc.’s (M) of ions(aq) at equilibrium
R
I
C
E
XaYb
#
–#
0M

aX+ + bY–
0M
+a#
a#
0M
+b#
+]a[Y–]b
K
=
[X
b#
sp
HW p. 763 #48a
Ksp Calculations
If solubility (or molar solubility) is known,
solve for Ksp . [PbBr2] is 0.010 M at 25oC .
(maximum that can dissolve)
R PbBr2(s)  Pb2+ + 2 Br–
Ksp =
– 2
2+
[Pb ][Br ]
I 0.010 M
0M
0M
C –0.010
+0.010 +0.020
E 0M
0.010 M 0.020 M
(all dissolved = saturated) (any excess solid is irrelevant)
Ksp = (0.010)(0.020)2
1 PbBr2 dissociates into…
–6
–
2+
K
=
4.0
x
10
1 Pb ion & 2 Br ions
sp
Ksp Calculations
If only Ksp is known, solve for x (M).
Ksp for PbCl2 is 1.6 x 10–5 .
R PbCl2(s)  Pb2+ + 2 Cl– Ksp = [Pb2+][Cl–]2
I x
0M
0M
2
K
=
(x)(2x)
sp
C –x
+x
+2x
3
K
=
4x
E 0M
x
2x
sp
(molar solubility)
1.6 x 10–5 = 4x3
[PbCl2] = 0.016 M
3
–6 = x
√4.0
x
10
[Pb2+] = 0.016 M
0.016 = x
[Cl–] = 0.032 M
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)
Ba2+(aq) + SO42−(aq)
BaSO4(s)
BaSO4 would be least soluble in which of these
1.0 M aqueous solutions?
Na2SO4
BaCl2
Al2(SO4)3
most soluble?
NaNO3
Common Ion  less soluble
(in pure H2O)
LaF3(s)  La3+ + 3 F–
x
0
0
–x
+x
+3x
0
x
3x
(in 0.010 M KF)
LaF3(s)  La3+ + 3 F–
0.010
x
0
–x
+x
+3x
0
x 0.010 + 3x
≈
0.010
sol’n w/ common ion
Solubility is lower in _________________
b/c K <<<1
Ksp = [La3+][F–]3
Ksp = [La3+][F–]3
Ksp = (x)(3x)3
Ksp = (x)(0.010 + 3x)3
2 x 10–19 = 27x4
x = 9 x 10–6 M LaF3
2 x 10–19 = (x)(0.010)3
x = 2 x 10–13 M LaF3
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)
more soluble
by forming
complex ions
Adding :NH3 causes…
shift  ,
more soluble.
Ag(NH3)2+
NH3
AgCl(s)  Ag+(aq) + Cl−(aq)
Will a Precipitate Form?
(OR…is Q > K ?)
XaYb(s)  aX+(aq) + bY−(aq)
• In a solution,
Ksp = [X+]a [Y–]b Q = [X+]a [Y−]b
– 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←)
Will a Precipitate Form?
p. 764 #62b
(is Q > K ?)
AgIO3(s)  Ag+ + IO3–
Ksp = [Ag+][IO3–]
Ksp = 3.1 x 10–8
100 mL of 0.010 M AgNO3
10 mL of 0.015 M NaIO3
+][IO –]
Q
=
[Ag
3
(mixing changes M and V)
Q = (0.0091)(0.0014)
Q=
M1V1 = M2V2
[Ag+] = ________
0.0091 M
Q = 1.3 x 10–5
(0.010 M)(100 mL) = M2(110 mL) Q > K , so…
[IO3–] = ________
0.0014 M
rxn shifts left
(0.015 M)(10 mL) = M2(110 mL) prec. will form
p. 764 #66a
When Will a Precipitate Form?
(BaSO4)
BaSO4(s)  Ba2+ + SO42–
(SrSO4)
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
Ba2+ will precipitate first b/c…less SO42– is
needed to reach equilibrium (Ksp).