Transcript 17 Heterogeneous and complex equilibria

19 Heterogeneous
and complex equilibria
Formation of crystals such as AgCl in a
solution is a heterogeneous equilibrium,
because there are more than one phase,
AgCl (s) = Ag+ (aq) + Cl– (aq)
Species such as Ag(NH3)2+ & Ag(CN)2–
are complexes (or complex ions).
Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2+ (aq)
CaCO3 exists as
aragonite 
 calcite
19 Heterogeneous & complex equilibria
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Beauty due to heterogeneous equilibria
There are many natural
heterogeneous equilibria.
Please think of some!
19 Heterogeneous & complex equilibria
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The solubility product
For the dissolution,
CaCO3 (s) = Ca2+ (aq) + CO32- (aq)
Ksp = [Ca2+] [CO32-] = 3.8e-9 (a constant) solubility product
Substance
Formula
Ksp
Aluminum hydroxide
Al(OH)3
4.6e-35
Barium chromate
BaCrO3
1.2e-10
Calcium phosphate
Ca3(PO4)2
1e-26
Iron sulfite
FeS
6e-18
Lead sulfite
PbS
2.5e-27
Mercury sulfite
HgS
1.6e-52
19 Heterogeneous & complex equilibria
Ksp = [Al3+] [OH-]3
Ksp = [Ca2+]3 [PO4]2
Which is the
most and least
soluble?
3
Compound
Ksp
Compound
Ksp
AgBr
5.0 x 10-13
Fe(OH)3
4.0 x 10-38
AgCl
1.8 x 10-10
FeS
6.3 x 10-18
AgI
8.3 x 10-17
HgS
1.6 x 10-52
AgIO3
3.1 x 10-8
Mg(OH)2
1.8 x 10-11
Ag3PO4
1.3 x 10-20
MgC2O4
8.6 x 10-5
Al(OH)3
2.0 x 10-32
Mn(OH)2
1.9 x 10-13
Ba(OH)2
5.0 x 10-3
MnS
2.5 x 10-13
BaSO4
1.1 x 10-10
NiS
1.0 x 10-24
Bi2S3
1.0 x 10-97
PbCl2
1.6 x 10-5
CaCO3
4.8 x 10-9
PbSO4
1.6 x 10-8
CaC2O4
4.0 x 10-9
PbS
8.0 x 10-28
CaSO4
1.2 x 10-6
SrSO4
3.2 x 10-7
CdS
8.0 x 10-27
Zn(OH)2
3.3 x 10-17
CoS
2.0 x 10-25
ZnS
1.6 x 10-23
CuS
6.3 x 10-36
19 Heterogeneous & complex equilibria
Solubility
product
constants
Table like this
is available in
handbooks or
data bases.
Know where
to find them
when you
need them.
Write Ksp
expressions.
4
Ksp and solubility
19-2
The Ksp = 1.0e-6 for BaF2, what are [Ba2+] and [F¯]?
Solution:
BaF2 = Ba2+ + 2 F¯
assume  x
2x
Ksp = x (2x)2 = 1.0e-6
x = 3 (1.0e-6 / 4) = 6.3e-3 M
Molar mass of BaF2=175 g/mol
[Ba2+] = x = 6.3e-3 M  molar solubility = 6.3e-3 *175 = 1.1g/L
[F¯] = 2 x = 0.013 M
Checking:
6.3e-3 *0.0132 = 1e-6 = Ksp
Calculate solubility of BaCrO3, Ksp = 1.2e-10
19 Heterogeneous & complex equilibria
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Concentrations of ions in solution
CaF2(s) = Ca2+(aq) + 2 F- (aq), Ksp know where to find
Ksp = 5.3e-9
19 Heterogeneous & complex equilibria
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Perception of a saturate solution
Ag2S = 2 Ag+ + S2–
Ksp = [Ag+]2[ S2–]
19 Heterogeneous & complex equilibria
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Calculate Ksp from Solubility
When 0.50 L saturated CaC2O4 solution was dried, 0.0030 g dry salt
was obtained. Evaluate the Ksp.
Solution:
3.0e-3 g 1 mol
———— ——— = 4.7e-5 mol / L
0.50 L
128 g
= [Ca2+]
CaC2O4 = Ca2+ + C2O424.7e-5 4.7e-5
Figure out how to calculate
solubility from Ksp .
Ksp = 4.7e-5 * 4.7e-5 = 2.2e-9
How many grams of HgS will
dissolve in 1 L?
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Precipitation of AgCl
Goto URL:
http:// dl.clackamas.cc.or.us/ch105-05/solubili.htm
for a demonstration if you have not seen the experiment
Ag+ (aq) + Cl- (aq) = AgCl(s)
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Common-ion effect on solubility
19-3
Like acid-base equilibria, presence of common ions from more than
one electrolyte affects the solubility, since Ksp remains constant.
For example, the Ksp = 1.8e-10 for AgCl. The maximum [Ag+] is governed
by the condition,
NaCl = Na+ + Cl? Should we consider
0.10 0.10
AgCl = Ag+ + ClAgCl = Ag+ + Clx
0.10+x
[Ag+] 0.10
Thus [Ag+] * 0.10 = 1.8e-10
[Ag+] = 1.8e-9 M
Solubility of AgCl = 1.8e-10 = 1.3e-5 M in pure water is 7,454 times more.
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Graph the common ion effect
AgCl = Ag+ + ClNaCl = Na+ + Cl-
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Condition for precipitation
19-5
Recall: Predicting reaction directions by comparing Qc and Kc.
Same principle applies to precipitation (ppt)
Q < Ksp, unsaturated (solution)
Q = Ksp, saturated (usually two phases are present)
Q > Ksp, super-saturated
(unstable, often needs a seed to start the ppt)
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Separation by precipitation
19-6
A 10-mL solution contains 0.10 M each of Cl–, Br–, and I– ions. Micro
amounts of 0.10 M AgNO3 solution is added to the system. What are the
Ag+ concentrations before AgI, AgBr, and AgCl precipitate?
Solution: Data sheet Ksp:AgCl 1.8e-10, AgBr 5.0e-13, and AgI 8.3e-17.
[Ag+] for AgI, AgBr, & AgCl solids to form:
[I–] = 0.10 M [Ag+] = 8.3e-17/ 0.10 = 8.3e-16
AgI(s) appears
[Br–] = 0.10 M [Ag+] = 5.0e-13/ 0.10 = 5.0e-12
AgBr(s) appears
[Cl–] = 0.10 M [Ag+] = 1.8e-10 / 0.10 = 1.8e-9
AgCl(s) appears
As [Ag+] increases, how do [Cl-], [Br-] and [I-] vary ?
[Ag+]
[Cl–]
[Br–]
[I–]
0  8.3e-16
0.1
0.1
0.10 AgI(s)
 5.0e-12
0.1
0.10 AgBr(s)
 1.7e-5
 1.8e-9 0.10 AgCl(s)
 2.8e-4
 4.6e-8
 1e-7
 1.8e-3
 5.0e-6
 8.3e-10
 0.1
____?
____? equilibria ____?
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pH H2S and solubility of metal ions
19-7
The pH affects the equilibrium of many species, for example:
H2S = H+ + HS–
Ka1 = 1e-9
HS– = H+ + S2–
Ka2 = 1e-14 (doubtful but adopt)
H2S = 2 H+ + S2– Koverall = 1e-23
[S2–] = [H2S] *1e-23 / [H+]2 = [H2S] *1e(2*pH-23), strongly affected by pH
If [H2S] = 1.0 M
pH
[S2–]
Ksp of MS
[S2–] for [M2+] = 0.001
1
1e-21
1.6e-52 HgS
1.6e-49 (ppt)
2
1e-19
2.5e-27 PbS
2.5e-24 (ppt)
2.52
1.1e-18 1.1e-21 ZnS
1.1e-18 (no ppt pH < 2.5)
4.39
6e-15
6e-18 FeS
6e-15 (no ppt pH < 4.4)
7
1e-9
2.5e-13 MnS _____ (no ppt pH < ____)
10
1e-3
11
0.1 Recalculate
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& complex
equilibria
at various
pH if [H S] = 0.10 M
2
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pH, and CO2 on CaCO3 solubility
The [CO32–] in a 0.0010 M H2CO3 solution is determined by,
H2CO3 = H+ + HCO3– Ka1 = 4.3e-7
HCO3– = H+ + CO32– Ka2 = 5.6e-11
H2CO3 = 2 H+ + CO32– Koverall = Ka1*Ka2 = 2.4e-17
= [H+]2 [CO32–] / [H2CO3]
[CO32–] = (0.0010*2.4e-17) [H+]–2
= 2.4e-20 * 1e(2*pH)
= 2.4e(2*pH-20)
Decrease pH by 1 increases
2–
2+
[CO3 ] affects [Ca ] due to equilibrium
2+] by 2 order of magnitude
[Ca
CaCO3 = Ca2+ + CO32– Ksp = 8.7e–9
2+]
pH
[Ca
[Ca2+] = 8.7e–9 [CO32–]–1
8
3.6e-5
8.7e–9
=
/ 2.4e(2*pH–20)
7
3.6e-3
= 3.6e(20 – 9–2*pH)
6
0.36
= 3.6e(11-2*pH)
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Stalactites and
stalagmites
stalactites
Rain dissolves limestone and when water drops form
stalactites and stalagmites in caves. They grew
about 2 cm per 1000 years.
The slightly acidic rain dissolves lime stone:
CaCO3 (s) + H+ = Ca2+ (aq) + HCO3- (aq)
When acidity is reduced, solid forms:
Ca2+ (aq) + HCO3- (aq) + OH- (aq) = CaCO3 (s) + H2O
stalagmites
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Stalactites
hanging down
from the ceiling
formed over
hundreds of
years in Mercer
Cavern
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The "Angel Wings" are two
delicate and translucent crystalline
formations, over 9 feet long and
2.5 feet wide in Mercer Caverns
Aragonite formations
found at Mercer
Caverns, California
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Equilibrium of complexes
19-8
Metal ions tend to attract Lewis bases forming coordinated complexes,
or complex ions. These formation is governed by equilibrium,
metal ion ligand
Step-wise formation constant
Ag+ (aq) + NH3 (aq) = Ag(NH3)+ (aq) K1
Ag(NH3)+ (aq) + NH3 (aq) = Ag(NH3)2+ (aq) K2
formation constant
Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2+ (aq) Kf = K1 K2 = 1.7e7
Ag+ (aq) + 2 S2O32– (aq) = Ag(S2O3)23– (aq) Kf = 2.9e13
Ag+ (aq) + 2 CN– (aq) = Ag(CN)2– (aq) Kf = 5.6e18
Dissociation constant
Ag(NH3)2+ (aq) = Ag+ (aq) + 2 NH3 (aq) Kd = 1 / Kf = 5.9e-8
Ag(S2O3)23– (aq) = Ag+ (aq) + 2 S2O32– (aq) Kd = 1/2.9e13 = 3.4e-14
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Logical, but impractical method!!!
Evaluate [Ag+] in a solution containing 1.0 M NH3 and 0.10 M AgNO3.
Solution:
Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2+ (aq)
0.1-y
1-2y
y
y
——————— = 1.7e7
(1-2y)2(0.10-y)
y
——— = (1-2y)2 (0.1-y)
1.7e7
[Ag+] = 0.10 – y ~ 0
Kf = 1.7e7 (find data)
(think  way)
5.9e-8y = 0.1 –0.8y –2y2
By trial an error, y = 0.10
0.10-y = 6e-9, too small, impractical!
(cannot evaluate [Ag+])
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Concentration of ions in presence of ligands
Evaluate [Ag+] in a solution containing 1.0 M NH3 and 0.1 M AgNO3.
Solution:
Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2+ (aq)
x
0.8+2x
0.1-x
0.1-x
————— = 1.7e7
(0.8+2x)2 x
0.1 - x ~ 0.1;
0.1
——— = 1.7e7
0.82 x
Kf = 1.7e7 (find data)
(think  way)
(treated as dissociation)
3.4e7 x2 + 1.7e7 x – 0.1 = 0
x =[ -1.7e7+{(1.7e7)2 + 4*0.1}] / 2 = 0 (cannot solve)
0.8+2x ~ 0.8
(x is very small) then
0.1
x = ————— = 9.2e-9 = [Ag+]
0.82 *1.7e7 Small indeed
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Complex and precipitate formation-1
What is the maximum [Cl-] before AgCl(s) forms in a solution containing
1.0 M NH3 and 0.1 M AgNO3?
Solution:
Previous slide showed [Ag+] = 9.2e-9 M in a solution when [NH3] = 1.0 M
Ksp = 1.8e-10 for AgCl (know where to look up)
Thus, the max. [Cl-] = 1.8e-10 / 9.2e-9 = 0.02 M
When no NH3 is present, max. [Cl-] = 1.8e-10/0.1 = 1.8e-9
What is the maximum [Br-] before AgBr(s) forms in a solution containing
1.0 M NH3 and 0.1 M AgNO3?
0.02
Ksp = 5e-13 for AgBr,
-------- = 368 times
Thus, the max. [Br-] = 5e-13 / 9.2e-9 = 5.4e-5 M.
5.4e-5
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Complex and precipitate formation-2
What is the maximum [Br-] in a solution suppose to containing
0.10 M AgNO3 and 0.20 M Na2S2O3 before AgBr(s) (Ksp = 5e-13)
forms?
Solution:
Kf = 2.9e13 for Ag(S2O3)23–;
Ag(S2O3)23– = Ag+ + 2 S2O32– , K = 1/2.9e13 = 3.4e-14
0.1-x
x
2x
4x3
——— = 3.4e-14; x = (3.4e-14*0.1)1 / 3 = 1.5e-5 = [Ag+]
(0.1–x) small
max [Br–] = Ksp/[Ag+] = 5e-13/1.5e-5 = 3.3e-8 (very low)
What is [Br–] = ? If [Na2S2O3] is increased to 1.0 M?
See next page
19 Heterogeneous & complex equilibria
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Complex and precipitate formation-3
What is the maximum [Br-] in a solution suppose to containing 0.10 M
AgNO3 and 1.0 M Na2S2O3 before AgBr(s) (Ksp = 5e-13) forms?
Solution:
Kf = 2.9e13 for Ag(S2O3)23– K = 1/2.9e13 = 3.4e-14
Ag(S2O3)23– = Ag+ + 2 S2O32–
0.1-x
x
2x+0.8
(0.8+2x)2x
———— = 3.4e-14,
(0.1-x) small
x = 3.4e-14*0.1/0.82 = 5.3e-14 = [Ag+]
max [Br –] = 5e-13/5.3e-14 = 94 M, unrealistically large
What is the maximum [I – ] in a solution containing 0.10 M AgNO3
and 1.0 M Na2S2O3 before AgI(s) (Ksp = 8e-17) forms?
[I-] = 8e-17 / 5.3e-14 = 0.0015 M; small compare to 0.1 M
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19-9
Chemical casserole
Experiment
AgNO3 solution
|  Cl –
white AgCl (s)
|  6.0 M NH3
Ag(NH3)2+ + Cl –
|  Br–
ivory AgBr (s) + NH3, Cl –
|  S2O32–
Ag(S2O3)23– + Br – + NH3, Cl–
|  I–
brown AgI (s)
June 29
Data
Solid
AgCl
AgBr
AgI
Ksp
1.8e-10
5.0e-13
8.0e-17
Complex Kf
Ag(NH3)2+ 1.7e7
Ag(S2O3 ) 23– 2.9e13
19 Heterogeneous & complex equilibria
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Some complexes
Ag(NH3)2+,
Cu(CN)32- (Cu(I)), Cu(PPh3)2Br,
(NH3)2PtCl2 (cis & trans),
Ni(CN)53-,
Ligands (Lewis bases):
Negative ions, F-, Cl-, Br-, I-, OH-, CN-, SO42-, RCOO-, …
Neutral molecules: H2O, NH3, NR3, ROR, CO, C5H5, PR3, C5H5N, …
Multidentate ligands: H2NCH2CH2NH2, H2NCH2CH2NHCH2CH2NH2,
19 Heterogeneous & complex equilibria
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The heme and hemoglobin
(a) The structure of heme is a planar porphoryin ring with
iron at the center. (b) Four heme units and four coiled
polypeptide chains are bonded together in a molecule of
hemoglobin.
19 Heterogeneous & complex equilibria
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Amphoteric hydroxides
Amphoteric metal hydroxides react with both acids and bases.
Examples: (M = Fe, Zn)
M(H2O)4(OH)2 + H+ = M(H2O)5(OH)+
M(H2O)5(OH)+ + H+ = M(H2O)62+
M(H2O)4(OH)2 + OH– = M(H2O)3(OH)3 –
M(H2O)3(OH)3 – + OH– = M(H2O)2(OH)42 –
Aluminum hydroxide behave similarly:
Al(H2O)3(OH)3 + H+ = Al(H2O)4(OH)2+
Al(H2O)4(OH)2+ + H+ = Al(H2O)5(OH)2+
Al(H2O)5(OH)2+ + H+ = Al(H2O)63+
Al(H2O)3(OH)3 + OH – = Al(H2O)2(OH)4 –
Al(H2O)2(OH)4– + OH – = Al(H2O)(OH)52–
Al(H2O)(OH)52– + OH – = Al(OH)63 –
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Qualitative analysis of metal ions
group I
Ag+, Hg22+, Pb2+ 
chloride |
group II
Cu2+, Cd2+, Hg2+, Pb2+ 
As3+, Sb3+, Bi3+, Sn4+ |
acid insoluble sulfide |
group III Mn2+, Fe2+, Co2+, Ni2+, Zn2+ 
base insoluble sulfide |
Al(OH)3, Cr (OH)3 |
hydroxide |
group IV
Mg2+, Ca2+, Sr2+, Ba2+ 
ppt as carbonates or phosphates |
Mixture of metal ions
 add HCl
filtrate of soluble chloride
 add 0.3 M H+ & H2S
filtrate of soluble metal sulfide
| add OH- & H2S
 (NH4OH)
filtrate of soluble metal ions
| add CO32- or PO43|

filtrate of soluble metal ions
K+, Na+ group V
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Heterogeneous and complex ion equilibria
Summary
Calculate Ksp
Evaluate molar solubility (or in other units) from Ksp
Discuss concentrations of species in solution with two or more
electrolytes (common ion effect, pH effect, separation by ppt)
Predicts ppt (heterogeneous equilibria)
Describe ligands, metal ions, complexes (ions), and formation constants,
and dissociation constant
Apply complex formation to explain solubility
Be able to get out of traps (think in both directions)
19 Heterogeneous & complex equilibria
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