“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Introduction/
Context
Roman pipes at Pompei
Solution Equilibria: Solubility
In homes older than 1980 most of the plumbing
is lead pipe (Pb = plumbous). Even in newer
homes with copper pipe, solder joints are a
lead/tin alloy. Even without solder joints, many
of the faucet heads are machined with a 10-20%
lead content brass.
The limit on lead is set to be
<5 g Pb/109 g water
< 5 x10--9 g Pb/g water
< 5 ppb
What to do?
Solutions:
1.
2.
3.
Take out all plumbing
Place water filtration devices at
all outlets (sinks, showers, hoses).
Have the water department take
care of it somehow.
What do you think the average homeowner
prefers?
Coat the pipes from the inside out with
a dense impermeable quasi-permanent layer
= insoluble salt
What makes insoluble salts?
hint: same concepts as govern
what ions are or are not
spectators.
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Review Charge
Density
Jm
k  8.99 x10
C2
9
Energy electrostatic
r1
It’s all about charge
 q1 q 2 
 k

 d 
Charge on object 1 or 2, in coulombs
Distance between the objects
r2
d
r1  r2
 q1q2 

E el  k 
 r1  r2 
Review: Module 5
Ion
hydrogen
lithium
sodium
potasium
cesium
Symbol
D+
Li+
Na+
K+
Cs+
beryllium Be2+
magnesiumMg2+
calcium
Ca2+
strontium Sr2+
barium
Ba2+
oxide
sulfide
selenide
telluride
fluoride
chloride
bromide
iodine
O2S2Se2Te2FClBrI-
Review: Module 5
Are there differences in predicted electrostatic effect?
Charge
Radius (pm) Charge/radius
4
1
1 89.66666667
1 127.4285714
1
164
1 192.8333333
2
59
2
85
2
129.5
2 143.3333333
2
149
-2
-2
-2
-2
-1
-1
-1
-1
0.25
0.011152416
0.007847534
0.006097561
0.005185825
0.033898305
0.023529412
0.015444015
0.013953488
0.013422819
H+ should
behave
differently
Be2+ and Mg2+
should behave
differently
124.2 -0.01610306
O2- and, maybe, S2170 -0.011764706 should behave
184 -0.010869565 differently
207 -0.009661836
116.625
167
182
206
-0.008574491
-0.005988024
-0.005494505
-0.004854369
F- should
behave
differently
 q1q2 

E el  k 
 r1  r2 
D+
0.04
Be2+
Charge/(average ionic radius)
0.03
Mg2+
0.02
Li+
0.01
0
-2.5
-2
-1.5
-1
-0.5
F-
0
-0.01
O2 -0.02
Some of the points don’t fall
Within a “cluster”
Review: Module 5
Charge on ion
0.5
1
1.5
2
2.5
Low charge
Density
NOT an
Alpha dog
Review: Module 5
Charge is
Distributed
Throughout
The volume
Who precipitates and who stays soluble?
Depends upon who reacts with whom?
No
NO3Group 1 cations (1+)
NH4+
Clean
Cl-
Socks
SO42-
Oh
Card me
OH-
CO32-
PleeeeaSe!!
PO43-
S2-
Weak Electrostatic
Interaction = Soluble
Group 2 lg cations (2+)
Transition metal cations
(usually sm size 2+)
Low Charge Density
Intermediate Charge Density
High Charge Density
BaSO4 Mg(OH)2
AgCl
Strong Electrostatic Interaction
results in precipitation
Extremes
1. Low -Low charge density ion interactions - weak electrostatic energy Stay soluble
2. High -High charge density ion interactions - strong electrostatic energy Precipitate
3. High- Intermediate charge density ion interactions – generally strong electrostatic energy – precipitate
In Between
1. Low – High charge density ion interactions – generally weak electrostatic energy: Soluble with exceptions
2. Low – Intermediate charge density ion interactions – generally weak electrostatic energy: Soluble with
exceptions
3. Intermediate – Intermediate – charge density ion interactions – generally weak electrostatic energy: soluble
with exceptions
Who gives up and who holds onto a hydroxide?
No
Clean
NO3-
H+
Cl-
Socks
SO42-
Low Charge Density
Intermediate Charge Density
High Charge Density
Oh
Card me
OH-
CO32-
STRONG acids
PleeeeaSe!!
PO43-
WEAK acids
Group 1 cations (1+)
Strong
Bases
Group 2 cations (2+)
Weak bases are produced by an alternative manner
Review Acid Base Definitions: Module 6
S2-
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Solubility Product, Ksp
Mathematically Express These Concepts:
(Memorization Table is a short hand)
MX s  H2 O
Keq 
M aq  X aq


 M  X 

aq

aq
 MX  H O 
s

2


K so lub ility product  Keq H2 O  55Keq
  
Didn’t we learn
A trick about this?




  M  X 
 MX 
M aq X aq
s
K sp  M aq X aq
The smaller K, the less aquated ions, the
less soluble the material

aq

aq
Solubility Constants
Numbers are determined from measuring
the amount of stuff in solution.
Mm+ Xx-
Example 1: Calculate the Ksp of Bismuth sulfide
if there is 1.0x10-15 mol/L of the compound
in solution at 25 oC.
Bismuth is a post transition metal with
the electronic configuration of?:
Bi
s2d10p3
What do you think it will do to become
a cation?
Calculate the Ksp of Bismuth sulfide
if there is 1.0x10-15 mol/L of the compound
in solution at 25 oC.
Lose three e
Bi
s2d10p3
Bi3+
s2d10p0
The electron configuration on S is:
S
s2p4
What will it do to get to the noble gas?
Calculate the Ksp of Bismuth sulfide
if there is 1.0x10-15 mol/L of the compound
in solution at 25C.
Gain two e:
S
s2p4
S2-
s2p6
Formula?: Bi3+ with S2-
Bi2S3
Calculate the Ksp of Bismuth sulfide
if there is 1.0x10-15 mol/L of the compound
in solution at 25C.
Reaction?:
3
2
Bi2 S s 
2
Bi

3
S

aq
aq
Ksp   A  B
a
b
  S 
K sp  Bi
3
aq
2
2
aq
3
Now What?
What do we know/don’t know/want?
so lub ility of solid  s  10
. x10
Calculate the Ksp of Bismuth sulfide
if there is 1.0x10-15 mol/L of the compound
in solution at 25C.
 15
mole
L
2Bi3+aquated + 3S2-aquated
2
3
0
0
+2x
+3x
2.0x10-15 3.0x10-15
Bi2S3(solid)
1
solid
-x
1.0x10-15
stoic
Init
Change
Equil.
  S 
K sp  Bi
3
aq
2
2
aq
Ksp   2 x  3x
2
OJO!
3
Or you can do it:
3
Ksp  4 x 27 x  108x
2
3
Ksp  10810
. x10

15 5
5
Ksp   2.0x10
  30. x10 
15 2
15 3
 108
. x10 73
Calculate the Ksp of Bismuth sulfide
if there is 1.0x10-15 mol/L of the compound
in solution at 25C.
Ksp  108
. x10
 73
For an enormous list of Ksp:
http://www.northland.cc.mn.us/chemistry/solubility_products.htm
Example Calculation 2 Calculate Solubility,s, and
ion concentrations from Ksp
Candidates for water treatment for lead?
Ksp
PbF2
4x10-8
Criteria you will
PbCl2
1.6x10-5
use?
PbI2
1.4x10-8
PbSO4
1.3x10-8
PbCrO4 2x10-16
Solubility = s=
PbCO3
1.5x10-15
amount of compound
Pb(OH)2 1.2x10-15
that is soluble in
PbS
7x10-29
water
Pb3(PO4)2 1x10-54
Do our “rules” clue us to Ksp values?
stoic
Init
Change
Equil?

3Pb2+ + 2PO433
2
0
0
3x 2x
3x 2x
Pb3(PO4)2
1
solid
-x
-
Ksp  10
. x10  54  Pbaq2
  PO 
3
3
aq
2
x  s  6.2 x1012
Ksp  10
. x10
 54
  3x  2 x
Ksp  10
. x10
 54
 27 x 4 x  108x
5
5
3
10
. x10  54
 x s
108
9.27 x10  57  x  s
3
2
OJO!
2
Are we done?
No, need
[Pb2+]
5
 Pb   3x  36.2x10
 Pb   18. x10
2
aq
2
aq
11
12

Is this below the federal standards (5 ppb)?
Yes:


2
Pbaq
 18
. x10 11

  207 gPb 
1L
gPb

 11 molePb 
 12
. x10
. x10
 18
 3
  385


L   10 gwater   mole 
gwater
9
3




gPb
10
385
.
x
10
gPb
12
3
. x10

385
.
x
10
ppb
 385
 9 
9
gwater   10 

10 gwater
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Common Ion Effects
Will the actual amount of lead be more or
less than this value?
water
plant
PO43Pb3(PO4)2
LeC principle: Common ion effect
add continuously PO43
2
3
Pb3  PO4  2 ,s 
3
Pb

2
PO

aq
aq
Constant flow of phosphate will
suppress lead dissociation, value should be
even lower.
http://www.caruschem.com/phosphate_zpoly.htm
Name
formula
Zincorthophosphate

3
Zn3 POaq

2
Zn3  PO4  2
MM
386.05
K


spZn3  PO3 

aq 
2
 9.0x10  33
 Zn3 P2 O6
Zinc pyrophosphate
Zn2 P2 O7
304.685
 3lb   1gal   453.6 g   1mole  0.9336moles
s 

 0.9336 M







L
 gal   3.785 L   1lb   386.05 
sZn2 P2O7
3lbs

1gal
Water quality plants use either polyphosphate
Or pyrophosphate (NOT orthophosphate)
5
10
P3 O 
 P2 O74   2 H   PO43
4
7
P2 O
5
10
H2 O

 2 PO43  2 H 
H2 O
P3 O 
 3 PO43  4 H 
H2 O
polyphosphate
pyrophosphate
Example Calculation 3Common Ion What will be the solubility of lead if a constant
stream of 0.001 M phosphate is fed through the water system?
Notice the set up here
1. Looks like accounting for H+ from water
2. Called a “common ion” (ion from another rx)
Zn2P2O7
stoic
Init
Change
Equil?
Assume
Pb3(PO4)2
1
solid
-x
-
2Zn2+ 2PO430.001
3Pb2+ + 2PO433
2
0
0.001
+3x +2x
3x 0.001+2x
3x 0.001
What will be the solubility of lead if a constant stream of 0.001 M
phosphate is fed through the water system?
Zn2P2O7
Pb3(PO4)2
1
solid
-x
-
stoic
Init
Change
Equil?
Assume
  PO
K sp  1x10
 54
  Pb
Ksp  1x10
 54
  3x 10
2 3
3
3
aq


2
3
3 2
Ksp  1x10 54  27 x 3 10  6 
3
2Zn2+ 2PO430.001
3Pb2+ + 2PO433
2
0
0.001
3x 2x
3x 0.001+2x
3x 0.001
1x10  54
6  x
27 x10
3.7 x10  50  x
x  347
. x10 17
2 x  2347
. x1017   0.001?
 Pb   3x
 Pb   33.47 x10
. x10
 Pb   104
2
aq
2
aq
2
aq
 Pb 
2
aq
no phosphate
 17

 16
 18
. x10 11
Wilmette, Ill.
Proprietary license
Western suburbs
Chicago Tribune, 2001
Phosphate coating
1. lowers total volume
2. Creates friction
3. Increases energy
cost
4. Lowers life span
Suburbs want
compensation from
Chicago
LeC principle: Common ion effect
Can help ppt
Can prevent ppt
Addition of phosphate should suppress solubility
2
3
add continuously  Zn2 P2 O7  2Znaq
 2 POaq
 2H 
complete
2
3
Pb3  PO4  2 ,s 
3
Pb

2
PO

aq
aq
3
POaq
 H3O 
Removal of lead
2
Pb
aq
Under basic conditions
Should also enhance solubility


2
HPOaq
 H2 O
Removal of phosphate
Under acidic conditions
Should enhance
solubility

 
 OHaq   Pb OH  aq
Pb(OH)3PbCO3
s
American Waterworks
Association
Hydroxide (pH) vs Carbonate
Pb2+
[CO32-]
At intermediate pH
Pb(OH)2
Pb(OH)2 drops out soln
At high carbonate
Solubility, s,
PbCO3 should drop,
Depends on pH
BUT
not
very
insoluble
is high at High pH, (Pb(OH)3-)
Suppress effects
and high at low pH, Pb2+
With phosphate
If little dissolved carbonate NOTE
Scale change!
pH
MARCH 20/21, 2004
260 ppb Lead.
1200
Plot of amount of lead
In morning water
Number of observations
First Draw1000
800
600
400
200
Pb pipes
0
0
20
40
60
80
100
120
140
160
ppb Pb in water
1200
Second Draw
Pb pipe
Cu pipe
Brass fixture
Unknown pipes
Other
Number of observations
1000
800
600
400
200
0
0
20
40
60
80
100
120
140
160
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Qualitative
Analysis: Intro
PbS
Muriatic acid
“of or pertaining to brine
Or salt” HCl
Qualitative Analysis
First known text on chemical
(as opposed of alchemical) analysis
Qualitative Analysis
Oh
Card
me
Plea
S
hydroxide
Carbonate
OHCO3-2
Phosphate
Sulfide
PO43S2-
M 2  X x  
 precipitate

ClNH3
OH-
nLx 

M  OH  2 ,s
MS
MCO3
M 3  PO4  2
MLbn ,aq
Move back and forth between precipitate
And soluble species
Preceding example was:
How to get rid of Pb2+ by precipitation
as a phosphate
Next example:
How to bring it back into solution
a. With acid
b. With a ligand
1.
Dissolving Precipitates
Strong acid
HCr2 O7

H

 2
2 2 


CO
PbCO
Pb 2Pb

Cr
O
PbC
aqueous 2 7 3aqueous
aqueous

solid))
2 O73((solid
aqueous
LeChatlier’s
Acid dissolves: chromates; carbonates, sulfides,
1.
Dissolving Precipitates
Strong acid
HS 

H

LeChatlier’s
Pb 2  aqueous  S 2  aqueous 
 PbS ( solid )
Acid dissolves: chromates, carbonates; sulfides
Dissolving Precipitates
Strong acid
Complex forming reagents (:NH3, :OH-)
1.
2.
2
aq
Pb
 OH
 
aq 



Pb OH aq
Pb OH  aq  OHaq 
 Pb OH  2 aq

o
o


Pb OH 2 aq 
 Pb OH  2 ( s )
Pb OH  2 aq  OHaq 
 Pb OH  3aq
o

Pb OH  3aq  OHaq 
 Pb OH  4 aq

2
Complexes
Dissolving Precipitates
1.
Strong acid
2.
Complex forming reagents (:OH-)
1
0.9
Pbaq2
PbOHaq
Pb OH  2 aq
Pbtotal
Pbtotal
Pbtotal
o
Pb OH  4 aq
2
Pbtotal
Fraction
0.8
0.7
Pb OH  3aq
0.6
Pbtotal

0.5
0.4
0.3
Pb(OH) o2( aqueous)  Pb(OH) 2( solid )
0.2
Neutral can result in a solid
0.1
0
0
2
4
6
8
pH
10
12
14
Ligands like :NH3 and
:OH- can bring various precipitates into solution
Cation radius,CN
CN: 4tet;6oct
Ag+
Cu2+
71, 87
Zn2+
74, 88
Cd2+
92, 109
Ni2+
Al3+
67.5
Sb3+
90
Sn4+
Complex
Kf
Ag(NH3)2+
Cu(NH3)42+
Zn(NH3)42+
Cd(NH3)42+
Ni(NH3)62+
1.8x107
2x1012
3.6x108
2.8x107
9x108
Hold Coordination
Number constant
What do you observe?
Complex Kf
Zn(OH)42Cd(OH)42-
3x1014
1.2x109
Al(OH)4Sb(OH)4Sn(OH)62-
1x1033
Size matters: the larger
The cation (less charge
dense) the smaller the
Kf
ligand: :NH3
LeChatlier’s
Principle

Bring
Silver
Into solution
Ag aqueous  Cl

2 NH3,aqueous

aqueous


3 2 ( aqueous )
Ag ( NH )


AgCl( solid )
Example 4: Calculate moles AgCl dissolved in 1 L of
6.0 M NH3 at a temperature of 298 K
Don’t Know
s
Know
6.0 M NH3
reaction


AgCl( solid ) 
 Ag aqueous  Cl
aqueous
Ag  aqueous  2 NH3,aq 
 Ag NH3  2

Red Herring
298K
Krx
Ksp  18
. x1010
K f  17
. x107

AgCl( solid )  2 NH3,aq 
 Ag NH3  2 ,aq  Cl
aqueous

Kreaction  Ksp K f  18
. x10 10 17
. x107   31
. x10 3
stoic
Init
Change
Equil
AgCls
n.a.
[NH3]
2
6.0 M
-2x
6.0-2x
[Ag(NH3)2+]
1
0
x
x
[Cl-]
1
0
x
x
Example: Calculate moles AgCl dissolved in 1 L of
6.0 M NH3 at a temperature of 298 K
AgCls
n.a.
stoic
init
change
equil
[NH3]
2
6.0 M
-2x
6.0-2x
Ag NH   Cl 



Kreaction  31
. x10
31
. x10
3

3
x x
 6.0  2 x 2
31
. x10  3 
x
6.0  2 x


3 2
NH3

[Ag(NH3)2+]
1
0
x
x
[Cl-]
1
0
x
x
X= moles complex
= moles AgCl dissolved
2
0.0556(6.0  2 x)  x
0.334  01112
.
x x
0.334  11112
.
x
x  0.300216
Alanah:
Check
Moles vs
molarity
We can calculate the solubility of AgCl as a function of the ammonia concentration
Kreaction  31
. x10


3

Ag NH3  2  Cl  

 NH 

2
3
 NH 
3 init
Krx 
 2x

3 init
Krx
 x
init

5
4
x
 NH 
1  2

6
2
 2x

s (M)
Krx 
x x 

Krx NH 3
3
2

Krx
 NH 
3 init

 2x  x
1
0

Krx NH3


 x  2x
Krx


Krx NH3


 x 1  2 Krx

init
init


-4
-2
0
2
4
6
8
p[NH3]
AgCl becomes soluble
Around 1 M ammonia
10
12
14
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Qualitative Analysis:
Cl to separate Pb, Hg22+, Ag
Using pH and
complexation to
Separate Ions
For Qualitative
Analysis
Separation in lab is not always the same
As in the text.
Text starts with 6 M Cl- = pCl = -.778
The separation below
Corresponds to
What goes on in lab
Questions:
1. How does Cl- do the first step?
 Pb
aq ,all forms
1
4 
0.9
 PbCl 
2
4 aq
 Pb
aq ,all forms
   Pb
2
aq

   PbCl
 
3 
0.8
Fraction
0.7
1 
 PbCl 
1
aq
 Pb
aq ,all forms
0.6


aq



  PbCl o  PbCl   PbCl 2
2 aq
3 aq
4 aq

 PbCl 

3 aq
Pbaq ,all
forms

o 

 Pb 
 Pb
2
aq
aq ,all forms


0.5
0.4
0.3
0.2
2 
0.1
 PbCl 
 Pb
0
2 aq
aq ,all forms

0
-2
-1
0
pCl  0
[Cl  ]  1
1
2
pCl
3
4
5
Questions:
1. How does Cl- do the first step?
2. Why can we get Pb2+ from Hg and Ag
with hot water?
Hot water
AgCl(s); Hg2Cl2(s)
Pb2+
2

Pbaq
 Claq 
 PbClaq
 
0
PbClaq  Claw
 PbCl2 ,aq
Kf1

PbCl20,aq  Claq 
 PbCl3,aq
Kf 3
2
PbCl3,aq  Claq 
 PbCl4 ,aq
Kf 4

0
Agaq
 Claq 
 AgCl,aq
1
AgCl,0aqs  Claq 
 AgCl2 ,aq
2
AgCl2,aqs  Claq 
AgCl

3,aq
PbCl20,s
Kf 2
Kf 1
Kf 2
Kf 3
AgCl,0s
1
 Ag 
 Ag 
2
0.9
total
0.8
0.7AgCl 
 Ag 
0.6
 AgCl 
 Ag 
fraction
2
3

2
total
total
0.5
0.4
0.3
 AgCl 
 Ag 
o
0.2
0.1
total
0
-2
0
2
4
6
pCl
pCl  2;  Cl

  0.01
8
10
12
14
pCl  0
For lead it was:
[Cl  ]  1
2

Pbaq
 Claq 
 PbClaq
 
0
PbClaq  Claw
 PbCl2 ,aq
Kf1
PbCl20,s
Kf 2

PbCl20,aq  Claq 
 PbCl3,aq
Kf 3
2
PbCl3,aq  Claq 
 PbCl4 ,aq
Kf 4

0
Agaq
 Claq 
 AgCl,aq
1
AgCl,0aqs  Claq 
 AgCl2 ,aq
2
AgCl2,aqs  Claq 
AgCl

3,aq


2 Hgaq
 Claq 
 Hg 2 Claq
Hg2 Claq  Claq 
 Hg 2 Cl2

Hg2 Cl2  Claq 
 Hg 2 Cl3,aq
2
Hg2 Cl3,aq  Claq 
 Hg 2 Cl4 ,aq
Kf 1
AgCl,0s
Kf 2
Kf 3
Kf1
Kf 2
Kf 3
Kf 4
Hg2 Cl2( solid )
1
0.9
 Hg Cl 
0.8
2
4 ,aq
2
 Hg 


2
total
2
total

0.6

0.5
0.4
Hg 2 Cl3,aq
 Hg 

 Hg 
2
total
 Hg 
0.7
fraction

Hg 2 Cl20,aq
2
Hgaq

Hg2 Cl,0aq
 Hg 

2
total
2
total
0.3
0.2
0.1
0
-2
0
2
4
6
pCl
pCl  35
. ;  Cl    0.00035
8
10
12
14
1
Pb/Cl plot
0.9
0.8
Fraction
0.7
0.6
0.5
soluble
0.4
0.3
0.2
0.1
0
-2
0
2
4
6
8
10
12
14
pCl
1
0.9
Ag/Cl plot
0.8
Lower chloride concentration
by placing ppt in hot water.
PbCl2 dissolves;
AgCl and Hg2Cl2 remain
insoluble
fraction
0.7
0.6
0.5
0.4
0.3
0.2
insoluble
0.1
0
-2
0
2
4
6
8
10
12
14
pCl
1
Hg/Cl plot
0.9
0.8
fraction
0.7
insoluble
0.6
0.5
0.4
0.3
0.2
0.1
0
-2
0
2
4
6
pCl
8
10
12
14
Questions:
1. How does Cl- do the first step?
2. Why can we get Pb2+ from Hg and Ag
with hot water?
3. How do we ppt. Pb2+?
Hot water
AgCl(s); Hg2Cl2(s)
Pb2+
CrO42,aqueous
ppt
2
Pbaqueous
 CrO42,aqueous 
 PbCrO4 , solid
Candidates for precipitating Pb2 from supernatent?
PbF2
PbCl2
PbI2
PbSO4
PbCrO4
PbCO3
Pb(OH)2
PbS
Pb3(PO4)2
Ksp, Pb
4x10-8
1.6x10-5
1.4x10-8
1.3x10-8
2x10-16
1.5x10-15
1.2x10-15
7x10-29
1x10-54
Need very insoluble species,
that doesn’t bring along Cd
or Zn
Cd
-
?
5.5x10-13
6.45x10-6
_
_
Zn
- OH
, CO32- bring Cd2+
_
S2-, PO43- bring Zn2+
_
6.3x10-17
1.99x10-25
3.8x10-36
Makes a very
nice, older,
chrome pigment!!
Example Calculation 5. Before lead in paint was discontinued, lead
chromate was a common pigment in yellow paint. A 1.0 L solution is
prepared by mixing 0.50 mg of lead nitrate with 0.020 mg of
potassium chromate. Will a precipitate form?
P is like Q
P   Pb
2
CrO   K
2
4
sp
(2 x10
 16
P  Ksp ; ppt forms
)?


 1g 
mole

 0.50mgPb NO3   3 
2  10 mg 
 331gPb NO3  2 
 151
. x10 6 MPb 2 
1L




 1g 
mole

 0.020mgK2 CrO4  3 
 10 mg 
 194.2 gK2 CrO4 
 1029
.
x10 7 MCrO42 
1L
P  151
. x10 6 1029
. x10 7   1554
. x10 13  Ksp (2 x10 16 )
Questions:
1. How does Cl- do the first step?
2. Why can we get Pb2+ from Hg and Ag
with hot water?
3. How do we ppt. Pb2+?
4. What is the purpose of NH3 to get Ag+?
Hot water
AgCl(s); Hg2Cl2(s)
Pb2+
CrO42,aqueous
NH3
HgNH2Cl(s)
ppt
Ag(NH3)6+


Agaqueous
 NH3aqueous 
 AgNH3aqueous
AgNH3aqueous  NH3aqueous 
 Ag NH3  2 aqueous

We did an example in which we calculated the
Solubility of AgCl in the presence of ammonia

( solid ) 
AgCl
Ag

aqueous
Ag
 2 NH
AgCl( solid )  2 NH

aqueous

3,aq 

3,aq 
 Cl
Ksp  18
. x1010

Ag NH3  2
aqueous

Ag NH3  2,aq  Cl

K f  17
. x107
Kreaction  31
. x10  3

aqueous
Which number is larger?
Which has the higher solubility?
water or ammonia
6
5
s (M)
4

3
2

Krx NH 3
1  2
Krx

init

 x
1
0
-4
-2
0
2
4
6
p[NH3]
8
10
12
14
1.2
Ag(NH3)2+
1
Ag+
fraction
0.8
0.6
Hg22+ does not
form complexes
with NH3
0.4
Ag(NH3)+
0.2
0
0
2
4
6
pNH3
8
10
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Qualitative Analysis:
S to separate Cu 2+, Hg+,
Bi3+, Cd2+
Questions:
1. How does H2S do the second step?
Cu2+
Bi3+
Hg2+
Cd2+
Al3+
Fe3+
Cr3+
Mn2+
Zn2+
Ba2+
Ca2+
logKsp
-48.5
-52.7
-27.0
S2-
Strong precipitators with S2-,
will ppt at very low [S2-]
Weak Precipitators with [S2-]
Will ppt only at higher [S2-]
How can we control [S2-]
Easily?


H2 Saq 
 Haq  HS aq
-10.5
-24.7
-

2
HSaq 
 Haq  Saq

2
H2 Saq 
 2 Haq  S aq
Ka 1
Ka 2
Ka 1 Ka 2
Control [S2-] through pH
H S
2
Fraction H2S, HS-, and S2-
1
H S
0.8
2
H S 
   HS   S 
2
aq
H S
aq

aq
2
aq
2
S 
   HS   S 
2
aq
aq

aq
2
aq
 HS 
   HS   S 

aq
aq

aq
2
aq
0.6
0.4
0.2
0
0
2
4
6
8
10
pH
Who ppts here?
Very insoluble
MS species
12
14
16
18
Who ppt here?
More soluble MS species
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Qualitative Analysis:
S to separate Zn 2+, Co+2,
Mn2+, Ni2+
Questions:
1. How does (NH4)2S do the first step?
2. Why is the solution basic (pH 8) in
next step?
Cu2+
Bi3+
Hg2+
Cd2+
Al3+
Fe3+
Cr3+
Mn2+
Zn2+
Ba2+
Ca2+
logKsp S2-48.5
-52.7
Strong precipitators with S2-,
will ppt at very low [S2-]
-27.0
-10.5
-24.7
-
Intermediate: need relatively
larger values of [S2-] to
ppt
Do not ppt at any value of
[S2-]
2M NH4+; 5 MNH3; (NH4)2S
Ppt out sulfides
Buffer pH to basic solution of 9.599
to ppt hydroxides

NH4 
 NH3  H
pH  pKa  log
 A
 HA
pH   log(5.6 x10
 10
5M
)  log
2M
pH  9.25  0.397  9.65
Ka  5.6x10 10
Al3+, Cr3+ Fe3+ come along
as a hydroxides
 Al 
1.2
 Al
1
3
aq
total ,aq
 Al OH  
 Al 
 Al OH   Al OH   Al OH 
 Al   Al 
 Al 

2,aq
2
aq

total ,aq
total ,aq
o
3,aq
total ,aq

4,aq
total ,aq
Al OH  3,s
o
Fraction
0.8
0.6
0.4
0.2
0
4
6
8
10
pH
12
14
Re-dissolve
In acid
1.
2.
Dissolving Precipitates
Strong acid
Complex forming reagents (:NH3, :OH-)
HCO3

H

LeChatlier’s
Pb 2  aqueous  CO32  aqueous 
 PbCO3( solid )
Acid dissolves: carbonates; sulfides, hydroxides
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Qualitative Analysis:
Separating Zn2+,
And Al3+, from the sulfides
By OH complexation
Re-dissolve
In acid
Increase pH
Example Calculation 6:
Ag+
Zn2+
Mn2+
Cr3+
Al3+
Fe3+
Sn2+
Kf
1x104
4.6x1017
2.51x103
6.3x103
1x1033
1.99x1022
2.5x1025
Ag(OH)2Zn(OH)42Mn(OH)+
Cr(OH)2+
Al(OH)4Fe(OH)2+
Sn(OH)3-

AgOH s  Ag aq
 OHaq
KspKf
1.99x10-5
145
3.96x10-10
1.28
4.6
7.96x10-16
3.0x108
K sp  199
. x10  9

Ag aq
 2OHaq  ag OH  2 ,aq

AgOHs  OHaq  Ag OH  2,aq

Ksp
Ag(OH)
1.99x10-9
Zn(OH)2
3.16x10-16
Mn(OH)2(s) 1.58x10-13
Cr(OH)3(s) 6.3x10-31
Al(OH)3(s) 4.6x10-33
Fe(OH)3(s) 4x10-38
Sn(OH)2(s) 1.2x10-17
K f  1x10 4
Knet  Ksp K f  199
. x10  9 1x104   199
. x10 5
Which compounds can be brought into solution by raising the pH?
 Al 
1.2
 Al
1
3
aq
total ,aq
 Al OH  
 Al 
 Al OH   Al OH   Al OH 
 Al   Al 
 Al 

2,aq
2
aq

total ,aq
total ,aq
o
3,aq
total ,aq

4,aq
total ,aq
Al OH  3,s
o
Fraction
0.8
0.6
0.4
0.2
0
4
6
8
10
pH
12
14
Zn 
1
0.9
Zn
2
aq
total ,aq
Zn OH 
Zn 
2
4 ,aq
Zn OH 
Zn 
o
2,aq

total ,aq
total ,aq
Fraction Zn species
0.8
Zn OH  2,s
0.7
o
0.6
0.5
0.4
Zn OH 
Zn 

3,aq
0.3
total ,aq
Zn OH  
Zn 

aq
0.2
0.1
total ,aq
0
0
2
4
6
8
pH
10
12
14
How do we distinguish Zn from Al?
Between Al(OH)4- and Zn(OH)4only Zn has complexation with
:NH3
Drop pH to re-ppt Al(OH)3, while
simultaneously forming soluble Zn(NH3)42+
NH 4


NH 3  H 
pH  pKa  log
Ka  5.6 x10 10
 A
 HA
3M
pH   log(5.6 x10 )  log
12 M
pH  9.25   0.602  8.65
 10
3
12
M NH4+
M NH3
1.
2.
Dissolving Precipitates
Strong acid
Complex forming reagents (:NH3, :OH-)
Zn 2  aqueous  2OH  aqueous 
 Zn OH  2 ( solid )

4 NH3

Zn( NH3 ) 24 ( aqueous)
1.2
Zn2+
fraction Zn complexes
Zn(NH13)42+
0.8
0.6
0.4
0.2
0
-2
-1
0
1
2
pNH3
12 M NH3
3
4
5
6
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Qualitative Analysis:
PO43- to separate Ba 2+,
Ca+2, Mg2+
CN6
Na+
K+
NH4+
Mg2+
Ca2+
Ba2+
r(pm)
116
152
q/r
0.0086
0.0065
Ksp
-
86
114
149
0.023
0.0175
0.013
Ca3(PO4)2
Ba3(PO4)2
1.2x10-29
3.4x10-23
“A” students work
(without solutions manual)
~7 problems/night.
Solubility
What you need
To know
Summary Points
Complexation vs Solubility
Complexation based on electrostatic attraction
lone pairs for central cation
Size matters!; Charge matters!
Can result in multiple points of binding
Can result in charge -, o, +
Based on charge can result in precipitation
Use to manipulate and separate elements (biology too!)
Ksp = solubility product; large - # implies not soluble
Kf = formation constant; large +# implies strong binding
Calculations proceed similarly to Ka
EXCEPT – STOICHIOMETRY is trickier
Ba3(PO4)2
“A” students work
(without solutions manual)
~7 problems/night.
END