Roman pipes at Pompei
Solution Equilibria: Solubility
“A” students work
(without solutions manual)
~ 10 problems/night.
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.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
The limit on lead is set to be
<5 g Pb/109 g water
< 5 x10--9 g Pb/g water
< 5 ppb
Introduction/
Context
Solutions:
1.
2.
3.
What to do?
Coat the pipes from the inside out with
a dense impermeable quasi-permanent layer
Take out all plumbing
Place water filtration devices at
all outlets (sinks, showers, hoses).
Have the water department take
care of it somehow.
= insoluble salt
What makes insoluble salts?
hint: same concepts as govern
what ions are or are not
spectators.
What do you think the average homeowner
prefers?
1
k = 8.99 x10 9
“A” students work
(without solutions manual)
~ 10 problems/night.
Jm
C2
Energyelectrostatic
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
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
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
r1 + r2
Review Charge
Density
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
Coulomb’s Law
Review: Module 5
Are there differences in predicted electrostatic effect?
Charge
Radius (pm) Charge/radius
1
4
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
D+
0.04
H+ should
behave
differently
Be2+
0.03
Charge/(average ionic radius)
Ion
⎛ q1 q 2 ⎞
⎟
E el = k ⎜
⎝ r1 + r2 ⎠
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
Mg2+
0.02
Li+
0.01
0
-2.5
-2
-1.5
-1
-0.5
F-
0
-0.01
O2 -
F- should
behave
differently
-0.02
Some of the points don’t fall
Within a “cluster”
⎛ qq ⎞
E el = k ⎜ 1 2 ⎟
⎝ r1 + r2 ⎠
Review: Module 5
2
Charge on ion
0.5
1
1.5
2
2.5
Who precipitates and who stays soluble?
Depends upon who reacts with whom?
No
NO3Group 1 cations (1+)
NH4+
Clean
Cl-
Low charge
Density
NOT an
Alpha dog
Charge is
Distributed
Throughout
The volume
No
Clean
Cl-
NO3H+
Socks
SO42-
Low Charge Density
Intermediate Charge Density
High Charge Density
Oh
Card me
OH-
CO32-
STRONG acids
Card me
OH-
CO32-
PleeeeaSe!!
PO43-
S2-
BaSO4 Mg(OH)2
AgCl
Strong Electrostatic Interaction
results in precipitation
“A” students work
(without solutions manual)
~ 10 problems/night.
PleeeeaSe!!
PO43-
SO42-
Oh
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
Review: Module 5
Who gives up and who holds onto a hydroxide?
Socks
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
S2-
WEAK acids
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Group 1 cations (1+)
Strong
Bases
Office Hours Th&F 2-3:30 pm
Group 2 cations (2+)
Module #19:
Precipitation Reactions
Weak bases are produced by an alternative manner
Solubility Product, Ksp
Review Acid Base Definitions: Module 6
3
Mathematically Express These Concepts:
(Memorization Table is a short hand)
MX s + H2 Ol
Keq =
M aq+ + X aq−
→
←
[ M ][ X ]
+
aq
−
aq
Didn’t we learn
A trick about this?
[ MX ][ H O ]
l
2
s
[
]
K so lub ility product = K eq H 2 Ol = 55K eq =
[ ][ X ]
K sp = M
+
aq
Solubility Constants
Numbers are determined from measuring
the amount of stuff in solution.
[ M ][ X ] = [ M ][ X ]
[ MX ]
+
aq
−
aq
+
aq
−
aq
s
−
aq
Mm+ Xx-
The smaller K, the less aquated ions, the
less soluble the material
Example 1: 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
Bismuth is a post transition metal with
the electronic configuration of?:
Bi
s2d10p3
Bi3+
s2d10p0
The electron configuration on S is:
Bi
s2d10p3
S
What do you think it will do to become
a cation?
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 25 oC.
Calculate the Ksp of Bismuth sulfide
if there is 1.0x10-15 mol/L of the compound
in solution at 25C.
4
Reaction?:
Gain two e:
S
s2p4
S2-
s2p6
3+
2−
Bi2 S s →
← 2 Bi aq + 3S aq
a
b
K sp = [ A] [ B]
[ ][ ]
3+
K sp = Biaq
Formula?: Bi3+ with S2-
3
Now What?
so lub ility of solid ≡ s = 10
. x10 −15
Calculate the Ksp of Bismuth sulfide
if there is 1.0x10-15 mol/L of the compound
in solution at 25C.
stoic
Init
Change
Equil.
[ ] [S ]
2
2−
aq
2Bi3+aquated + 3S2-aquated
2
3
0
0
+2x
+3x
-15
2.0x10
3.0x10-15
K sp = 4 x 27 x = 108 x
http://www.northland.cc.mn.us/chemistry/solubility_products.htm
Or you can do it:
K sp = [ 2.0 x10 −15 ] [ 3.0 x10 −15 ]
2
3
For an enormous list of Ksp:
OJO!
3
2
3
K sp = [ 2 x ] [ 3x ]
5
K sp = 108(10
. x10 −15 ) = 108
. x10 − 73
3
5
Calculate the Ksp of Bismuth sulfide
if there is 1.0x10-15 mol/L of the compound
in solution at 25C.
mole
L
Calculate the Ksp of Bismuth sulfide
if there is 1.0x10-15 mol/L of the compound
in solution at 25C.
Bi2S3(solid)
1
solid
-x
1.0x10-15
2
S aq2 −
What do we know/don’t know/want?
Bi2S3
3+
K sp = Biaq
2
K sp = 108
. x10 − 73
5
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
-8
use?
PbI2
1.4x10
-8
PbSO4
1.3x10
PbCrO4 2x10-16
Solubility = s=
-15
PbCO3
1.5x10
amount of compound
Pb(OH)2 1.2x10-15
that is soluble in
PbS
7x10-29
water
-54
Pb3(PO4)2 1x10
stoic
Init
Change
Equil?
K sp = 10
. x10 − 54 = Pbaq2 +
2+
aq
3
3−
aq
2
x = s = 6.2 x10 −12
OJO!
5
5
10
. x10 − 54
= x= s
108
9.27 x10 − 57 = x = s
Are we done?
No, need
[Pb2+]
K sp = 10
. x10 − 54 = 27 x 3 4 x 2 = 108 x 5
Is this below the federal standards (5 ppb)?
[ Pb ] = 18. x10
] [ PO ]
3
2
K sp = 10
. x10 − 54 = [ 3x ] [ 2 x ]
Do our “rules” clue us to Ksp values?
Yes:
[
3Pb2+ + 2PO433
2
0
0
3x 2x
3x 2x
Pb3(PO4)2
1
solid
-x
-
[ Pb ] = 3x = 3(6.2 x10
[ Pb ] = 18. x10
2+
aq
2+
aq
− 12
)
− 11
“A” students work
(without solutions manual)
~ 10 problems/night.
− 11
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
⎞ ⎛ 207 gPb ⎞
molePb ⎞ ⎛
1L
gPb
⎛
. x10 −11
. x10 −12
⎟ = 385
⎜ 18
⎟⎜
⎟⎜
⎝
L ⎠ ⎝ 10 3 gwater ⎠ ⎝ mole ⎠
gwater
Office Hours Th&F 2-3:30 pm
⎛
gPb ⎞ ⎛ 10 9 ⎞ 385
. x10 − 3 gPb
. x10 −12
. x10 − 3 ppb
= 385
⎜ 385
⎟⎜ 9⎟ =
gwater ⎠ ⎝ 10 ⎠
⎝
10 9 gwater
Module #19:
Precipitation Reactions
Common Ion Effects
6
Will the actual amount of lead be more or
less than this value?
LeC principle: Common ion effect
add continuously PO43−
water
plant
2+
3−
Pb3 ( PO4 ) 2 ,s →
← 3 Pbaq + 2 POaq
PO43Pb3(PO4)2
Constant flow of phosphate will
suppress lead dissociation, value should be
even lower.
http://www.caruschem.com/phosphate_zpoly.htm
Name
formula
Zinc orthophosphate
(
Zn3 POaq3−
)
2
Zn3 ( PO4 ) 2
MM
386.05
K
⎛
⎞
spZn3 ⎜ PO 3− ⎟
⎝
aq ⎠
2
= 9.0 x10 − 33
Water quality plants use either polyphosphate
Or pyrophosphate (NOT orthophosphate)
= Zn3 P2 O6
2
P3 O105− ⎯⎯⎯
⎯→ P2 O74 − + 2 H + + PO43−
H O
2
P2 O74 − ⎯⎯⎯
⎯→ 2 PO43− + 2 H +
H O
Zinc pyrophosphate
Zn2 P2 O7
304.685
sZn2 P2 O7 =
2
P3 O105− ⎯⎯⎯
⎯→ 3 PO43− + 4 H +
H O
3lbs
1gal
⎡ 3lb ⎤ ⎡ 1gal ⎤ ⎡ 453.6 g ⎤ ⎡ 1mole ⎤ 0.9336moles
= 0.9336 M
s= ⎢
⎥⎢
⎥⎢
⎥⎢
⎥=
L
⎣ gal ⎦ ⎣ 3.785 L ⎦ ⎣ 1lb ⎦ ⎣ 386.05 ⎦
7
polyphosphate
pyrophosphate
What will be the solubility of lead if a constant stream of 0.001 M
phosphate is fed through the water system?
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?
Zn2P2O7
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
-
stoic
Init
Change
Equil?
Assume
2Zn2+ 2PO430.001
2+
3Pb + 2PO433
2
0
0.001
+3x +2x
3x 0.001+2x
3x 0.001
Pb3(PO4)2
1
solid
-x
-
[
K sp = 1x10 − 54 = [ Pb 2 + ] POaq3−
3
3
K sp = 1x10 − 54 = [ 3x ] (10 − 3 )
]
2
3
2Zn2+ 2PO430.001
2+
3Pb + 2PO433
2
0
0.001
3x 2x
3x 0.001+2x
3x 0.001
1x10 − 54
= x
27 x10 − 6
2
K sp = 1x10 − 54 = 27 x 3 (10 − 6 )
3
3.7 x10 − 50 = x
x = 3.47 x10 −17
2 x = 2(3.47 x10 −17 ) < 0.001?
Wilmette, Ill.
[ 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
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
Proprietary license
8
PbCO3
Addition of phosphate should suppress solubility
s
2+
add continuously → Zn2 P2 O7 ⎯⎯⎯⎯⎯→ 2 Znaq
+ 2 POaq3− + 2 H +
complete
[CO32-]
At intermediate pH
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!
2+
3−
Pb3 ( PO4 ) 2 ,s →
← 3 Pbaq + 2 POaq
Removal of lead
Pbaq2+
Under basic conditions
Should also enhance solubility
Hydroxide (pH) vs Carbonate
Pb2+
pH
Pb(OH)2
POaq3− + H 3 O +
→
←
American Waterworks
Association
Pb(OH)3-
LeC principle: Common ion effect
Can help ppt
Can prevent ppt
HPOaq2 − + H 2 Ol
Removal of phosphate
Under acidic conditions
Should enhance
solubility
+
+ OHaq− →
← + Pb( OH ) aq
1200
Plot of amount of lead
In morning water
Number of observations
First Draw1000
MARCH 20/21, 2004
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
260 ppb Lead.
20
40
60
80
100
ppb Pb in water
9
120
140
160
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Qualitative
Analysis: Intro
PbS
Qualitative Analysis
Muriatic acid
“of or pertaining to brine
Or salt” HCl
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
First known text on chemical
(as opposed of alchemical) analysis
Move back and forth between precipitate
And soluble species
10
Preceding example was:
How to get rid of Pb2+ by precipitation
as a phosphate
Dissolving Precipitates
Strong acid
1.
HCr2 O7−
Next example:
How to bring it back into solution
a. With acid
b. With a ligand
↑↓
H+
+
−
+ 2+
→→
+ 2CO
PbCO
Pb 2Pb
O72 −32aqueous
+ Cr
aqueous
aqueous
← ←PbC
solid))
2 O73((solid
aqueous
LeChatlier’s
Acid dissolves: chromates; carbonates, sulfides,
1.
Dissolving Precipitates
Strong acid
Dissolving Precipitates
Strong acid
Complex forming reagents (:NH3, :OH-)
1.
2.
Pbaq2+ + OH aq− →
← Pb( OH ) aq
+
Pb( OH ) aq + OH aq− →
← Pb( OH ) 2 aq
+
HS −
↑↓
aqueous
Pb( OH ) 2 aq →
← Pb( OH ) 2 ( s )
Pb( OH ) 2 aq + OHaq− →
← Pb( OH ) 3aq
o
H+
+
Pb 2 + aqueous + S 2 −
o
Pb( OH ) 3aq + OH aq− →
← Pb( OH ) 4 aq
−
LeChatlier’s
→
←
−
PbS ( solid )
2−
Complexes
Acid dissolves: chromates, carbonates; sulfides
11
o
Dissolving Precipitates
1.
Strong acid
2.
Complex forming reagents (:OH-)
1
0.9
Pbaq2+
PbOH aq+
Pb( OH ) 2 aq
Pbtotal
Pbtotal
Pbtotal
Ligands like :NH3 and
:OH- can bring various precipitates into solution
o
Pb( OH ) 4 aq
−2
Pbtotal
Fraction
0.8
0.7
Pb( OH ) 3aq
0.6
Pbtotal
−
0.5
0.4
0.3
Cation radius,CN
CN: 4tet;6oct
Ag+
Cu2+
71, 87
Zn2+
74, 88
Cd2+
92, 109
Ni2+
3+
Al
67.5
Sb3+
90
Sn4+
Pb(OH ) 2o ( aqueous ) ⇔ Pb(OH ) 2 ( solid )
0.2
Kf
Ag(NH3)2+
Cu(NH3)42+
Zn(NH3)42+
Cd(NH3)42+
Ni(NH3)62+
1.8x107
2x1012
3.6x108
2.8x107
9x108
Complex Kf
Hold Coordination
Number constant
Neutral can result in a solid
0.1
Complex
2
4
6
8
10
12
3x1014
1.2x109
Al(OH)4Sb(OH)4Sn(OH)62-
1x1033
Size matters: the larger
The cation (less charge
dense) the smaller the
Kf
What do you observe?
0
0
Zn(OH)42Cd(OH)42-
14
pH
Example 4: Calculate moles AgCl dissolved in 1 L of
6.0 M NH3 at a temperature of 298 K
ligand: :NH3
Know
6.0 M NH3
LeChatlier’s
Principle
Bring
Silver
Into solution
Ag ( NH 3 ) 2+( aqueous )
Ag
+
aqueous
+ 2 NH
→
3,aq ←
Red Herring
298K
Krx
reaction
+
−
AgCl( solid ) →
← Ag aqueous + Cl
Ag + aqueous + Cl − aqueous →
← AgCl( solid )
+
2 NH 3,aqueous
↑↓
Don’t Know
s
Ag( NH 3 ) 2
aqueous
+
−
AgCl( solid ) + 2 NH 3,aq →
← Ag ( NH 3 ) 2 ,aq + Cl
K sp = 18
. x10 −10
K f = 17
. x10 7
+
aqueous
Kreaction = K sp K f = (18
. x10 −10 )(17
. x10 7 ) = 31
. x10 −3
stoic
Init
Change
Equil
12
AgCls
n.a.
[NH3]
2
6.0 M
-2x
6.0-2x
[Ag(NH3)2+]
1
0
x
x
[Cl-]
1
0
x
x
AgCls
n.a.
stoic
init
change
equil
[NH3]
2
6.0 M
-2x
6.0-2x
[ Ag( NH ) ][ Cl ]
=
+
Kreaction = 31
. x10 − 3
31
. x10 − 3 =
[
x[ x ]
[ 6.0 − 2 x]
31
. x10 − 3 =
x
6.0 − 2 x
2
−
3 2
NH 3
]
[Ag(NH3)2
1
0
x
x
+]
[Cl-]
We can calculate the solubility of AgCl as a function of the ammonia concentration
[ Ag( NH ) ][ Cl ]
+
−
(
3 2
Kreaction = 31
. x10 − 3 =
[ NH ]
2
3
1
0
x
x
K )[ NH ]
(1 + 2 K )
rx
3 init
= x
rx
Krx =
x[ x]
[[ NH ]
3 init
Krx =
X= moles complex
= moles AgCl dissolved
− 2x
]
3 init
5
4
x
[[ NH ]
6
2
− 2x
]
s (M)
Example: Calculate moles AgCl dissolved in 1 L of
6.0 M NH3 at a temperature of 298 K
3
2
(
2
0.0556(6.0 − 2 x ) = x
Krx
)[[ NH ]
3 init
]
− 2x = x
1
0
0.334 − 01112
.
x= x
(
Krx NH 3
)[
0.334 = 11112
.
x
x = 0.300216
(
Krx NH 3
)[
]
= x + 2x
Krx
)
]
= x 1 + 2 Krx
)
init
init
(
(
Using pH and
complexation to
Separate Ions
For Qualitative
Analysis
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Qualitative Analysis:
Cl to separate Pb, Hg22+, Ag
13
-4
-2
0
2
4
6
8
p[NH3]
AgCl becomes soluble
Around 1 M ammonia
10
12
14
Questions:
1. How does Cl- do the first step?
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
[ Pb
aq ,all forms
1
α4 ≡
0.9
] = ⎡⎣⎢ Pb
2+
aq
[ PbCl ]
[ Pb
]
⎤ + ⎡ PbCl
⎦⎥ ⎢⎣
2−
4 aq
aq ,all forms
α3 ≡
[
[ PbCl ]
[ Pb
0.7
α1 ≡
][
][
⎤ + PbCl o + PbCl − + PbCl 2 −
2 aq
3 aq
4 aq
⎦⎥
−
3 aq
aq ,all forms
0.8
Fraction
+
aq
αo ≡
]
Questions:
1. How does Cl- do the first step?
2. Why can we get Pb2+ from Hg and Ag
with hot water?
]
[ Pb ]
[ Pb
]
2+
aq
aq ,all forms
[ PbCl ]
[ Pb
]
1+
aq
Hot water
aq ,all forms
0.6
0.5
AgCl(s); Hg2Cl2(s)
0.4
0.3
0.2
α2 ≡
0.1
[ PbCl ]
[ Pb
]
0
2 aq
aq ,all forms
0
-2
-1
0
pCl = 0
1
2
3
4
5
pCl
[Cl − ] = 1
14
Pb2+
+
Pbaq2 + + Claq− →
← PbClaq
Kf1
0
− →
PbClaq+ + Claw
← PbCl2 ,aq
PbCl
−
3,aq
+ Cl
+ Cl
− →
aq ←
− →
aq ←
−
3,aq
Kf 3
2−
4 ,aq
Kf 4
PbCl
PbCl
[ Ag ]
[ Ag ]
2+
0.9
total
0.8
+
0
Ag aq
+ Claq− →
← AgCl,aq
Kf1
1−
AgCl,0aqs + Claq− →
← AgCl2 ,aq
AgCl,0s
Kf 2
2−
AgCl2−,aqs + Claq− →
← AgCl3,aq
[0.7AgCl ]
[ Ag ]
0.6
[ AgCl ]
[ Ag ]
2−
3
fraction
PbCl
0
2 ,aq
1
PbCl20,s
Kf 2
−
2
total
total
0.5
0.4
Kf 3
0.3
[ AgCl ]
[ Ag ]
o
0.2
0.1
total
0
-2
0
2
4
6
pCl
pCl = 2; [ Cl − ] = 0.01
+
Pbaq2 + + Claq− →
← PbClaq
PbCl + Cl
PbCl
0
2 ,aq
PbCl
−
3,aq
− →
aw ←
+ Cl
+ Cl
PbCl
− →
aq ←
− →
aq ←
PbCl
PbCl
−
3,aq
Kf 4
+
+
2 Hg aq
+ Claq− →
← Hg 2 Claq
Hg 2 Claq+ + Claq− →
← Hg 2 Cl2
−
3,aq
−2
Hg 2 Cl3−,aq + Claq− →
← Hg 2 Cl4 ,aq
2
Kf 2
AgCl,0s
−2
4 ,aq
14
[Cl ] = 1
2+
total
[ Hg ]
[ Hg ]
2+
aq
]
[ Hg ]
2+
total
[ Hg Cl ]
0.6
[ Hg Cl ]
0.5
2
2
0
,aq
[ Hg ]
−
3,aq
2+
total
[ Hg ]
0.4
Kf 3
Hg 2 Cl20,aq
2+
total
[ Hg ]
0.7
Kf1
[
[ Hg Cl ]
0.8
2−
AgCl2−,aqs + Claq− →
← AgCl3,aq
Hg 2 Cl
0.9
Kf 3
2−
4 ,aq
1−
AgCl,0aqs + Claq− →
← AgCl2 ,aq
Hg 2 Cl2 + Cl
12
For lead it was:pCl− = 0
1
PbCl20,s
Kf 2
+
0
Ag aq
+ Claq− →
← AgCl,aq
− →
aq ←
10
Kf1
0
2 ,aq
fraction
+
aq
8
2+
total
0.3
0.2
Kf1
0.1
Kf 2
Hg 2 Cl2( solid )
0
-2
Kf 3
0
2
4
6
pCl
Kf 4
pCl = 35
. ; [ Cl − ] = 0.00035
15
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
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
fraction
0.7
0.6
0.5
Pb2+
AgCl(s); Hg2Cl2(s)
0.4
0.3
0.2
CrO42,−aqueous
insoluble
0.1
0
-2
0
2
4
6
8
10
12
14
ppt
pCl
1
Hg/Cl plot
0.9
0.8
fraction
0.7
insoluble
0.6
0.5
2+
Pbaqueous
+ CrO42,−aqueous →
← PbCrO4 ,solid
0.4
0.3
0.2
0.1
0
-2
0
2
4
6
8
10
12
14
pCl
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?
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!!
P is like Q
P = [ Pb
2+
][CrO ] > K
2−
4
sp
(2 x10
− 16
P > K sp ; 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 > K sp (2 x10 −16 )
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+?
We did an example in which we calculated the
Solubility of AgCl in the presence of ammonia
+
−
AgCl( solid ) →
← Ag aqueous + Cl
Ag
aqueous
+ 2 NH
→
3,aq ←
Ag( NH 3 ) 2
aqueous
+
−
AgCl( solid ) + 2 NH 3,aq →
← Ag ( NH 3 ) 2 ,aq + Cl
Hot water
K sp = 18
. x10 −10
K f = 17
. x10 7
Kreaction = 31
. x10 − 3
+
CrO42,−aqueous
NH3
aqueous
Which number is larger?
Which has the higher solubility?
water or ammonia
Pb2+
AgCl(s); Hg2Cl2(s)
+
6
5
HgNH2Cl(s)
ppt
s (M)
4
Ag(NH3)6+
(
3
2
Ag
+
aqueous
+ NH
→
3aqueous ←
AgNH
+
3aqueous
)[
(1 + 2
Krx NH 3
Krx
]
init
)
= x
1
AgNH3+aqueous + NH3aqueous →
← Ag( NH 3 ) 2 aqueous
+
0
-4
-2
0
2
4
6
8
10
12
14
p[NH3]
“A” students work
(without solutions manual)
~ 10 problems/night.
1.2
Ag(NH3)2+
1
Ag+
fraction
0.8
0.6
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Hg22+ does not
form complexes
with NH3
0.4
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Ag(NH3)+
0.2
0
0
2
4
6
8
Qualitative Analysis:
S to separate Cu 2+, Hg+,
Bi3+, Cd2+
10
pNH3
17
Questions:
1. How does H2S do the second step?
Cu2+
Bi3+
Hg2+
Cd2+
Al3+
Fe3+
Cr3+
Mn2+
Zn2+
Ba2+
Ca2+
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
[ HS ]
] + [ HS ] + [S ]
−
aq
-52.7
-27.0
Weak Precipitators with [S2-]
Will ppt only at higher [S2-]
How can we control [S2-]
Easily?
+
−
H 2 S aq →
← H aq + HS aq
-10.5
-24.7
2−
+
HS aq− →
← H aq + S aq
2−
+
H 2 S aq →
← 2 H aq + S aq
K a1
Ka 2
K a1 K a 2
-
“A” students work
(without solutions manual)
~ 10 problems/night.
2−
aq
−
aq
aq
S2-,
logKsp S2- Strong precipitators with
will ppt at very low [S2-]
-48.5
2−
aq
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
0.6
0.4
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
0.2
0
0
2
4
6
8
10
pH
Who ppts here?
Very insoluble
MS species
12
14
16
Qualitative Analysis:
S to separate Zn 2+, Co+2,
Mn2+, Ni2+
18
Who ppt here?
More soluble MS species
18
Questions:
1. How does (NH4)2S do the first step?
2. Why is the solution basic (pH 8) in
next step?
logKsp S2-48.5
Cu2+
Bi3+
Hg2+
Cd2+
Al3+
Fe3+
Cr3+
Mn2+
Zn2+
Ba2+
Ca2+
2M NH4+; 5 MNH3; (NH4)2S
-10.5
-24.7
Intermediate: need relatively
larger values of [S2-] to
ppt
-
Do not ppt at any value of
[S2-]
-
[ Al
1
Ppt out sulfides
-27.0
[ Al ]
1.2
Strong precipitators with S2-,
will ppt at very low [S2-]
-52.7
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
Buffer pH to basic solution of 9.599
to ppt hydroxides
NH 4+
→
←
NH 3 + H +
pH = pKa + log
[ A]
[ HA]
pH = − log(5.6 x10
− 10
5M
) + log
2M
Ka = 5.6 x10 −10
0.6
0.4
Al3+, Cr3+ Fe3+ come along
as a hydroxides
0.2
0
4
pH = 9.25 + 0.397 = 9.65
6
8
10
pH
19
12
14
1.
2.
Dissolving Precipitates
Strong acid
Complex forming reagents (:NH3, :OH-)
HCO3−
↑↓
H+
+
Re-dissolve
In acid
Pb
2+
aqueous
LeChatlier’s
2−
→
3 aqueous ←
+ CO
PbCO3( solid )
Acid dissolves: carbonates; sulfides, hydroxides
“A” students work
(without solutions manual)
~ 10 problems/night.
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Re-dissolve
In acid
Qualitative Analysis:
Separating Zn2+,
And Al3+, from the sulfides
By OH complexation
Increase pH
20
Example Calculation 6:
[ Al ]
1.2
Ag(OH)2Zn(OH)42Mn(OH)+
Cr(OH)2+
Al(OH)4Fe(OH)2+
Sn(OH)3-
+
AgOH s ⇔ Ag aq
+ OH aq−
+
aq
Ag + 2OH
−
aq
⇔ ag( OH )
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
KspKf
1.99x10-5
145
3.96x10-10
1.28
4.6
7.96x10-16
3.0x108
3+
aq
total ,aq
[ Al( OH ) ]
[ Al ]
[ Al( OH ) ] [ Al( OH ) ] [ Al( OH ) ]
[ Al ] [ Al ]
[ Al ]
2+
aq
]
+
2,aq
total ,aq
o
3,aq
total ,aq
total ,aq
−
4,aq
total ,aq
Al( OH ) 3,s
o
0.8
K sp = 199
. x10 − 9
−
2 ,aq
[ Al
1
Fraction
Ag+
Zn2+
Mn2+
Cr3+
Al3+
Fe3+
Sn2+
Kf
1x104
4.6x1017
2.51x103
6.3x103
1x1033
1.99x1022
2.5x1025
0.6
0.4
K f = 1x10 4
0.2
Knet = K sp K f = (199
. x10 − 9 )(1x10 4 ) = 199
. x10 − 5
AgOH s + OH aq− ⇔ Ag( OH ) 2 ,aq
−
0
4
Which compounds can be brought into solution by raising the pH?
6
8
10
12
pH
How do we distinguish Zn from Al?
1
0.9
[Zn ]
[Zn ]
[Zn( OH ) ]
[Zn ]
2−
4 ,aq
[Zn( OH ) ]
[Zn ]
2+
aq
o
2,aq
total ,aq
Between Al(OH)4- and Zn(OH)4only Zn has complexation with
:NH3
total ,aq
total ,aq
Fraction Zn species
0.8
Zn( OH ) 2,s
0.7
o
0.6
Drop pH to re-ppt Al(OH)3, while
simultaneously forming soluble Zn(NH3)42+
0.5
0.4
[Zn( OH ) ]
[Zn ]
−
3,aq
0.3
NH 4+
total ,aq
[Zn( OH ) ]
[Zn ]
+
aq
0.2
0.1
→
←
NH 3 + H +
pH = pKa + log
total ,aq
Ka = 5.6 x10 −10
[ A]
[ HA]
pH = − log(5.6 x10 − 10 ) + log
0
0
2
4
6
8
10
12
14
pH = 9.25 + − 0.602 = 8.65
pH
21
3M
12 M
3
12
M NH4+
M NH3
14
1.2
Zn2+
Zn(NH13)42+
fraction Zn complexes
1.
2.
Dissolving Precipitates
Strong acid
Complex forming reagents (:NH3, :OH-)
Zn 2 + aqueous + 2OH − aqueous →
← Zn( OH ) 2 ( solid )
+
4 NH 3
0.8
0.6
0.4
↑↓
0.2
Zn( NH 3 ) 24 +( aqueous )
0
-2
-1
0
1
2
pNH3
12 M NH3
“A” students work
(without solutions manual)
~ 10 problems/night.
CN6
Na+
K+
NH4+
Mg2+
Ca2+
Ba2+
Dr. Alanah Fitch
Flanner Hall 402
508-3119
[email protected]
Office Hours Th&F 2-3:30 pm
Module #19:
Precipitation Reactions
Qualitative Analysis:
PO43- to separate Ba 2+,
Ca+2, Mg2+
22
r(pm)
116
152
q/r
0.0086
0.0065
86
114
149
0.023
0.0175
0.013
Ksp
-
Ca3(PO4)2
Ba3(PO4)2
1.2x10-29
3.4x10-23
3
4
5
6
Summary Points
“A” students work
(without solutions manual)
~7 problems/night.
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
Solubility
What you need
To know
Ba3(PO4)2
“A” students work
(without solutions manual)
~7 problems/night.
END
23