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```Chapter 18
Solubility and Complex-Ion Equilibria
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1051.php
Solubility equilibria
MmXx (s)  m Mn+ (aq) + x Xy- (aq)
The equilibrium is
established when we have a
saturated solution of ions
forming the solid and solid
is dissociating to form the
ions in solution. The rates
of these processes must be
equal. (eq’m definition)
2
Solubility equilibria
For a dissolution process, we give the
equilibrium constant expression the
name solubility product (constant)
Ksp. For
MmXx (s)  m Mn+ (aq) + x Xy- (aq)
Ksp = [Mn+]m [Xy-]x
3
Ksp is an equilibrium constant
Since Ksp is an equilibrium
constant we MUST
refer to a specific balanced equation
(by definition this balanced equation
is one mole of solid becoming
aqueous ions)
at a specific temperature.
4
Problem
Write the expressions of Ksp of:


K sp  Ag Cl 
a) AgCl
2
 2
b) PbI2
K sp  P b I 
c) Ca3(PO4)2 K sp  Ca 2 3 P O4 3 2
d) Cr(OH)3
3
 3
K  Cr OH 
sp
5
6
Problem
If a saturated solution of BaSO4 is
prepared by dissolving solid BaSO4 in
water, and [Ba2+] = 1.05 x 10-5 molL-1,
what is the Ksp for BaSO4?
2

2
BaSO 4 (s) 
Ba
(aq)  SO 4 (aq)



K sp  Ba 2 SO 4

2

 
K sp  x 2 if x  Ba 2  SO 4

K sp  1.05 x 10-5

2
  1.05 x 10
-5
M
2
K sp  1.1 x 10-10
7
Molar solubility
If we know the Ksp value for a solid, we can
calculate the molar solubility, which is the
number of moles of the solid that can
dissolve in a given amount of solvent
before the solution becomes saturated.
The molar solubility leads to the solubility
(by using the molar mass) which is the
mass of the solid that can dissolve in a
given amount of solvent before the solution
becomes saturated.
8
Molar solubility
Alternatively, if we know the
molar solubility
OR
the molar mass and the solubility
we can calculate
the Ksp for the solid.
9
Molar solubility
A good (but NOT physically correct) way to think
about molar solubility is to treat our solid dissolution
AS IF there are two separate processes:
M m X x (s) 
 M m X x (aq)
H 2O
M m X x (aq)  m M
dissociation
n
(aq)  x X
y
(aq)
The molar solubility has the same value as the
concentration of aqueous MmXx after the first step.
We can figure out this concentration based on
concentrations of ions in the second step.
10
Problem
A handbook lists the aqueous solubility of
AgOCN as 7 mg per 100 mL at 20 C.
What is the Ksp of AgOCN at 20 C? The
molar mass of AgOCN is 149.885 gmol-1.
Answer: Ksp = 2 x 10-7
11
Problem
A handbook lists the aqueous solubility of
lithium phosphate (Li3PO4) as 0.034 g per
100 mL at 18 C. What is the Ksp of lithium
phosphate at 18 C? The molar mass of
Li3PO4 is 115.794 gmol-1.
Answer: Ksp = 2.0 x 10-9
12
Problem
Which has the greater molar solubility:
AgCl with Ksp = 1.8 x 10-10
or
Ag2CrO4 with Ksp = 1.1 x 10-12?
Answer: The molar solubility of AgCl is 1.3 x 10-5 M
while the molar solubility of Ag2CrO4 is 6.5 x 10-5 M.
Silver chromate has a higher molar solubility.
13
Problem
How many milligrams of BaSO4 (molar
mass is 233.391 gmol-1) are dissolved in a
225 mL sample of saturated aqueous
barium sulphate? Ksp = 1.1 x 10-10 at 25 C.
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The common-ion effect
MmXx (s)  m Mn+ (aq) + x Xy- (aq)
If we have dissolved a solid in pure water and
we add to this solution another solution
containing one of the common ions, then Le
Chatalier’s Principle tells us what will happen:
The presence of the common-ion in
the added solution will force the
dissolution reaction to the left,
meaning more solid will form!
15
The common-ion effect
16
Figure
17
The common-ion effect
MmXx (s)  m Mn+ (aq) + x Xy- (aq)
If instead of dissolving a solid in pure water we
try and dissolve it into a solution that already
contains one of the common ions, then Le
Chatalier’s Principle tells us what will happen:
The presence of the common-ion
already in solution will force the
dissolution reaction to the left,
meaning less solid will dissolve than
would dissolve in pure water!
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19
Molar
Problem
Calculate the molar solubility
of MgF2 (Ksp = 7.4 x 10-11) in
pure water and in 0.10 molL-1
MgCl2 at 25 °C.
Answer: The molar solubility is 2.6 x 10-4 M in
pure water and 1.4 x 10-5 M in 0.10 M
magnesium chloride.
20
Problem
What is the the molar solubility of
Fe(OH)3 (Ksp = 4 x 10-38) in a
buffered solution with pH = 8.20 at
25 °C.
Answer: The molar solubility is 1 x 10-20 M in
the buffered solution.
21
Limitations of Ksp
If our solid is more than slightly soluble
then we really should use the activities of
our ions in solution rather than
concentrations.
These two measures are nearly the same
for very dilute ion concentrations, but
can become quite different at higher
concentrations!
22
The diverse (“uncommon”) ion effect
The activities of ions tend
to be LESS than the
concentration value as
the total ionic
concentration increases.
23
The diverse (“uncommon”) ion effect
Adding a salt that does NOT feature a
common ion to the solution will tend to
decrease the activity (the “effective”
concentration) of the ions in solution.
The ion concentrations appear smaller
than they should be at equilibrium so
more solid dissolves to reach the
appropriate equilibrium
“concentrations”.
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25
Molar
Ion-pair formation and Ksp
We assume in Ksp calculations that the
solid dissociates completely into ions
in solution.
If this is not true, then the ionic
concentrations we measure do not
include dissolved but undissociated
molecules or ion pairs which come from
solid dissolution.
26
Ion-pair formation and Ksp
Positive ions and
negative ions are
attracted to each
other and so they
can form an ion pair
that has a chemical
identity different
from each of the
individual ions!
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Ion-pair formation and Ksp
Say we measure a molar
solubility for magnesium fluoride
to be 4 x 10-3 M and assume that
[Mg2+] = 4 x 10-3 M and
[F-] = 8 x 10-3 M
to give a
calculated Ksp of 3 x 10-7
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Ion-pair formation and Ksp
IN REALITY the presence of
undissociated MgF2 (aq)
and MgF- ion pairs
means that not all of the solid that has
dissolved is found as free ions, so our
ionic concentrations are LOWER than
we assumed, and so Ksp is actually
smaller than we calculated.
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Simultaneous equilibria
We’ve seen in reference to Le Chatalier’s
Principle that if more than one reaction
can take place in a container, then the
reactions might not be able to be treated
independently. Other equilibrium
processes may affect the solubility of
the solid and lead to miscalculated Ksp
values.
30
Simultaneous equilibria
For example, we’ve seen that AgI
becomes more soluble when we
formation of a complex of silver
ions and ammonia.
We’ll look more closely at complex
formation a little later.
31
Assessing the limitations of Ksp
Most tabulated Ksp values are actually
based on activities, and not
concentrations.
We use concentrations in our examples,
so our calculations represent an ideal,
and not reality.
Generally we could be in error by over
100 times!
32
Criteria for precipitation and its completeness
Can we predict if a solid will form if we mix
two solutions of different ions?
Consider the mixing of two different
solutions, one with Ca2+ ions and one with
F- ions. A formation of solid is the
dissolution reaction in reverse, so we can
express the reaction using the dissolution
equation
CaF2 (s)  Ca2+ (aq) + 2 F- (aq) Ksp = [Ca2+ ][F-]2
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Criteria for precipitation and its completeness
When we mix the solutions
(BE CAREFUL – mixing ALWAYS changes
the concentrations of both our ions!)
the system is most likely not at equilibrium.
Like in other equilibrum problems, we can use
a reaction quotient Qsp (often called the ion
product) to tell us in which direction the
system must go to reach equilibrium
Qsp = [Ca2+ ][F-]2
34
Criteria for precipitation and its completeness
If Qsp > Ksp, the solution is supersaturated, so
the system is not at equilibrium. The
concentration of the ions is greater than it
would be at equilibrium, and so the reaction
wants to shift from ions towards the solid.
We expect precipitation to occur!
If Qsp = Ksp, the solution is saturated, and the
system is at equilibrium.
No precipitation occurs!
35
Criteria for precipitation and its completeness
If Qsp < Ksp, the solution is unsaturated, so
the system is not at equilibrium. The
concentration of the ions is less than it
would be at equilibrium, and so the reaction
wants to shift from solid towards the ions.
No precipitation can occur!
36
Mixing and equilibrium take time!
We must wait
until dilution
is completed
and
equilibrium is
established
BEFORE
we say
precipitation
occurred!
37
Problem
Will a precipitate form when 0.150 L of
0.10 molL-1 Pb(NO3)2 and 0.100 L of 0.20
molL-1 NaCl are mixed?
Ksp of PbCl2 is 1.2 x 10-5
Answer: Qsp = 3.8 x 10-4 > Ksp so precipitation
should occur.
38
Problem
How many drops (1 drop = 0.05 mL) of
0.20 M KI must we add to 100.0 mL of
0.010 M Pb(NO3)2 to get precipitation of
Ksp of PbI2 is 7.1 x 10-9
Answer: We require at least 9 drops.
39
Complete precipitation
Generally we treat precipitation as
complete if 99.9% of the original ion
concentration has been lost to the
precipitate.
For example, if our initial [Pb2+] is 0.10 M,
then precipitation by adding I- is
complete when our solution contains a
[Pb2+] less than 1 x 10-4 M.
40
Problem
A typical Ca2+ concentration in seawater is
0.010 M. Will the precipitation of Ca(OH)2
be complete from a seawater sample in
which [OH-] is maintained at 0.040 M?
Ksp of Ca(OH)2 is 5.5 x 10-6
Answer: Since the final [Ca2+] is 3.4 x 10-3 M,
which is 34 % of 0.010 M, the precipitation is not
complete.
41
Problem
What [OH-] should be maintained in a
solution if, after precipitation of Mg2+ as solid
magnesium hydroxide, the remaining [Mg2+]
is to be at a level of 1gL-1?
Molar mass Mg is 24.305 gmol-1
Ksp of Mg(OH)2 is 1.8 x 10-11
Answer: [OH-] needed is 1.6 x 10-2 M.
42
Fractional precipitation
If we have a solution with
both CrO42- ions and Br- ions
and add a large amount of Ag+ ions
at once,
then both Ag2CrO4 and AgBr
will precipitate
in our container at the same time.
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Fractional precipitation
the solid with the significantly lower
molar solubility (AgBr in this case – do
the calculations to check this for yourself)
will precipitate first
preferentially.
44
Fractional precipitation
In other words,
the concentration of Ag+
CAN NOT become large enough
to precipitate Ag2CrO4
until the AgBr
precipitation is complete.
45
Fractional precipitation
46
Problem
AgNO3 is slowly added to a solution with
[Cl-] = 0.115 M and [Br-] = 0.264 M. What
percent of the Br- remains unprecipitated
at the point at which AgCl (s) begins to
precipitate?
Ksp values
AgCl = 1.8 x 10-10 AgBr = 5.0 x 10-13
Answer: 0.12 % of Br- remains.
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Solubility and pH
If a solid dissolves to give a basic anion in
solution, addition of strong acid will
increase the solubility of the solid.
CaCO3 (s)  Ca2+ (aq) + CO32- (aq) Ksp = 2.8 x 10-9
Carbonate, CO32-, is a basic anion that
will react with a proton to give HCO3CO32- (aq) + H3O+ (aq)  HCO3- (aq) + H2O (l)
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Solubility and pH
CO32- (aq) + H3O+ (aq)  HCO3- (aq) + H2O (l)
This reaction will shift to the right (products)
as the pH becomes more acidic which means
CO32- (aq) decreases. However, in our first
equilibrium IT ALSO MUST DECREASE so the
first equilibrium will also shift to the right to
compensate.
More solid will dissolve!
49
50
Molar
A better way to state the effect of pH on
solubility comes when we add dissolution
and weak base – strong acid reactions
together:
CaCO3 (s)  Ca2+ (aq) + CO32- (aq)
Ksp = 2.8 x 10-9
CO32- (aq) + H3O+ (aq)  HCO3- (aq) + H2O (l)
K = Kb x 1/Kw = (2.1 x 10-4) x (1.0 x 1014)
K = 2.1 x 1010
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CaCO3 (s) + H3O+ (aq) 
Ca2+ (aq) + HCO3- (aq) + H2O (l)
K’ = K x Ksp = (2.1 x 1010) x (2.8 x 10-9)
K’ = 59
The solubility of the solid will
increase in the presence of H3O+!
52
Problem
Will a precipitate of Fe(OH)3 form from a
solution that is 0.013 M Fe3+ in a buffer
solution that is 0.150 M acetic acid – 0.250
M acetate?
Ksp Fe(OH)3 = 4 x 10-38 Ka = 1.8 x 10-5
Answer: Since Qsp is 1 x 10-29, then
precipitation will occur since Qsp>Ksp.
53
Problem
What minimum [NH4+] must be present to
prevent precipitation of Mn(OH)2 (s) from a
solution that is 0.0050 M MnCl2 and 0.025
M NH3? For Mn(OH)2 Ksp = 1.9 x 10-13 and
Kb for NH3 is 1.8 x 10-5.
54
Formation of complex ions
Solubility of a solid increases if there is the
ability to form a complex ion.
An example of a complex ion is Ag(NH3)2+.
Such complexes affect solubility by reducing
the concentration of the cation so that the
dissolution reaction must shift to the
products to replace the cation concentration
to re-establish equilibrium.
55
Formation of complex ions
Ag+ (aq) + 2 NH3 (aq)  Ag(NH3)2+ (aq) Kf = 1.7 x 107
AgCl (s)  Ag+ (aq) + Cl- (aq)
Ksp = 1.8 x 10-10
In the presence of ammonia, the dissolution of
AgCl can be expressed by the sum of these two
reactions
2 NH3 (aq) + AgCl (s)  Ag(NH3)2+ (aq) + Cl- (aq)
K = Kf x Ksp = 1.7 x 107 x 1.8 x 10-10 = 3.1 x 10-3
56
Formation of complex ions
We see the dissolution of AgCl
occurs to a greater level of
completion
in the presence of ammonia
(K = 3.1 x 10-3)
than it does in pure water
(Ksp = 1.8 x 10-10).
57
58
Molar
Qualitative cation analysis
Qualitative analysis is concerned with
“what do we have?” and NOT “how
much do we have?”
If we want to identify what cations we have
in a solution, we can use a series of
precipitation reactions in a certain order
to tell us.
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60
Groups of precipitated ions
1) Chloride group –
Pb2+, Ag+, Hg22+
2) Hydrogen Sulphide groupPb2+, Hg2+, Bi3+, Cu2+, Cd2+, As3+, Sn2+, Sb3+
3) Ammonium sulphide group –
Mn2+, Fe2+, Fe3+, Ni2+, Co2+, Al3+, Zn2+, Cr3+
4) Carbonate group –
Mg2+, Ca2+, Sr2+, Ba2+
5) Soluble group –
Na+, K+, NH4+
61
Reactions with hydrogen chloride
Most metal ions form soluble salts with
chloride EXCEPT Pb2+, Hg22+, and Ag+.
Adding aqueous HCl to our unknown
solution will let us now if we have one or
more of these ions because we will get
white precipitate(s). If we want to know if
we have more than one of these ions, we
do further tests on the precipitated solids…
62
Further tests for insoluble chlorides
In a) we have a mixture of
AgCl, Hg2Cl2, and PbCl2.
AgCl should dissolve
because of complex ion
formation
AgCl (s) + 2 NH3 (aq) 
[Ag(NH3)2]+ (aq) + Cl- (aq)
63
Further tests for insoluble chlorides
(b), any Hg2Cl2 will give us a
grey solid that is a mixture of
black liquid Hg and white
solid HgNH2Cl:
Hg2Cl2 (s) + 2 NH3 (aq) 
Hg (l) + HgNH2Cl (s) + NH4Cl (aq)
black
white
64
Further tests for insoluble chlorides
to a Pb2+ solution derived by
heating the precipitate
solutions (it’s the most
soluble) will give a yellow
precipitate - PbCrO4
Pb2+ (aq) + CrO42- (aq) 
PbCrO4(s)
65
Reactions with hydrogen sulfide
S2- is capable of giving precipitates
of many ions.
H2S is a potential source of S2- in solution
because it is a diprotic acid
H2S (aq) + H2O (l)  H3O+ (aq) + HS- (aq)
Ka1 = 1.0 x 10-7
HS- (aq) + H2O (l)  H3O+ (aq) + S2- (aq)
Ka2 = 1 x 10-19
66
Reactions with hydrogen sulfide
However, in acidic solution (with HCl),
some of the precipitates dissolve, leaving
behind
PbS, HgS, Bi2S3, CuS, CdS, As2S3,
SnS, Sb2S3
In basic solution (by adding ammonia)
these precipitates dissolve, leaving behind
MnS, FeS, Fe(OH)3, NiS, CoS, Al(OH)3,
ZnS, Cr(OH)3
67
Reactions with carbonate
Addition of carbonate ion (CO32-) in basic
solution (usually with an ammoniaammonium buffer) will precipitate the
alkali earth metal carbonates (see Chapter
21)
CaCO3, MgCO3, SrCO3,
BaCO3
68
The soluble group
Any ions left in solution after the first four
reaction groups are tested for are the
cations of soluble salts
+
Na ,
+
K,
and
+
NH4
69
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