Solubility and Complex-Ion
Equilibria
Solubility
• Solubility is the ability of a substance to dissolve
in water. The solubility is measured in terms of
concentration of an ion that is present in a
smaller ratio in solution.
• On the other hand, solubility equilibrium refers to
the equilibrium between the dissolved salt (ions)
and undissolved salt that usually exists in a
saturated solution or a solution of a sparingly
soluble salt.
2
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)
3
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
4
Solubility Product
• Consider the saturated solution of silver chloride
(AgCl), where the equilibrium exists between
dissolved ions and undissolved silver chloride
according to the following equation.
5
Solubility Product
6
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.
7
Problem
•
•
•
•
•
Write the expressions of Ksp of:
a) AgCl




K sp  Ag Cl 
b) PbI2
2
 2
K sp  Pb I 
c) Ca3(PO4)2
3 2
2 3
K sp  Ca  PO4 
d) Cr(OH)3

K sp  Cr
3
OH 
 3
8
Common Ksp Values
9
What do the Ksp values tell us?
• Ksp values are derived from the concentrations
of ions in equilibrium reactions; the higher the
concentrations of ions, greater the Ksp.
• Higher the concentrations of ions means greater
the solubility of the salt. Therefore, the
magnitude of Ksp directly indicates the solubility
of the salt in water.
10
Example
Arrange the following salts in order of increasing
solubility.
(a) Aluminum hydroxide [Al (OH)3] (Ksp=1.8x10 )
(b) Calcium hydroxide [Ca (OH)2] (Ksp=8.0x10 )
(c) Chromium (III) hydroxide [Cr (OH)3] (Ksp = 3.0x10 )
(d) Copper (II) hydroxide [Cu (OH)2] (Ksp=2.2x10 )
-33
-6
-29
-20
Al(OH)3 < Cr(OH)3 < Cu(OH)2 < Ca(OH)2
11
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
2
  1.05 x 10
-5
M
K sp  1.1 x 10-10
12
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.
13
Molar solubility
• Alternatively, if we know the
• molar solubility
• OR
• the molar mass and the solubility
• we can calculate
• the Ksp for the solid.
14
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.
15
Molar solubility
• Suppose, you want to • Suppose, you want to
convert solubility (g/L)
convert molar
to molar solubility
solubility (mol/L) to
(mol/L), you divide
solubility(g/L), you
solubility by molar
multiply molar
mass (g/mol):
solubility by molar
mass (g/mol):
16
Relationship between Ksp and solubility
• Solubility is the molar concentration and Ksp is
the product of molar concentrations, and hence
both are related to one another.
• In the following, we develop relationships
between these two for various salts.
• Let us illustrate the concept using a simple
example of AgCl that dissociates into one cation
(Ag+) and one anion (Cl-) at equilibrium:
17
Relationship between Ksp and solubility
• The Ksp expression for this is:
• Let s be the molar solubility(mol/L) of Ag ion.
This will also be the molar solubility of Cl ion
because number of Ag+ ions is equal to number
of Cl- ions that can be seen by the above
equilibrium reaction. That is,
• s = [Ag+] and s = [Cl-]
+
-
18
Relationship between Ksp and solubility
• Substituting these into Ksp equation, we get
• Thus, the molar solubility is just square root of
the Ksp.
19
Relationship between Ksp and solubility
• We can extend similar procedure to develop the
relationship between Ksp and s for more
complicated compounds:
20
Relationship between Ksp and solubility
21
Calculating Ksp from solubility
• Example
• Calculate the Ksp of calcium sulfate (CaSO4) if the
solubility is 0.56 g/L.
• Answer
• Solubility is the concentration of the ion present
in a smaller proportion. Here both Ca2+ ion and
SO4 ion are present in an equal proportion
according to the following equation.
2-
22
23
Calculating solubility from Ksp
• Example
• Calculate the molar solubility of Cu(OH)2 if Ksp =
2.2 x10 .
-20
24
25
Calculating Ksp From Molar Solubility
It is found that 1.2x10-3 mol of lead (II) iodide, PbI2,
dissolves in 1.0 L of aqueous solution at 25 oC.
What is the Ksp at this temperature?
PbI2 
Pb2+ + 2I-
Ksp = [Pb2+][I-]2
Ksp = [1.2 x 10-3] [2(1.2 x 10-3)]2
Ksp = 6.9 x 10-9
2 iodide ions
form, so you
must multiply
molarity by 2!
Calculating Molar Solubility From Ksp
Calculate the molar solubility of silver chromate,
Ag2CrO4, in water from Ksp = 1.1x10-12 for Ag2CrO4.
Ag2CrO4(s)  2Ag+ + CrO42At equilibrium
2x
x
Ksp = [Ag+]2[CrO4-2]
1.1 x 10-12 = (x)(2x)2
1.1 x 10-12 = 4x3
x = 6.5 x 10-5 M = [Ag2CrO4] = [CrO42-]
[Ag+] = 2(6.5 x 10-5)
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
28
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mol1.
Answer: Ksp = 2.0 x 10-9
29
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.
30
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.
Answer: mass = 0.55 mg
31
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!
32
The common-ion effect
33
Figure
34
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!
35
36
Molar
The common ion effect “Le Chatelier”
Why is AgCl less soluble in sea water than in fresh water?
AgCl(s)  Ag+ + Cl–
Seawater contains
NaCl
37
Problem: The solubility of AgCl in pure water is 1.3 x 10–5
M. What is its solubility in seawater where the [Cl–] = 0.55
M? (Ksp of AgCl = 1.8 x 10–10)
Ksp= [Ag+][Cl–]
AgCl(s)  Ag+ + Cl–
I.
C.
E.
N/A
N/A
0
+x
0.55
+x
N/A
+x
0.55 + x
Ksp= [x][0.55 + x]
try dropping this x
Ksp = 0.55x
1.8 x 10–10 = 0.55x
x = 3.3 x 10–10 = [Ag+]=[AgCl]
“AgCl is much less soluble in seawater”
38
more Common ion effect:
a. What is the solubility of CaF2 in 0.010 M Ca(NO3)2?
Ksp(CaF2) = 3.9 x 10–11
CaF2(s)  Ca2+ + 2F–
I.
C.
E.
Ksp=[Ca2+][F-]2
[Ca2+]
[F–]
0.010
0
+x
+2x
0.010 + x
2x
Ksp= [0.010 + x][2x]2
 [0.010][2x]2 = 0.010(4x2)
3.9 x 10–11 = 0.010(4x2)
x = 3.1 x 10–5 M Ca2+ from CaF2 so = M of CaF2
Now YOU determine the solubility of CaF2 in 0.010 M NaF.
39
Answer:
3.9 x 10–7 M Ca2+
CaF2(s)  Ca2+ + 2F–
0
+x
x
Ksp = [x][0.010 + 2x]2
0.010
2x
0.010 + 2x
 x(0.010)2
3.9 x 10-11 =x(0.010)2
x = 3.9 x 10-7
What does x tell us
40
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.
41
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.
42
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
43
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
44
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!
45
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!
46
Mixing and equilibrium take time!
We must wait
until dilution
is completed
and
equilibrium is
established
BEFORE
we say
precipitation
occurred!
47
Problem: A solution is 1.5 x 10–6 M in Ni2+. Na2CO3 is added
to make the solution 6.0 x 10–4 M in CO32–. Ksp(NiCO3) = 6.6
x 10–9. Will NiCO3 ppt?
We must compare Q to Ksp.
NiCO3  Ni2+ + CO32–
Ksp = [Ni2+][CO32–]
Q = [Ni2+][CO32–]
Q = [1.5 x 10–6][6.0 x 10–4] = 9.0 x 10–10
Q < Ksp
no ppt.
48
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.
49
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 lead iodide to
start?
• Ksp of PbI2 is 7.1 x 10-9
Answer: We require at least 9 drops.
50
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.
51
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.
52
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.
53
Effect of pH on Solubility
• The solubility of an ionic solute may be greatly
affected by pH if an acid-base reaction also occurs
as the solute dissolves.
• In other words, some salts will not dissolve well in
pure water, but will dissolve in an acid or a base.
• If the anion (A-) of the salt/precipitate is that of a
weak acid, the salt/precipitate will dissolve more
when in a strong acid (H+ ions will form HA with A)
• However, if the anion of the precipitate is that of a
strong acid, adding a strong acid will have no
effect on the precipitate dissolving more.
Effect of pH on Solubility
• How would the addition of HCl affect the solubility of PbCl2?
– Cl- is the conjugate base of a strong acid, thus it is a weak base.
– It will not react with H+ ions, so there is no effect.
• How would the addition of HCl affect the solubility of FeS?
– S2- is a strong base (conjugate base of weak acid)
– Thus, it will react with H+ ions to form H2S,
– This will shift the equilibrium to make more FeS dissolve!
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 HCO3• CO32- (aq) + H3O+ (aq)  HCO3- (aq) + H2O (l)
56
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!
57
58
Molar
Adding equilibria
• 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
59
Adding equilibria
• 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+!
60
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.
61
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.
Answer: [NH4+] > 0.073 M
62
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.
63
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
64
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).
65
66
Molar
Complex ion formation:
AgCl(s)  Ag+ + Cl– Ksp= 1.8 x 10–10
Ag+(aq) + NH3(aq)  Ag(NH3)+(aq)
Ag(NH3)+(aq) + NH3(aq)  Ag(NH3)2+(aq)
Ag+(aq) + 2NH3(aq)  {Ag(NH3)2}+(aq)
formation or stability constant:
Kf

Ag ( NH ) 

 1.7 x 10
Ag NH 

3 2

2
3
For Cu2+:
Cu2+ + 4NH3  [Cu(NH3)4]2+(aq)
K1 x K2 x K3 x K4 = Kf = 6.8 x 1012
67
7
Solubility and complex ions:
Problem: How many moles of NH3 must be added to dissolve 0.050 mol of
AgCl in 1.0 L of H2O? (KspAgCl = 1.8 x 10–10 ; Kf[Ag(NH3)2]+ = 1.6 x 107)
AgCl(s)  Ag+ + Cl–
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
AgCl(s) + 2NH3  Ag(NH3)2+(aq) + Cl–
sum of RXNS:


Ag ( NH )  Cl 

 K

K overall

3 2
NH 3 
2
–3
=
2.9
x
10
x
K
sp
f
Now use Koverall to solve the problem:
Koverall= 2.9 x
10–3
 Ag (NH )  Cl   0.0500.050

=
3 2
NH 3 
2

NH 3  2
[NH3]eq = 0.93 but ..... How much NH3 must we add?
[NH3]total= 0.93 + (2 x 0.050) = 1.03 M
2 ammonia’s for each Ag+
68
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.
69
70
Groups of precipitated ions
• 1) Chloride group –
• Pb2+, Ag+, Hg22+
• 2) Hydrogen Sulphide group• Pb2+, 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+
71
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…
72
Further tests for insoluble chlorides
• In a) we have a mixture of AgCl,
Hg2Cl2, and PbCl2.
• If we add ammonia, any AgCl
should dissolve because of
complex ion formation
• AgCl (s) + 2 NH3 (aq) 
• [Ag(NH3)2]+ (aq) + Cl- (aq)
73
Further tests for insoluble chlorides
• Also after adding ammonia (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
74
Further tests for insoluble chlorides
• Adding chromate (CrO42-) (c) 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)
•
75
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
76
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
77
Reactions with carbonate
• Addition of carbonate ion (CO32-) in basic
solution (usually with an ammonia-ammonium
buffer) will precipitate the alkali earth metal
carbonates (see Chapter 21)
• CaCO3, MgCO3, SrCO3, BaCO3
78
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+
79