Kinetics & Equlibria
5.
SOLUBILITY PRODUCT
Solubility Product only applies to substances that are almost insoluble, but not completely. Solubility
product may be defined as the equilibrium constant for the dissolution of sparingly soluble salt.
When excess of the sparingly soluble salt silver chloride is placed in water the following equilibrium
is set up:
Ag+ (aq) + Cl- (aq)
AgCl (s)
The equilibrium constant expression is
[Ag+ (aq)] [Cl- (aq)]
Kc =
[AgCl (s)]
But [AgCl (s)] is constant in solution so
Kc [AgCl (s)] = [Ag+ (aq)] [Cl-(aq)]
i.e.
Kc [AgCl (s)] = Ks.p = [Ag+(s)] (Cl-(aq)]
Where, Ks.p is the new constant called the solubility product. For silver chloride
Ks.p (AgCl) = 2.0 * 10-10 mol2 dm-6
[At 250C]
Problem
A saturated solution of silver chloride contains 1.46 * 10-3 g dm3 at 180C. What is the solubility
product of silver chloride at this temperature?
Mass conc’n of solute (g dm-3)
Now, mass of one mol
of solute
=
Concentration (mol dm-3)
1.46 * 10-3
So AgCl (143.5)
=
Concentration
Concentration =
1.46 * 10-3
mol dm-3
143.5
= 1 * 10-5 mol dm-3
At equilibrium: AgCl(s) =
AgCl (s)
So [Ag+ (aq)]
Ag+ + Cl-
[Ag+ (aq)] [Cl- (aq)]
=
1* 10-5 mol dm-3
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Kinetics & Equlibria
And [Cl- (aq)]
1* 10-5 mol dm-3
=
= [Ag+ (aq)] [Cl- (aq)]
So Ks.p [AgCl]
= (1.0 * 10-5) (1.0 * 10-5)
= 1 * 10-10 mol2 dm-6
= 1 * 10-10 mol2 dm-3
Ks.p
Example
The solubility product of silver carbonate at 250C is 8 * 10-12 mol3 dm-9. What is its solubility at this
temperature?
Solution
2Ag+ (aq) + CO32- (aq)
Ag2CO3 (s)
Let the solubility of Ag2CO3 be 5 mol dm-3
So
And
[Ag+ (aq)]
[CO3 (aq)]
Ag2CO3
= 2s
= 2s
= s
+ s
= [Ag+(aq)]2 [CO32-(aq)]
Ksp [Ag2CO3]
= (2s)2 (s)
= 4s3
4s3
= 8 * 10-12
s3
= 2 * 10-12
s
= 1.5 * 10-4 mol dm-3
s
=
3
2.0 * 1012
Common Ion Effect
The lowering of the solubility of an ionic compound by the addition of a common ion to the solution
is known as the common ion effect e.g.
CaCO3 (s)
Ca2+ (aq) + CO32- (aq)
Addition of common ion (Ca2+)
CaCl2 (s)
Ca2+ (aq) +
-
-
(A)
(B)
Cl- (aq)
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Kinetics & Equlibria
In the mixture Ca2+ (aq) is greatly increased. The ion is common to both calcium carbonate and calcium
chloride. According to Le Chatelier’s principle the Calcium Carbonate equilibrium will shift to reduce
the calcium ions. The shift will be to the left of the equilibrium A, so that solubility is reduced, hence
a lowering of the carbonate ions.
Problem
Common ion used (BaSO4/Na2SO4)
Solubility of BaS04 in water
= 1.0 * 10-5 mol dm-3
BaSO4 (s)
Ba2+ (aq) + SO42- (aq)
Na2SO4 (s)
2Na+ (aq) + SO42- (aq)
Example
What is the solubility of BaS04 in 0.1 mo dm-3 Na2504?
Solution
The solubility of BaSO4 in water is 1.0 * 10-5 mol dm-3
In Na2SO4 the solubility will be changed to say, 5 mol dm-3
In this case
[Ba2+ (aq)]
= s1 mol dm-3
But, [S042-(aq)]
= (s1 + 0.1) mol dm-3
Ks.p (BaS04)
= 1 * 10-10 mol dm-3
Ks.p (BaS04)
= [Ba2+] [S042-]
= (s1) (s1 + 0.1)
= 51 (s1 + 0.1)
=
51(51 + 0.1)
Now 51 « 0.1;
= 1 * 10-10
(51 + 0.1) ~ 0.1
51
+ 0.1 = 1 * 10-10
51
= 1 * 10-9
So that solubility of BaS04 in 0.1 mol dm-3 Na2S04 = 1 * 10-9 mol dm-3
Note that:
Ks.p (BaS04)
[Ba2+ (aq)]
= 1 * 10-10
= 1 * 10-5
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Kinetics & Equlibria
[SO42-(aq)]
= 1 * 10-5
In Na2SO4 [SO42- (aq)]
= 1 * 10-9
The solubility is greatly reduced and is due to the common ion SO42-
Precipitation
A precipitate will appear if the solubility product is exceeded i.e. precipitation of an insoluble salt
occurs when
Ionic product
> Solubility product
Example
What will happen if 0.030 mol dm-3 NaCl and 0.30 mol dm-3 Lead (II) Nitrate are mixed?
Solution
First thing to note. The concentration is halved.
[NaCl (aq)] = 0.015 mol dm-3
[PbNO3 (aq)]
= 0.15 mol dm-3
Ks.p (PbCl2) = 1.7 * 10-5 mol3 dm-9
PbCl2(s) = [Pb2+ (aq)] [Cl-(aq)]2
[Pb2+ (aq)] [Cl- (aq)]2 = [0.15] [0.015]2
Ionic Product
= 3.4 * 10-5 mol3 dn-9
Here ionic product
> Solubility product
Hence precipitate will form.
Selective Precipitation
In selective precipitation the ionic product < Ks.p e.g.
[Pb2+] [Cl-]2
< Ks.p
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Kinetics & Equlibria
Solubilities of Ionic Compounds in Water
Anion
NO3- (Nitrate)
CH3COO- (Acetate)
ClO3- (Chlorate)
ClO4- (Perchlorate)
F- (Fluoride)
Cl- (Chloride)
Br- (Bromide)
I- (Iodide)
SO42- (Sulphate)
S- (Sulphide)
CO3- (Carbonate)
SO32- (Sulphite)
PO43- (Phosphate)
OH- (Hydroxide)
Soluble
Slightly Soluble
Insoluble
All
Most
All
Most
Group I, AgF, BeF
Most
Most
Most
Most
Group I and II, (NH4)2S
Group I, (NH4)2CO3
Group I, (NH4)2SO3
Group I, (NH4)3PO4
Group I, Ba(OH-)2
KClO4
SrF2, BaF2 PbF2
PbCl2
PbBr2, HgBr2
CAS04, Ag2S04
Sr(OH)2 Ca(OH)2
Be(CH3COO)2
MgF2, CaF2
AgCl, Hg2Cl2
AgBr, Hg2Br2
AgI, Hg2I2, PbI2, HgI2
SrSO4, BasO4, PbsO,Ag2SO4
Most
Most
Most
Most
Most
Soluble compounds are defined as those that dissolve to the extent of 10 g dm-3, slightly, soluble as 1
to10 g dm-3, and insoluble as less than 0.1g/dm3 at room temperature.
The analysis of a mixture of elements usually requires that the mixture be separated into its
components. One way to do this is to exploit the differences in the Solubilities of compounds of the
elements. To separate silver from lead for example, a search is made for compounds of these elements
that
1) Have common ion
2) Have widely different solubilities
The chlorides AgCl and PbCl2 are two such compounds for which the solubility equilibria are:
AgCl(s)
Ag+(aq) + Cl-(aq); Ks.p = 1.6 * 10-10
PbCl2(s)
Pb2+ (aq) + 2Cl-(aq); Ks.p = 2.4 * 10-4
Lead chloride is far more soluble in water than silver chloride. Consider a solution that is 0.10 mol
dm-3 in both Ag+ and Pb2+. It is possible to add enough Cl- to precipitate almost all the Ag+ ions but to
leave all the Pb2+ ions to remain in solution its ionic product must remain SMALLER than Ks.p.
[Pb2+] [Cl-]2 < Ks.p
(0.1) (0.1)2 < 2.4 * 10-4
1 *
10-3
< 2.4 * 10-4
Now inserting Ks.p and the concentration of Pb2+ gives
Ks.p = 2.4 * 10-4
-
= 2.4 * 10-3
[Cl ] <
2+
[Pb ]
[Cl-]
=
0.1
2.04 * 10-3
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Kinetics & Equlibria
= 4.9 * 10-2
Or
= 0.049 mol dn-3
So long as the chloride in concentration remains smaller than 0.049 mol dn-3, no PbCl2 should
precipitate. In order to reduce the Silver concentration in solution as far as possible (i.e., to precipitate
out as much silver chloride as possible) while not exceeding 0.049 mol dn-3. If exactly this
concentration of Cl-(aq) is chosen, then the equilibrium
1.6 * 10-10
Ks.p
[Ag+]
=
=
[Cl-]
0.049
= 3.3 * 10-9
At that concentration of Cl-, the concentration of Ag+ will have been reduced to 3.3 * 10-9 from the
original concentration of 0.10 mol dm-3. In other words, only about three Ag+ ions in 108 will remain
in solution, but all the Pb2+ will be left in solution. A nearly perfect separation of two ionic species.
The chloride ion is the selected species.
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