Experiment #9
Solubility Product
and
Common ion effect
What are we doing in
this experiment?
Determine the molar solubility and solubility
product constant (Ksp) of potassium hydrogen
tartarate (KHT).
Study the effect of common ion on the Ksp
of KHT and the molar solubilites of its ions.
Remember!!
In this experiment, we are dealing with compounds
that are very slightly soluble that they are called
“Insoluble compounds”.
Our bones and teeth are mostly calcium phosphate,
Ca3(PO4)2, a very slightly soluble compound.
Solubility product
In general, the solubility product expression for a compound
is the product of the concentration (molar solubility) of its
constituent ions, each raised to the power that corresponds to
the number of ions in one formula unit of the compound. The
quantity is constant at constant temperature for a saturated
solution of the compound.This statement is called the
solubility product principle
yMZ+ (aq)
MyXz (s)
K sp  M
Solubility product constant
+ zXY-(aq)
 X 
z y
y z
Molar solubility of the ions
Solubility product
2Bi3+ (aq)
Bi2S3 (s)
K sp  Bi
Solubility product constant
+ 3S2-(aq)
 S 
3 2
2 3
Molar solubility of the ions
Remember that we are dealing with molar solubilities and
not concentration.
For a saturated solution, molar solubility is equal to molar
concentration.
Different types of solution
Unsaturated solution: More solute can be dissolved
in it.
Saturated solution: No more solute can be dissolved
in it. Any more of solute you add
will not dissolve. It will precipitate
out.
Super saturated solution: Has more solute than can
be dissolved in it. The
solute precipitates out.
Molar solubility
Let us say, we try to dissolve 1 g of Bi2S3 in 1 L of water. If
only 8.78×10-13 g out of the 1.0 g dissolves, we can make the
following conclusions:
1. The solution is saturated with Bi2S3.
2. If we filter out the undissolved Bi2S3, the amount of solute
that dissolved (soluble) in 1. 0 L of water is 0.0025 g.
2Bi3+ (aq)
Bi2S3 (s)
K sp  Bi
+ 3S2-(aq)
 S 
3 2
2 3
So we can say, the solubility of Bi2S3 is 8.78 ×10-13 g per liter
Molar solubility
Molar solubility is solubility in moles per liter
So lub ility in moles So lub ility in grams
1mol


1L
1L
Molar mass( g )
8.78  10 13 g
1mol
Molar so lub ility of Bi2 S3  Bi2 S3  

1L
513.96( g )
1.708 1015 mol
Bi2 S3  
 1.708 1015 M
1L
Solubility product constant
2Bi3+ (aq)
Bi2S3 (s)
K sp  Bi
Bi2S3 (s)
+ 3S2-(aq)
 S 
3 2
2 3
2Bi3+ (aq)
+ 3S2-(aq)
If we wanted to figure out the Ksp of Bi2S3, then we need to
know the molar solubilities of Bi3+ and S2-. The molar
solubilities of the ions are usually figured out from the
solubility of the parent compound.
Solubility product constant
If the molar solubility of Bi2S3 is “s”, the molar solubility
of Bi2+ is “2s” and the molar solubility of S2- is “3s”.
2Bi3+ (aq) + 3S2-(aq)
Bi2S3 (s)
s
2×s
3×s
This is because, there are 2 ions of Bi3+ produced for
Each molecule of the parent, Bi2S3 and 3 ions of S2produced for each molecule of the parent.
K sp  Bi
 S 
3 2
2 3
K sp  2  s  3  s 
2
3
K sp  2 s  3s 
2
3
Solubility product constant
K sp  2 s  3s 
2
3
K sp  2s  2s  3s  3s  3s
K sp  (2  2  3  3  3)  ( s  s  s  s  s)
K sp  (4  27)  ( s )
5
K sp  (108)  ( s )
5
Since we already know the value of molar solubility for
Bi2S3, which is 1.708×10-15 M
K sp  (108)  (1.708 10 )
15 5
K sp  1.569 10
72
How to find the molar solubility if
we know is Ksp?
Find the molar solubility of Ca (OH)2, if the Ksp of
Ca(OH)2 is 7.9 ×10-6.
Ca(OH)2 (s)
Ca2+ (aq) + 2OH-(aq)
s
1×s
2×s
Let the molar solubility of Ca(OH)2 be “s”. So, the molar
Solubility of Ca2+ should be “1s” and the molar solubility
of OH- should be “2s”.
This is because, there are 1 ion of Ca2+ produced for
each molecule of the parent, Ca(OH)2 and 2 ions of OHproduced for each molecule of the parent.
How to find the molar solubility if
we know is Ksp?
Ca2+ (aq) + 2OH-(aq)
Ca(OH)2 (s)
s
1×s
K sp  Ca
 OH 
2 1
 2
K sp  1 s  2  s 
1
2
K sp  (1s)  (2s)  (2s)
K sp  (1  2  2)  ( s  s  s)
K sp  4s 3
2×s
How to find the molar solubility if
we know is Ksp?
K sp  4s 3
But , K sp  7.9 10
7.9  10  4 s
6
6
3
7.9 10
 s3
4
6
1.97  10  s
6
3
s  1.97 10  1.25 10 M
3
6
2
How to find the molar solubility if
we know is Ksp?
s  1.97 10  1.25 10 M
3
6
2
So the molar solubility of Ca(OH)2 = s = 1.25 ×10-2 M
The molar solubility of Ca2+ = [Ca2+] =1s = 1.25 ×10-2 M
The molar solubility of OH- = [OH-] = 2s = 2×1.25 ×10-2 M
2.50 ×10-2 M
Is it possible to find the pH of the
Ca(OH)2 solution at 25C?
Yes
We know that [OH-] = 2.5 ×10-2 M
Ionic product of water , Kw  H 3O  OH

K w (25 C )  H 3O  OH   110



14
H O  2.5 10   110
110
H O  
 4 10 M

3
2
14

3
14
13
2.5 10
2


Is it possible to find the pH of the
Ca(OH)2 solution at 25C?
110
13
H 3O  

4

10
M
2
2.5 10
14

pH   Log H O 

3
pH   Log 4 10
13

pH  12.39
The Ca(OH)2 solution is basic.
Experiment- To determine the Ksp of
Potassium Hydrogen Tartarate, KHT
KHT is also called cream of tartar
K+ (aq) + HT-(aq)
KHT (s)
COOK
H-C-OH
H-C-OH
s
1×s
K sp  K
1×s
 HT 
 1
 1
COOH
KHT
If we want to determine the Ksp of KHT,
we need to know the molar solubilities of
K+ and HT-. Also remember that Ksp is
measured for a saturated solution.
How do we determine [K+] and [HT-]?
Firstly prepare a saturated solution of KHT.
3.0 g of KHT in 200 ml of water.
Filter out the undissolved KHT using gravity
filtration.
Now we have a saturated solution of KHT.
KHT (s)
K+ (aq) + HT-(aq)
COOK
H-C-OH
H-C-OH
s
1×s
K sp  K
1×s
 HT 
 1
 1
COOH
KHT
HT- can act an acid, so if we titrate it with a known
concentration of base (NaOH), we can find the [HT-]
How do we determine [K+] and [HT-]?
Once we know the concentration of HT-, based on
1 to 1 molar relationship between K+ and HT-,
[K+]= [HT-]
K sp  K
 HT 
 1
 1
NaOH is hygroscopic, so the NaOH solution needs to
be standardized by using KHP
LeChatelier’s Principle
If a stress (change of condition) is applied to a
system at dynamic equilibrium,the system shifts
in the direction that reduces the stress.
Common ion effect
Suppression of ionization of a weak electrolyte by
the presence in the same solution of a strong
electrolyte containing one of the same ions as
the weak electrolyte.
About Common ion effect
Common ion effect is a special case of
LeChatelier principle
Addition of a common ion is equivalent
to adding a stress to the system.
The system responds to the stress by
reducing the solubility of one of the ions
and keeping the Ksp constant.
Calculate the molar solubility of lead iodide PbI2,
from its Ksp in water at 25C
PbI2 (s)
Pb2+ (aq) + 2I-(aq)
s
1×s
K sp  Pb
2×s
 I 
2 1
 2
K sp  1 s  2  s 
1
2
K sp  (1s)  (2s)  (2s)
K sp  (1  2  2)  ( s  s  s)
K sp  4s
3
K sp  4s 3
But , K sp  7.9 10
7.9  10  4 s
9
9
3
7.9 10
 s3
4
9
1.97  10  s
9
3
s  1.97 10  1.3 10 M
3
6
3
Calculate the molar solubility of lead iodide PbI2,
in 0.1 M NaI solution
Pb2+ (aq) + 2I-(aq)
PbI2 (s)
s
K sp ( PbI 2 )  7.9 10
1×s
9
2×s
Na+ (aq) + I-(aq)
NaI (s)
0.1M
K sp  Pb
0.1M
 I 
2 1
 2
7.9 10  1 s (2  s)  0.1
9
1
2
Common
ion
Calculate the molar solubility of lead iodide PbI2,
in 0.1 M NaI solution
7.9 10  1 s (2  s)  0.1
9
1
2
7.9 10  (1s)2s  0.1
9
2
Because the Ksp of PbI2 is really small, the solubility s is going
to be really small. Hence we can make a simplification.
2s  0.1
 0.1
7.9 10  (1s)0.1
7.9 109
s
 7.9 107 M
0.01
9
2
A comparison of solubility of PbI2
With and without common ion
With common ion
Without common ion
s= 1.3 × 10-3 M
s= 7.9 × 10-7 M
Solubility decreases because of the presence of
common ion
To study the effect of common ion
on the solubility of KHT
K+ (aq) + HT-(aq)
KHT (s)
s
1×s
K+ (aq) + Cl-(aq)
KCl (s)
0.1 M
Common ion
1×s
K sp  K
0.1 M
 HT 
 1
 1
K sp  1s  0.1 1s 
1
1
HT- can act an acid, so if we titrate it with a known
concentration of base (NaOH), we can find the [HT-]