Common-ion effect?

Question

can some one explain this effect to me in SIMPLE terms ?

and explain why it happens?

Answer 1

To explain this let's take a hypothetical slightly soluble salt of XY.
XY(s) ----------> X+(aq) + Y-(aq)

Suppose when the ion - concentrations reach 4M each, the system reaches the equilibrium.

Now first let's look at the some basic principles;

1) Like all equilibrium constants, Ksp is also temperature dependent. Therefore, first we will assume a constant temperature.
2) At this constant temperature, what is the meaning of Ksp?
If we consider the above example, we arrive at the following conclusion:
a) The solution is at equilibrium with 4M X+ and 4M Y-.
b) Since the solution is at equilibrium, this means that the solution is saturated.
c) Since the solution is saturated, you cannot dissolve more solute in it. This means that the maximum solubility of XY is 4M at the given temperature.
XY (s) dissolves and produces 4M X+ and 4M Y-.
XY(s) --------------> X+(aq) + Y-(aq)
.............................. 4 M ......... 4 M
These ions can be obtained by dissolving 4M XY.
XY(aq) -------------> X+(aq) + Y-(aq)
4 M ....................... 4 M ......... 4 M
That is the concentration of XY (at the given temperature) cannot exceed 4M. Therefore 4M is the solubility of XY, or, the solubility of XY, at the given temperature, is 4M (4 moles per liter).
d) Since the solubility of XY is 4M and this amount of XY will produce 4M of each of X+ and Y- ions. The Ksp of XY at that temperature will be;
Ksp = [X+] [Y-] = (4)(4) = 16
e) As long as the temperature is constant, the ion product cannot exceed this value. If exceeds, the excess portion causes precipitation. If less than this amount, more XY dissolves to satisfy the value of 16.

All the cases examined above concern the solubility of XY in pure water. Now, suppose we have added more soluble salt XZ to the solution which produces extra 2M of X+ ion.

This effect disturbs the equilibrium and according to Le Chatelier's principle the system will try to compansate this stress. Then the equilibrium will shift to left to decrease the concentration of X+.
XY(s) -------------------> X+(aq) + Y-(aq)
Initial equilibrium ...... 4 M ......... 4 M
Stress : (addition of) ..2 M
Consumption :........... - x ............. - x
New equilibrium :.... (6 - x) ........ (4 - x)

Ksp = [X+] [Y-] = 16
(6 - x) (4 - x) = 16
x2 + 10x + 8 = 0
One of the roots of this equation will be 0.88

Equilibrium concentrations;
[X+] = 6 - 0.88 = 5.12 M
[Y-] = 4 - 0.88 = 3.12 M

Check : (5.12)(3.12) = 15.97 = 16.00

Now what happened?
Since 2M X+ ion is added, the X+ ion produced by dissolving XY salt will be; 5.12 - 2 = 3.12 M and Y- ion will also be 3.12.

This means that instead of 4M , the solubility of XY will be 3.12M. As the concentration of one ion increases, the concentration of other ion decreases to satisfy the Ksp value.
Therefore, the solubility of the salt will be less compared to the solubility in pure water. This phenomenon is called "the common-ion effect".

I hope this illustrates and explains your question.

Answer 2

For Eg, if u take NaCl solution and if u add solid Na2Co3, the dissolution of na2Co3 is suppressed due to the presence of the common ion Na. This is called common ion effect.

Answer 3

you have a salt AB where A is a strong electrolyte and B a weak base. So you have only a small aamount of AB which is soluble calculated by
[A-] *[B+]= Ksp (solubility product)
anb [B+] is given by [B+]= Ksp/[A-]
If you add a strong electrolyte containing A- the amount
of ion A- will increase as Ksp is constant the mount of [B+]is decreased and the solubility of [B+] is decreased. As for AB to have a dissociation you must have equal amount of A- and B+ , the amount of AB soluble is decreased