Common Ion Effect - Effect of concentration (Le Chatliers)

The solubility of an ionic compound is reduced if it is added to a solution already containing one of its ions ( a common ion).

if the solution being added contains either one of the ions in the original ions already present in the solution then this will reduce the solubility of the salt ice the presence of the common ion will moe the equilibrium toward the side of the precipitate salt.

Because Ks is a constant, the increase in concentration of one ion means that the concentration of the other must fall to maintain the value of Ks

Consider Le Chatliers Principle when solving these problems concerning the common ion effect.

Eg: AB(s) <-----> A+(aq) + B- (aq)

if the concentration of either A+ or B- (the products) is increased, the position of the equilibrium will shift to the LEFT, so if [A+] increases then [B-] will decrease and vice versa

EXAMPLE QUESTION

A small amount of the sparingly soluble ionic compound , Barium Sulphate (BaSO3), was added to 50mL of pure water at 25*C.
Equilibrium between the solid and its ions was allowed to form.
This solution and the undissolved solid were added to 50 mL of 0.100 mol L-1 Barium Iodate (Ba (IO3)2) solution.
Ks (BaSO3) = 5.00 x 10-10. Ks( Ba(IO3)2)= 1.5 x 10-9

The common ion between these 2 solutions is Barium (Ba) as they both contain this.

When these two solutions are mixed, the concentration of the sulphate and iodate ions will decrease, and the concentration of solid barium sulphite in the system will increase.

Explain the effect that adding Ba(IO3)2) solution has had on the equilibrium that has been established between solid BaSO3 and its ions in solution (calculations are not required) .

In terms of Le Chatliers:

BaSO <-----> Ba2+ (aq) + SO3 2- (aq)

Le Chatliers principle states that constraints placed on a system at equilibrium will force the equilibrium to shift in order to minimise the effects of these constraints. In this case the constraint is concentration. When the concentration of Barium (Ba2+) increases the position of the equilibrium will shift to the left hand side to decrease the concentration of Ba2+ being produced and also decreasing the concentration of SO32-. This means that the BaSO4 becomes less soluble.

In terms of Ks, the Ks is a constant. The common ion effect causes the Ks to be initially overtaken, causing precipitation until the new concentrations satisfy the value of Ks again

Common Ion Calculations

Note: It is important to note that calculation involving common ions Make major assumptions

"The common ion added to the solution containing the sparingly soluble compound will have a concentration very much higher than the concentration of the common ion in equilibrium solution. In calculations it is therefore acceptable to assume that all of the common ions come from the solution other than the one containing the sparingly soluble compound"

50.0mL of a 0.500 molL-1 solution of sodium chloride was added 50.0mL of a saturated solution of silver chloride. Calculate the [Ag +] after the two solutions were mixed and equilibrium allowed to re-establish. Ks (AgCl)= 1.77 x 10-10

Step 1- Assume all Cl- ions come from the sodium chloride solution, Include a dilution factor- 50.0 ml @ 0.500 made up to 100mL.

[Cl-] = 0.250 mol L-1
Ks= [Ag+][Cl-] = s2

Step 2- Rearrange the solubility product expression to save for the ion whose concentration is required.

Step 3- Substitute values and calculate (include unit)

Example 2

Dissolving of AgCl in pure water.

Equation: AgCl (s)<------> Ag2+ (aq) + Cl- (aq)

Ks (AgCl)= [Ag+][Cl-]

AgCl will be less soluble in sea water than in pure water because of the presence of the common Cl- ion dissolved in the seawater.
The value of Ks cannot change. So the [Cl-] Means that there is a lower [Ag+] at equilibrium. As the equilibrium shifts to minimise the effects .
This reduces the amount of solid AgCl that can dissolve

Example calculation - What is the solubility of AgCl in a -0.0025 molL-1 solution of NaCl?

Ks(AgCl) = [Ag+(aq)] x [Cl-(aq)]= 1.6x10 -10

Ks(AgCl) = [Ag+(aq)] x [Cl-(aq)]

1.6 x 10-10 = s x 0.0025 (rearrange formula to find s)

s= 1.6 x 10-10 / 0.0025

s= 6.4 x 10 -8 molL-1

	Created by Sophie Painter over 7 years ago

NCEA Level 2 Chemistry Note on L3- Equibrium Principles, created by Sophie Painter on 01/09/2018.

L3- Equibrium Principles