Common Ion Effect

The Common Ion Effect is based on Le Chatelier’s Principle for chemical equilibrium of the salts and other weak electrolytes and their ions in solutions. This effect is commonly used to manipulate the solubilities of salts and weak electrolytes, thereby also being used to precipitate salts from solutions.

History

Common Ion Effect is directly based on Le Chatelier’s Principle which was given by Henry Louis Le Chatelier. Le Chatelier was a French chemist who originally studied engineering. Le Chatelier and his father played important roles in the introduction and formation of French aluminium industries and the rise of railway transportation. Le Chatelier taught chemistry at the Ecole Polytechnique, even though he was originally an engineer and also had an interest in the issues of industries.

Le Chatelier’s most well-known works are Le Chatelier’s Principle, Principle of Chemical Equilibrium and the varying solubilities of salts in ideal solutions.

Common Ion Effect Explained

Common-Ion Effect is the phenomenon in which the solubility of a dissolved electrolyte reduces when another electrolyte, in which one ion is the same as that of the dissolved electrolyte, is added to the solution.

This is fundamentally based on Le Chatelier’s Principle, where if the concentration of any one of the reactants is increased then the equilibrium shifts in the direction in which these changes made to the equilibrium are nullified.

In common-ion effect, when a different salt or electrolyte with one common ion, with the already dissolved salt or electrolyte, is added to the solution, the concentration of common ion increases due to which equilibrium of the dissolved salt or electrolyte shifts to the initial salt or electrolyte side causing its solubility to decrease. 

For example, consider an imaginary salt AB and electrolyte CB. Here, CB is a weak electrolyte compared to AB. When the salt AB is added to a solution in which CB is already dissolved, due to the common ion effect BC’s ionization reduces as a result its solubility also reduces and the equilibrium shifts towards the left.

CB  ⇌  C⁺  +  B⁻
AB  ⇌  A⁺  +  B⁻

Limitations of Common Ion Effect

Transition elements usually violate this rule due to their ability to form complex compounds. As this is different from the equilibrium established when a salt is dissolved, the common-ion effect does not apply to complex compounds formed by these transition elements.

Applications

1. To reduce the sodium carbonate(Na₂CO₃) hardness of water using limestone and chalk.
2. Used in the precipitation of soaps in their manufacturing process.
3. Used to completely precipitate the salt of an ion with very low solubility product for gravimetric estimation.
4. It plays an important role in the regulation of buffer solutions.

Examples

Question 1. In a solution of acetic acid if we add 0.1M sodium acetate, then calculate the new pH of the solution. The initial concentration of the solution 0.1M(K_a of CH₃COOH = 1.8 × 10^-5).

 Solution.

CH₃COONa    ⇌    Na⁺  +  CH₃COO⁻
CH₃COOH     ⇌      H⁺  +  CH₃COO⁻

0.1M                    0M                  0.1M          : before equilibrium
0.1-x                      x                    0.1+x         : after equilibrium

Here, we know that the dissociated quantity of acetic acid x is very very less than 0.05, because the Kₐ value of a weak acid is very low. 

x << 0.05.

So, 
(0.1-x) ≈ (0.1+x) ≈ 0.1

We can calculate the pH of this new solution through Kₐ of acetic acid.

=> \(1.8 × 10^{-5} = \frac{(x)(0.1-x)}{(0.1+x)}\)
=> \(1.8 × 10^{-5} = (x)\frac{(0.1)}{(0.1)}\)
=> \(1.8 × 10^{-5} = x = [H^+]\)

We know that the pH of a solution is, 
\(pH= -\log[H^+]\)
\(pH = -\log(1.8 × 10^{-5})\)
=> \(pH = 4.74\)

Question 2. Calculate the pH of 100ml of 0.1M ammonia solution after it has been treated with 50ml of 0.1M HCl solution(K_b of ammonia = 1.77×10^-5).

Solution.  Up on the addition of 50ml of 0.1M HCl(i.e. 5mmol of HCl), 5mmol of the ammonia is neutralized. Now the left 5mmol of ammonia in 150ml of solution reacts with HCl to form ammonium ions and chlorine ions.

HCl  +  NH₃  ⇌  NH₄⁺  +  Cl^–

This ammonium ion formed reacts with water molecule and forms ammonium hydroxide, which ionises and is in an equilibrium with its ions. The molarities of neutralized 5mmol and unneutralized 5mmol ammonia is 0.033M.

NH₄OH           ⇌          NH₄⁺        +         OH⁻

0.033-a        0.033+a            a : At equilibrium

The equilibrium concentration of NH₄⁺ is (0.033+a)M because the already neutralized NH₃ is also present in the solution asNH₄⁺ which we need to include.

As given in the question, the dissociation constant Kb of ammonia is quite small, so the value of “a” is also very small.

So,
(0.033-a) ≈ (0.033+a) ≈ 0.033

We know that,

\(K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]}\)

=> \(1.77 × 10^{-5} = \frac{(a)(0.033+a)}{(0.033-a)}\)
=> \(1.77 × 10^{-5} = (a)\frac{(0.033)}{(0.033)}\)
=> \(1.77 × 10^{-5} = a = [OH^-]\)

We know that,

\(K_w = [H^+][OH^-] = 10^{-14}\)
=> \([H^+] = \frac{10^{-14}}{[OH^-]}\)
=> \([H^+] = \frac{10^-{14}}{1.77 × 10^{-5}}\)
=> \([H^+] = 0.56 × 10^{-9}\)

Since,
\(pH= -\log[H^+]\)
\(pH= -\log(0.56 × 10^{-9})\)
=> \(pH= 9.24\)

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