Common Ion Effect | Applications | Numericals

The common ion effect is the suppression in dissociation (decrease in the degree of dissociation) of weak electrolytes in the presence of a strong electrolyte, both having a common ion. When we dissolve a weak electrolyte such as weak acid, CH₃COOH, or a weak base as NH4OH in water, an equilibrium is established between the ions and the undissociated molecules.

CH₃COOH  ⇌ CH₃COO^– + H⁺

NH₄OH  ⇌ NH₄⁺ + OH^–

Now, a strong electrolyte containing a common ion is added to a solution of a weak electrolyte, then the concentration of the common Ion increases. For example, if NH₄Cl, a strong electrolyte, is added to the solution of NH₄OH, it dissociates and provides a large number of NH₄⁺ ions.

NH₄OH  ⇌ NH₄⁺ + OH^– [weak dissociation]

NH₄Cl  ⇌ NH₄⁺ + Cl^– [strong dissociation]

This increase in the concentration of NH₄⁺ ions causes the equilibrium of dissociation of NH4OH to be shifted towards the left, and has dissociation of ammonium hydroxide is suppressed. In other words, the degree of dissociation of weak electrolytes is suppressed by the addition of the solution of a strong electrolyte having a common ion; this is known as the common Ion effect.

Condition For Common Ion Effect

“Both the weak and strong electrolyte having common ions.”

Principle On Which Common Ion Effect Based

“Common Ion effect based on Le Chatelier’s principle”

According to Le Chatelier’s principle,

• If the concentration of the reactants increases, then the reaction is moved in the forward direction.
• If the concentration of the product increases, then the reaction is moved in the backward direction.
• If we increase the concentration of reactants, then the reaction is moved in the backward direction.
• If we decrease the concentration of the product, then the reaction is moved in the forward direction.

Explanation

Consider a weak electrolyte, HCN, which is partially dissociated into ions.

HCN ⇌ H⁺ + CN^–

In the same solution, a strong electrolyte, KCN, is present, which is 100% ionized and gives ions.

KCN ⇌ K⁺ + CN^–

So, in the solution, CN^– is a common ion.

Firstly, when HCl is only in the solution and the reaction is at equilibrium, but when we add KCN in the same solution, the concentration of CN^– Ion is increased in the solution, and according to Le Chatelier’s principle, if the concentration of product increases the reaction will move in the backward direction. So, CN^– ion reacts with H^+, and equilibrium will shift toward the backward direction and decrease the degree of dissociation of the weak electrolyte due to common ion.

Another example of a common ion effect is CH₃COOH and CH₃COONa. Consider a weak electrolyte, CH₃COOH, partially ionized and gives ions.

CH₃COOH  ⇌ CHCOO^– + H⁺

Another electrolyte, CH₃COONa, is strong and completely ionized in the solution and gives ions in the same solution.

CH₃COONa  ⇌ CHCOO^– + Na⁺

So, in the solution, CH₃COO^– is a common ion.

Firstly, CH₃COOH is at equilibrium, but when we add CH₃COONa (a strong electrolyte) in the same solution, it dissociates, and the CH₃COO^– ion concentration increases, and the OH^– ion concentration decrease. According to Le Chatelier’s principle, the reaction moves backward due to a common ion. As a result, the dissociation of weak electrolyte CH3COOH is suppressed due to common ions.

Importance of Common Ion Effect

1. In qualitative analysis, the conditions are created to precipitate certain cations while other cations remain in solution. This is done by the addition of an appropriate amount of the strong electrolyte having a common ion. The dissociation of a weak electrolyte is suppressed to the desired extent.
2. The gravimetric analysis also involved the common Ion effect. A gravimetric precipitate has slite solubility in the precipitating medium. The solubility of the precipitate can be decreased by using the common Ion effect. If the solubility is not very low, this method cannot be used for gravimetric estimation. This is achieved mainly through adding a slight excess of the precipitating agent having a common ion.

Example

If we want to determine Ag⁺ as AgCl, then a Cl^– is added to the solution of silver salt. If an exactly equivalent amount of Cl^– ion is added, then the resulting saturated solution of the AgCl will contain 0.001 g dm^-3 of AgCl in an aqueous medium. This is because it has 0.4% solubility. However, if a slight excess of Cl^– ion, say a 0.01 M, is added during precipitation, then the solubility of AgCl is reduced to 0.004%. Therefore, a very accurate estimation of Ag ⁺ can be done if a slight excess of Cl^– ion is added during precipitation.

Application of Common Ion Effect

1. Purification of Crude Common Salt

The common crude salt obtained by the evaporation of seawater contains sodium chloride contaminated with soluble impurities such as calcium chloride, magnesium chloride, potassium Bromide, etc. The pure sodium chloride can be obtained by passing HCl gas through a concentrated aqueous solution of common crude salt. The pure sodium chloride is precipitated, and the soluble impurities remain behind in the solution. The process may be represented as:

HCl + H₂O ⇌ H₃O⁺ + Cl^–

NaCl+ H₂O ⇌ Na⁺ + Cl^–

On passing HCl gas through the solution, the concentration of Cl^– ion increases, shifting the equilibrium to the left, thereby precipitation pure sodium chloride.

2. Salting Out of Soap

The recovery of dissolved salt by adding another salt to the solution is turned salting out. The soap is a sodium salt of higher fatty acids. Sodium stearate sodium oleate is the best example. When soap is prepared, it floats over spent-lye. Spent-lye is a residual aqua solution containing unused alkali, glycerole, etc. A significant amount of soap remains dissolved in the solution. Sodium chloride is added to the boiling soap solution to recover this soap. The recovery soap separates due to the common ion effect of Na⁺ following the reactions.

RCOONa ⇌ Na⁺ + RCOO^–

NaCl ⇌ Na⁺ + Cl^–

This increased concentration of Na⁺ in the solution due to the dissociation of sodium chloride, NaCl shifts the equilibrium towards the left, and the soap is precipitated.

Numericals

Why does the concentration of S^2- ions from H₂S decrease in the presence of HCl?

H₂S is a weak electrolyte while HCl is a strong electrolyte,

H₂S ⇌ 2H⁺ + S^2-

HCl ⇌ H⁺ + Cl^–

In this solution, the concentration of H⁺ ion increases, and H⁺ ion reacts with S^2- ion and reaction move in the backward direction so, in overall solution the H⁺ ion concentration increase and S^2- ion concentration decrease.

Calculate the concentration of H⁺ ions in a solution of HCN of 0.2 M and Ka = 4×10^-10 with KCN of 1 mol/L?

Let the concentration of HCN (0.2M) = c and the degree of dissociation of HCN is alpha α.

HCN ⇌ H⁺ + CN^–

	[HCN]	[H+]	[CN–]
Initially	c	0	0
At eq	c-cα	cα	cα

KCN ⇌ K⁺ + CN^–

	[KCN]	[K+]	[CN–]
Initially	1 M	0	0
At eq	0	1 M	1 M

Ka = [H⁺] [CN^–]/ [HCN]

This Ka = [H⁺] [CN^–]/ [HCN] not tells that the concentration of [CN^–] consider only from weak electrolyte. It means that the total [CN^–] ion is at equilibrium from the solution. So, CN^– ions come from HCN and KCN.

Ka = [H⁺] [CN^–]/ [HCN]

4×10^-10 = (cα) (cα+1) / c-cα

As α ≈ 0.05 (very small), c is also small and 1-α = 1

4×10^-10 = (α) (1) / 1

α = 4×10^-10

[H⁺] = cα

[H⁺] = (0.2) (4×10^-10)

[H⁺] = (2×10^-1) (4×10^-10)

[H⁺] = 8×10^-11 M

Calculate the concentration of H⁺ ion in a mixture of 0.02 M acetic acid and 0.1 M sodium acetate. Ka for acetic acid is 10^-5?

CHCOOH ⇌ CHCOO^– + H⁺

• Let the concentration of CH₃COOH = c
• Let D.O.D of CH₃COOH = α

	[CH3COOH]	[CH3COO–]	[H+]
Intially	c	0	0
Finally	c-cα	cα	cα

Ka = [CHCOO^–] [H⁺] / [CH₃COOH]

Ka = [cα] [cα + 1] / c[1-α]

10^-5 = [α] [0.02 α + 1] / [1-α]

10^-5 = (α) (0.1) / 1

α = 10^-5 / 10^-1

α = 10^-4

[H⁺] = cα

[H⁺] = (0.02) (10^-4)

[H⁺] = 2×10^-6