Common Ion Effect - Definition, Example, Exceptions, Application, FAQs

The Common Ion Effect was first discovered by the chemist William Hyde Wollaston in the starting of 19th century. However, this phenomenon was explained by the other scientist Gilbert N. Lewis in year of 1923. He provides very good and easy explanation of the in the 1923. The common ion effect was observed When the solubility od salt is change in the solution where the command ion were present. William observe that the solubility of the particularly is started decrease when the same ion are already present in the solution.

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This Story also Contains

1. Common Ion Effect
2. Applications of Common ion effect
3. Some Solved Examples
4. Summary

So the common idea effect idea is refer to as the change in the solubility of ionic compound in the solution which alredy ha the same ion. And this process occur fur to the change in the position of the equilibrium position according to Le Chatelier's principle explanation which is in a solution of a weak acid or base, adding a common ion shifts the equilibrium towards the formation of the undissociated compound, by which the concentration of the is reduced in that solution. For example, adding a salt that contains the common ion to a solution as that of weak acid decreases the dissociation of the acid and by which the concentration of the hydride ion is decreased.

Common Ion Effect

The iso-hydric effect and the common ion effect are both related to each in such a way that the presence of ions affects the equilibrium of chemical reactions, but they describe different phenomena. the iso-hydric effect is a specific application of the common ion effect, particularly use for the system of acids and bases . Both of these effects depict that how the addition of a common ion affects equilibrium, but the iso-hydric effect often focuses on pH-related changes in equilibrium involving weak acids and bases.

The Common Ion Effect is defined as the phenomenon where the solubility of an ionic compound decreases in a solution that already contains one of the ions present in the compound.In other words, the presence of a common ion in a solution reduces the dissociation of a weak acid or base and shifts the equilibrium of the solubility reaction, according to Le Chatelier's principle. This occurs because the increased concentration of one of the ions suppresses the dissociation of the ionic compound.

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The value of the degree of dissociation for a weak electrolyte is decreased by the addition of a strong electrolyte having a common ion. As a result of this effect, the concentration of the uncommon ion of the weak electrolyte decreases.
For example:

• ${\mathrm{NH}}_4\mathrm{OH}\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{-}$
  weak electrolyte
  
  ${\mathrm{NH}}_4\mathrm{Cl}\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{Cl}}^{-}$
  Common ion
  Here $\alpha$ for NH₄OH will be decreased by NH₄Cl
• $\begin{array}{rl}& {\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}^{+}\\ & {\mathrm{CH}}_3\mathrm{COONa}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{Na}}^{+}\end{array}$
  Here $\alpha$ for CH₃COOH will be decreased by CH₃COONa

Related Topics,

• Types of Titration
• Reversible and Irreversible Changes
• Ion Definition
• Neutralization

Applications of Common ion effect

• The solubility of a partially soluble salt decreases due to the common ion effect. For example, the presence of AgNO₃ or KCl decreases the solubility of AgCl in water.
• Salting out of soap by addition of NaCl.
• Purification of NaCl by passing HCl gas.

Isohydric Solution:

These are the solutions having the same concentration of common ions. An isohydric solution refers to solutions that maintain the same pH level, indicating an identical concentration of hydrogen ions. This can occur when solutions of different acids or bases are mixed in such a way that the resulting solutions have the same hydrogen ion concentration. When mixing solutions of different acids or bases, if the total concentration of hydrogen ions remains constant, the solutions can be considered isohydric. This principle is used in studying the buffering capacity of solutions.

Also, students can refer,

• NCERT notes Class 11 Chemistry Chapter 7 Equilibrium
• NCERT solutions for Class 11 Chemistry Chapter 7 Equilibrium
• NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium

Ph Of Weak Acid +Weak Base

The pH of a mixture of a weak acid and a strong acid can be understood using the following example.

Weak acid: H₂S (0.1M)
Strong acid: HCl (0.3M)
The K_a value for this mixture = 1.2 x 10^-20

The chemical equation for H₂S is given as follows:

${\mathrm{H}}_2\mathrm{S}\rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{S}}^{2-}$
Initial: c 0 0
Equil: c - c? c? c?

Thus, the equilibrium constant K_a is given as follows:

$\begin{array}{rl}& {\mathrm{K}}_{\mathrm{a}}=\frac{[\mathrm{c}\alpha {]}^2\cdot \mathrm{c}\alpha }{\mathrm{c}(1-\alpha )}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2\cdot \mathrm{c}\alpha }{\mathrm{c}(1-\alpha )}\\ & 1.2\times {10}^{-20}=\frac{[\mathrm{c}\alpha +0.3]\cdot \mathrm{c}\alpha }{\mathrm{c}(1-\alpha )}=(0.3{)}^2=0.09\alpha \\ & \text{ Thus, }\alpha =\frac{1.2\times {10}^{-20}}{0.09}=13.3\times {10}^{-20}\end{array}$

Again, the chemical equation for HCl is given as follows:
$\mathrm{HCl}\to {\mathrm{H}}^{+}+{\mathrm{Cl}}^{-}$
Initial: 0.3 0 0
Equil: 0 0.3 0.3

Thus, total [H⁺] for the mixture = c? + 0.3
[H⁺] = 3 x 10^-1 (? is very small for weak acids)

Now, the pH for the mixture is given as follows:

$\begin{array}{rl}& \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=-\log [3\times {10}^{-1}]=-\log 3+1=-0.47+1\\ & \mathrm{pH}=0.53\end{array}$

Recommended topic video on (Common ion effect )

Some Solved Examples

Example.1

1.The solubility product of Mg(OH)₂ is 1.2 x 10^-11. What minimum OH^- concentration must be attained (for example, by adding NaOH) to decrease the Mg²⁺ concentration in a solution of Mg(NO₃)₂ to less than 1.1 x 10^-10 M?

1) (correct)0.33

2)0.33-11

3)0

4)None of the above

Solution

K_sp expression:

K_sp = [Mg²⁺] [OH]²

We set [Mg²⁺] = 1.1 x 10^-10 and [OH] = S. Substituting into the K_sp expression:

1.2 x 10^-11 = (1.1 x 10^-10) (S)²

S = 0.33 M

Any sodium hydroxide solution greater than 0.33 M will reduce the [Mg²⁺] to less than 1.1 x 10^-10 M.

Hence, the answer is the option (1).

Example.2

2.The expression for solubility product of Al₂(SO₄)₃ is:

1)${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{A}}^{3+}\right]\left[{\mathrm{SO}}_4^{2-}\right]$

2) (correct)${\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Al}}^{3+}\right]}^2{\left[{\mathrm{SO}}_4^{2-}\right]}^3$

3)$\mathrm{Ksp}={\left[{\mathrm{Al}}^{3+}\right]}^3{\left[{\mathrm{SO}}_4^{2-}\right]}^2$

4)None of above

Solution

The solubility of Al₂(SO₄)₃.

Al₂(SO₄)₃ ⇌ 2Al³⁺ +3SO₄^2-

Hence,Ksp = [Al³⁺]² [SO₄^2-]³

Hence, the answer is the option (2).

Example.3

3.On addition of ammonium chloride to a solution of ammonium hydroxide:

1)Dissociation of NH₄OH increases

2) Concentration of OH⁻ increases

3) (correct) Concentration of OH⁻ decreases

4)Concentration of NH₄⁺ and OH⁻ increases

Solution

Common ion effect - The value of the degree of dissociation for a weak electrolyte is decreased by the addition of a strong electrolyte having a common ion. As a result of this effect, the concentration of the uncommon ion of the weak electrolyte decreases.
For example:

• ${\mathrm{NH}}_4\mathrm{OH}\leftrightarrow {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{-}$
  weak electrolyte
  ${\mathrm{NH}}_4\mathrm{Cl}\leftrightarrow {\mathrm{NH}}_4^{+}+{\mathrm{Cl}}^{-}$
  Common ion
  Here NH₄OH will be decreased by NH₄Cl
• Due to the common ion effect, the concentration of OH⁻ decreases.

Hence, the answer is the option (2)

Example.4

4.Find the pH of $0.004\mathrm{M}{\mathrm{NH}}_4\mathrm{OH}$ having $3.2\mathrm{\%}$ dissociation.

1)-3.8894

2)38.894

3)4.55

4) (correct)3.8894

Solution

Given,

$\begin{array}{ll}{\mathrm{NH}}_4\mathrm{OH}\to {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{-}(\text{weak base })& \\ \mathrm{c}00& \text{ conc. before ionization }\\ \mathrm{c}(1-\mathrm{a})0\text{ c.a }\text{ c.a }& \text{ conc. after ionization }\\ \left[{\mathrm{H}}^{+}\right]=\mathrm{c}\cdot a=4\times {10}^{-3}\times \frac{3.2}{100}& \\ & \\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=-\log [1.28\times {10}^{-2}]& \\ \mathrm{pH}=1.8927+2\approx 3.8894& \end{array}$
Hence, the answer is the option (4).

Example.5

5.How many grams of NaOH must be dissolved in one litre of solution to give it a pH value of 11?

1) (correct)0.04

2)0.08

3)0.4

4)4

Solution

$\begin{array}{rl}& \text{ Given }\mathrm{pH}\text{ of solution }=11\\ & \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\\ & 11=-\log \left[{\mathrm{H}}^{+}\right]\\ & \left[{\mathrm{H}}^{+}\right]={10}^{-11}\mathrm{M}\\ & \mathrm{Kw}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14}\\ & \left[{\mathrm{OH}}^{-}\right]=\frac{1\times {10}^{-14}}{{10}^{-11}}=1\times {10}^{-3}\mathrm{M}\\ & \text{ Strength }=\text{ Molarity }\times \mathrm{mol}.\text{ weight }\\ & ={10}^{-3}\times 40\\ & =0.04\mathrm{g}/\text{ litre. }\end{array}$
Hence, the answer is the option (1).

Example.6

6.What will happen to the degree of dissociation of ${\mathrm{CH}}_3\mathrm{COOH}$ if HCl is added to the solution?

1)It will increase because of the common ion

2) (correct)It will decrease because of the common ion

3)It will remain the same as HCl is a strong acid

4)It will first increase and then decrease

Solution

$\begin{array}{rl}& {\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}^{+}\\ & \mathrm{HCl}⟶{\mathrm{H}}^{+}+{\mathrm{Cl}}^{-}\end{array}$As HCl is a strong acid, it will produce ions which will cause the dissociation equilibrium to shift in the backward direction.

Hence, the degree of dissociation will decrease.

Hence, the answer is the option (2).

Summary

The common ion effect and the isohydric effect are important topics in chemistry, particularly in the relation of acid and base equilibrium and also in the buffer solution. The common ion Effect is very important in Knowing and for making rhe buffer Solutions. Adding that ion in the the solution Which is already present in the solution will change the position of the equilibrium. Which is helpfuk for maintaining the constant ph of Solution.

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• NCERT Exemplar Class 11th Chemistry Solutions
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NCERT Chemistry Notes:

• NCERT Notes Class 11th Chemistry
• NCERT Notes Class 12th Chemistry
• NCERT Notes For All Subjects