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

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

Edited By Team Careers360 | Updated on Oct 19, 2024 03:14 PM IST

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.

This Story also Contains
  1. Common Ion Effect
  2. Applications of Common ion effect
  3. Some Solved Examples
  4. Summary
Common Ion Effect - Definition, Example, Exceptions, Application,  FAQs
Common Ion Effect - Definition, Example, Exceptions, Application, FAQs

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
  • $\begin{aligned} & \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} \\ & \mathrm{CH}_3 \mathrm{COONa} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{Na}^{+}\end{aligned}$
    Here $\alpha$ for CH3COOH will be decreased by CH3COONa

Related Topics,

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,

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 Ka 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 Ka is given as follows:

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

Again, the chemical equation for HCl is given as follows:
$\mathrm{HCl} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}$
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:

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

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

Ksp expression:

Ksp = [Mg2+] [OH]2

We set [Mg2+] = 1.1 x 10-10 and [OH] = S. Substituting into the Ksp 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+ +3SO42-

Hence,Ksp = [Al3+]2 [SO42-]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 $3.2 \%$ dissociation.

1)-3.8894

2)38.894

3)4.55

4) (correct)3.8894

Solution

Given,

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

$\begin{aligned} & \text { Given } \mathrm{pH} \text { of solution }=11 \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & 11=-\log \left[\mathrm{H}^{+}\right] \\ & {\left[\mathrm{H}^{+}\right]=10^{-11} \mathrm{M}} \\ & \mathrm{Kw}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14} \\ & {\left[\mathrm{OH}^{-}\right]=\frac{1 \times 10^{-14}}{10^{-11}}=1 \times 10^{-3} \mathrm{M}} \\ & \text { Strength }=\text { Molarity } \times \mathrm{mol} . \text { weight } \\ & =10^{-3} \times 40 \\ & =0.04 \mathrm{~g} / \text { litre. }\end{aligned}$
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

$\begin{aligned} & \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} \\ & \mathrm{HCl} \longrightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}\end{aligned}$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.

Also check-

NCERT Chemistry Notes:

Frequently Asked Questions (FAQs)

1. What does it mean to have a common ion effect?

When a common ion (an ion that is already present in the solution) is added to a solution, the common ion effect defines the effect on equilibrium that happens. The solubility of a solution is generally reduced by the common ion effect.

2. Explain common ion effect with example.

The common ion effect is a phenomena in which adding a common ion to two solutes produces precipitation or decreases ionisation. When sodium chloride (NaCl) is introduced to a solution of HCl and water, the common ion effect occurs.

3. What is the common ion effect's application?

For gravimetric measurement, the common ion effect is employed to completely precipitate one of the ions as a sparingly soluble salt with a very low solubility product value. Silver ions precipitates to form silver chloride and barium ions precipitates to form barium sulphate.

4. What is the principle of Le Chatelier?

If a reaction at equilibrium is subjected to a change in parameters such as temperature, pressure, or the concentration of reactants and products, Le Chatelier's principle asserts that the reaction equilibrium shifts in a direction that eliminates the change.

5. What is the odd-ion effect?

Common ion effect is neutralised when any ion from other electrolyte absorbs any ion from interfering electrolyte. This effect is called the odd-ion effect.

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