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

Common Ion Effect

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.

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:

A weak base ammonium hydroxide dissociates as:

$\mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-}$

Ammonium chloride, a strong electrolyte, dissociates as:

$\mathrm{NH}_4 \mathrm{Cl} \rightarrow \mathrm{NH}_4^{+}+\mathrm{Cl}^{-}$

Both substances contain the common ion $\mathrm{NH}_4^{+}$. The addition of $\mathrm{NH}_4 \mathrm{Cl}$ increases the concentration of $\mathrm{NH}_4^{+}$, which suppresses the ionization of $\mathrm{NH}_4 \mathrm{OH}$. Therefore, the degree of dissociation $(\alpha)$ of $\mathrm{NH}_4 \mathrm{OH}$ decreases.

Similarly, acetic acid dissociates as:

$\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$

Sodium acetate dissociates as:

$\mathrm{CH}_3 \mathrm{COONa} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{Na}^{+}$

Here, $\mathrm{CH}_3 \mathrm{COO}^{-}$is the common ion. The addition of sodium acetate suppresses the dissociation of acetic acid, thereby decreasing its degree of dissociation ( $\alpha$ ).

Applications of Common ion Effect

1. Preparation of Buffer Solutions

The common ion effect is used in preparing acidic and basic buffer solutions that resist changes in pH.

Example:

• Acetic acid and sodium acetate form an acidic buffer.

2. Suppression of Ionization

The ionization of weak acids and weak bases decreases in the presence of a common ion.
Example:

$\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$

The addition of sodium acetate suppresses the ionization of acetic acid.

3. Selective Precipitation

The common ion effect is used in qualitative analysis for selective precipitation of ions.

Example

Addition of $C l^{-}$ions helps precipitate silver chloride ( $A g C l$ ).

4. Control of Solubility

The solubility of sparingly soluble salts decreases when a common ion is added.

Example:

$\mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-}$

Adding NaCl decreases the solubility of AgCl .

Related Topics

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

Isohydric Solution

Isohydric solutions are solutions having the same concentration of hydrogen ions (H⁺) or the same pH. When two weak acids having a common ion are mixed together, the ionization of each acid is suppressed due to the common ion effect. Such solutions are called isohydric solutions.

For example, a mixture of acetic acid and formic acid forms an isohydric solution because both acids produce H⁺ ions in solution.

The common ion suppresses the dissociation of each weak electrolyte according to Le Chatelier's Principle.

Also Read

Some Solved Examples

Question 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).

Question 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 $\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$.

$\begin{aligned}
& \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \rightleftharpoons 2 \mathrm{Al}^{3+}+3 \mathrm{SO}_4^{2-} \\
& \text { Hence, } \mathrm{Ksp}=\left[\mathrm{Al}^{3+}\right]^2\left[\mathrm{SO}_4^{2-}\right]^3
\end{aligned}$

Hence, the answer is the option (2).

Question 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)

Question 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).

Question 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).

Question 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).