Common Ion Effect: The solubility of a solute refers to the amount of solute that can be dissolved in a particular solvent. For example, table salt $\left(\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\right)$ in water dissolves easily. But, if more table salt is continuously added to the mixture, the resulting solution will reach a point at which no more salt can be dissolved. The solution hence is saturated, and the table salt has effectively reached its solubility limit.

Salts tend to ionize in a solution. For example- let us consider the ionization of a salt $\text{AB}\text{.}$

$\text{AB}\rightleftharpoons \left[{\text{A}}^{\text{ + }}\right]\text{ + }\left[{\text{B}}^{\text{ – }}\right]$

In a saturated solution, if the concentration of any one of the ions on the right-hand side is decreased, then according to Le chateliers principle, more salt dissolves, and the equilibrium shifts towards the right in a forward direction, until ${\text{K}}_{\text{sp}}\text{ = }{\text{Q}}_{\text{sp}}$

And if the concentration of any one of the ions is increased by adding an electrolyte, then according to Le chateliers principle, the equilibrium shifts towards the left, until ${\text{K}}_{\text{sp}}\text{ = }{\text{Q}}_{\text{sp}}$

If there are several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion.

Common Ions

When $\text{KCl}$ and $\text{NaCl}$ are dissolved in a particular solution, the anion $\text{C}{\text{l}}^{–}$ ions are common to both salts.

$\text{NaCl(s)}\to \text{N}{\text{a}}^{\text{ + }}\text{(aq) + C}{\text{l}}^{\text{ – }}\text{(aq)}$

$\text{KCl(s)}\to {\text{K}}^{\text{ + }}\text{(aq) + C}{\text{l}}^{\text{ – }}\text{(aq)}$

$\text{AgCl(s)}\rightleftharpoons \text{A}{\text{g}}^{\text{ + }}\text{(aq) + C}{\text{l}}^{\text{ – }}\text{(aq)}$

Similarly, when $\text{AgCl}$ is dissolved into a solution already containing $\text{NaCl},$ the $\text{C}{\text{l}}^{\text{ – }}$ ions come from the ionization of both $\text{AgCl}$ and $\text{NaCl}.$ Hence, the increase in the concentration of the $\left[\text{C}{\text{l}}^{\text{ – }}\right]$ ion interferes with the precipitation of an electrolyte.

Common Ion Effect

The common ion effect is an effect that causes suppression in the ionization of an electrolyte when another electrolyte (which contains an ion that is also present in the first electrolyte, i.e., a common ion) is added. It is a consequence of Le Chatlier’s principle (or the Equilibrium Law).

Illustration

In a saturated calcium sulfate solution, there is an equilibrium between the solid calcium sulfate and its constituting ions in the solution.

$\text{CaS}{\text{O}}_{\text{4}}\text{(s)}\rightleftharpoons \text{C}{\text{a}}^{\text{2 + }}{\text{(aq) + SO}}_{\text{4}}^{\text{2 – }}{\text{K}}_{\text{sp}}\text{ = 2}\text{.4 }\times \text{ 1}{\text{0}}^{\text{ – 5}}$

$\text{Ca}{\left(\text{N}{\text{O}}_{\text{3}}\right)}_{\text{2}}\text{(s)}\to \text{C}{\text{a}}^{\text{2 + }}{\text{(aq) + 2NO}}_{\text{3}}^{\text{ – }}$

The addition of calcium nitrate to the above-saturated solution would cause an increase in the concentration of the calcium ion in the solution. As a result, the ionic product given by ${\text{Q}}_{\text{sp}}.$

${\text{Q}}_{\text{sp}}\text{ = }\left[\text{C}{\text{a}}^{\text{2 + }}\right]\left[\text{S}{{\text{O}}_{\text{4}}}^{\text{2 – }}\right]$

would increase and now be greater than the ${\text{K}}_{\text{sp}}.$ 

According to LeChâtelier’s principle, the ionization equilibrium of calcium sulphate would shift to the left to relieve the stress of the added calcium ion. 

The shifting of the equilibrium direction would result in the association of the ions to form solid calcium sulfate and precipitate out of the solution. The process takes place until the ionic product ${\text{Q}}_{\text{sp}}$ once again becomes equal to the ${\text{K}}_{\text{sp}}.$ 

The new equilibrium would result in the increased concentration of the calcium ion than the sulfate ion concentration.

The above situation describes the common ion effect and results due to the addition of an electrolyte that has an ion common to an already dissolved electrolyte. The addition of a common ion causes an equilibrium imbalance between the electrolyte and its constituting ions. 

The common ion in the above example is $\text{C}{\text{a}}^{\text{2 + }}.$ 

Hence, the common ion effect is an effect that causes a decrease in the solubility of an ionic compound due to the addition of another ionic compound that has an ion common in it. 

In the above example, adding calcium nitrate that consists of calcium ions to the saturated solution of calcium sulfate causes a decrease in the solubility of $\text{CaS}{\text{O}}_4$ compound. Hence, additional $\text{CaS}{\text{O}}_4$ precipitates out of the solution, lowering its solubility. 

Similarly, adding an electrolyte containing sulfate ion, such as potassium sulfate, would result in the same common ion effect.

Common Ion Effect with Weak Acids and Bases

Adding a common ion to a weak ion or base prevents the ionization of the weak acid or weak base as it would occur without the added common ion. The common ion effect suppresses the ionization of a weak acid by shifting the equilibrium towards the reactant side.

1. The common ion effect of ${\text{H}}_3{\text{O}}^{+}$ on the ionization of acetic acid.

2. The common ion effect of $\text{O}{\text{H}}^{–}$ on the ionization of ammonia.

Adding the common ion, which is the hydroxide ion in the above example, shifts the reaction towards the left to decrease the stress (in accordance with Le Chatelier’s Principle). in product formation and causes more undissolved reactants. This decreases the reaction quotient because the reaction is being pushed towards the left to reach equilibrium.

Consequences of Common Ion Effect

1. The common ion effect can be used to obtain drinking water from aquifers contaminated with chalk or limestone. In this water treatment, Sodium carbonate (chemical formula $\text{N}{\text{a}}_2\text{C}{\text{O}}_3$) is added to the water to cause the precipitation of calcium carbonate. The common ion, $\text{C}{{\text{O}}_3}^{2–}$decreases the solubility of calcium carbonate. The $\text{CaC}{\text{O}}_3$ the precipitate is, therefore, a valuable by-product that can be used in the process of manufacturing toothpaste.
2. The common ion effect is used in the salting-out process of soap manufacture. Soaps are the sodium salts of carboxylic acids containing a long aliphatic chain (fatty acids). The soaps are precipitated by adding sodium chloride, which consists of the sodium cation as the common ion. The sodium ion reduces the solubility of the soap solution.

pH and the Common Ion Effect

The common ion effect causes a change in the pH of a buffer solution (solution containing a base and its conjugate acid, or acid and its conjugate base) when there is an addition of the conjugate ion present in the concerned buffer solution.

$\text{C}{\text{H}}_3\text{COOH}\rightleftharpoons {\text{H}}^{+}+\text{C}{\text{H}}_3\text{CO}{\text{O}}^{–}$

$\text{C}{\text{H}}_3\text{COONa}\to \text{N}{\text{a}}^{+}+\text{C}{\text{H}}_3\text{CO}{\text{O}}^{–}$

1. An example of such an effect can be observed when acetic acid and sodium acetate are dissolved in a given solution. Both acetic acid and sodium acetate generate acetate ions. Sodium acetate is a strong electrolyte that undergoes complete ionization to form sodium and acetate ions. The new acetate ions put forth by sodium acetate facilitate the suppression of the ionization of acetic acid. Hence, according to Le Chatelier’s principle, the addition of acetate ions results in shifting the equilibrium to the left. As the dissociation of acetic acid is reduced, the pH of the solution is increased $(>7).$ Therefore, the common ion solution containing acetic acid and sodium acetate will have an increased pH and be less acidic compared to an acetic acid solution.

Solved Example

Q.1. $0.0162\text{M}$ Lead(II) chloride solution is dissolved in some $0.100\text{M}$ sodium chloride solution. What is the equilibrium concentration of the lead(II) ions in the resulting solution? $\left({\text{K}}_{\text{sp}}\text{ = 1}\text{.7 }\times \text{ 1}{\text{0}}^{\text{ – 5}}\right)$
Ans: The complete balanced equation of ionization is-
$\text{PbC}{\text{l}}_{\text{2}}\text{(s)}\rightleftharpoons \text{P}{\text{b}}^{\text{2 + }}\text{(aq) + 2C}{\text{l}}^{\text{ – }}{\text{K}}_{\text{sp}}\text{ = 1}\text{.7 }\times \text{ 1}{\text{0}}^{\text{ – 5}}$
Given-
$\left[\text{C}{\text{l}}^{\text{ – }}\right]\text{ = 0}\text{.1M}$
${\text{K}}_{\text{sp}}\text{ = }\left[\text{P}{\text{b}}^{\text{2 + }}\right]{\left[\text{C}{\text{l}}^{\text{ – }}\right]}^{\text{2}}$
$\text{1}\text{.7 }\times \text{ 1}{\text{0}}^{\text{ – 5}}\text{ = }\left[\text{P}{\text{b}}^{\text{2 + }}\right]\text{[0}{\text{.1]}}^{\text{2}}$
$\left[\text{P}{\text{b}}^{\text{2 + }}\right]\text{ = }\frac{\text{1}\text{.7 }\times \text{ 1}{\text{0}}^{\text{ – 5}}}{\text{0}\text{.01}}\text{ = 1}\text{.7 }\times \text{ 1}{\text{0}}^{\text{ – 3}}\text{M}$
The original concentration of $\left[\text{P}{\text{b}}^{\text{2 + }}\right]$ ion is $0.0162\text{M}$ and after dissolving it in $\text{0}\text{.1}\text{M}\text{NaCl}$ solution, the concentration of $\left[\text{P}{\text{b}}^{\text{2 + }}\right]$ ions become $0.0017\text{M}\text{.}$

Summary

The presence of one or more common ions in a solution determines the precipitation of an ionic compound. Le Chatiliers’ Principle governs the precipitation of the compound. When the concentration of any one of the ionic species increases, the reaction proceeds in a backward direction to relieve the stress on the product side. In this article, we learned the concept, applications of the common ion effect. We also learned a few numericals on the common ion effect.

Frequently Asked Questions

Q.1. What is meant by the common ion effect?
Ans: The common ion effect is an effect that causes a decrease in the solubility of an ionic compound as a result of the addition of another ionic compound that has an ion common in it.

Q.2. Why is the common ion effect important?
Ans: The common ion effect plays an important role in reducing the solubility of solids. The solubility of a compound generally decreases due to a shift in equilibrium to the reactant side. The common ion effect also plays a role in the regulation of buffers.

Q.3. How common ion effect affects solubility?
Ans: Adding a solute that has a common ion with the dissolving solid decreases the solubility of the solute. This takes place as per Le Chatelier’s principle which states the reaction will shift toward the left (toward the reactants) to relieve the stress on the excess product.

Q.4. What is the net effect of common ion?
Ans: The net effect of the common ion is that it reduces the solubility of the solute in the solution. The common ion effect is observed when the addition of an ion common to two solutes causes precipitation or reduces ionization.

Q.5. What is Le Chatelier’s principle?
Ans: Le Chatelier’s principle, also called Chatelier’s principle, is a principle used to predict the effect of a change in conditions on chemical equilibria. If the concentration of a reaction species is increased (at constant T and V), the equilibrium system will shift in the direction that reduces the concentration of that species.

Practice question on Common Ion Effect here

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