Q.Explain common ion effect with an example.

The Common Ion Effect: An Intuitive First Look

Imagine you have a glass of water with some salt dissolved in it — say, sodium chloride (NaCl). The salt has dissociated into Na⁺ and Cl⁻ ions floating around. Now, if you add more salt, some of it will dissolve, but eventually the water becomes saturated and no more salt dissolves.

Now imagine a different scenario. You have a solution of silver chloride (AgCl) — a sparingly soluble salt. Very little of it dissolves, giving you a tiny concentration of Ag⁺ and Cl⁻ ions. What happens if you now add some sodium chloride (NaCl) to this solution? The NaCl will dissociate completely, flooding the solution with extra Cl⁻ ions.

Here's the key: the system tries to maintain its equilibrium. The solubility equilibrium for AgCl is:

When you add extra Cl⁻ from NaCl, Le Chatelier's principle kicks in. The equilibrium shifts to the left — more AgCl precipitates out of solution. The presence of a common ion (Cl⁻) suppresses the solubility of AgCl.

That's the common ion effect in a nutshell: the solubility of a salt decreases when you add another salt that shares a common ion with it.

The Precise Statement

In other words: when you have an equilibrium involving ions, adding more of one of those ions (from a different source) shifts the equilibrium away from the dissociated form.

Why It Matters (and Where You'll See It)

The common ion effect isn't just a textbook curiosity — it's used everywhere in chemistry:

• Controlling pH of buffer solutions: Adding a common ion (like acetate ion to acetic acid) suppresses the dissociation of the weak acid, keeping the pH stable.
• Qualitative analysis: In salt analysis, you selectively precipitate certain ions by adding a common ion. For example, to test for chloride, you add AgNO₃ — the Ag⁺ is common to AgCl, so even tiny amounts of Cl⁻ will precipitate.
• Industrial processes: In the Solvay process for making sodium carbonate, the common ion effect is used to precipitate sodium bicarbonate.

A Concrete Example with Numbers

Consider the solubility of silver chloride in pure water. The $K_{sp}$ of AgCl is $1.8 \times 10^{-10}$.

In pure water:

Now, what if the solution already contains 0.10 M NaCl (from a separate source)? The Cl⁻ concentration is now 0.10 M (plus a tiny bit from AgCl). Let $s$ be the new solubility of AgCl: …