A sparingly soluble salt having general formula Aₓp+ Byq- and molar solubility S is in equilibrium with its sa

NCERT Exemplar · Q53

Q.A sparingly soluble salt having general formula A_x^{p+} B_y^{q-} and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.

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Concept understanding — Solubility Product Constant

Solubility Product Constant: From Intuition to Precision

Imagine you drop a pinch of salt into a glass of water. The salt crystals disappear — they dissolve. But what if you keep adding salt, spoonful after spoonful? At some point, the water can't hold any more; the extra salt just sits at the bottom, undissolved. That's a saturated solution — the maximum amount of solute has dissolved at that temperature.

Now, here's the key question: even in that saturated solution, is everything static? Not at all. At the microscopic level, salt ions are constantly leaving the solid crystal and entering the solution (dissolving), while other ions in solution bump into the crystal and stick back (precipitating). At saturation, these two processes happen at exactly the same rate. The system is in dynamic equilibrium.

Note

A saturated solution is not "full" in a static sense — it's a busy, balanced dance between dissolving and precipitating.

The Intuition: A Crowded Dance Floor

Think of a dance hall with a capacity limit. The "dancers" are ions (like Na⁺ and Cl⁻ from table salt). The dance floor is the solution. When the floor is empty, dancers easily find space — dissolution is fast. As more dancers arrive, they start bumping into each other and some leave the floor (precipitate). At the maximum capacity, the number of dancers entering equals the number leaving. That equilibrium number of dancers is what we call solubility.

But here's the twist: for many salts, the "dancers" come in different types — say, positive ions and negative ions. The equilibrium isn't just about the total number; it's about the product of their concentrations. Why product? Because the chance of a positive and a negative ion meeting to form a solid depends on both their concentrations. If you double the concentration of positive ions, the chance of a collision doubles. If you double both, it quadruples.

That product — at equilibrium, for a saturated solution — is a constant. That's the solubility product constant, $K_{sp}$ .

The Precise Statement

For a sparingly soluble salt that dissociates in water as:

A_mB_n(s) \rightleftharpoons m A^{n+}(aq) + n B^{m-}(aq)

the solubility product constant is defined as:

K_{sp} = [A^{n+}]^m \cdot [B^{m-}]^n

where the square brackets denote molar concentrations (mol/L) at saturation (equilibrium with the solid).

Watch out

The solid $A_mB_n$ does not appear in the expression. Its concentration is constant (pure solid) and is absorbed into $K_{sp}$ . Never write $[A_mB_n]$ in the $K_{sp}$ expression.

What $K_{sp}$ Tells You

Small $K_{sp}$ (e.g., $10^{-30}$ ): The salt is very insoluble. Only a tiny amount dissolves.
Large $K_{sp}$ (e.g., $10^{-2}$ ): The salt is relatively soluble.
$K_{sp}$ is temperature-dependent — always quote the temperature (usually 25°C).

Important

$K_{sp}$ is an equilibrium constant. It only applies to saturated solutions in contact with undissolved solid. If no solid is present, the solution may be unsaturated ( $Q < K_{sp}$ ) or supersaturated ( $Q > K_{sp}$ ), but $K_{sp}$ itself doesn't change.

A Concrete Example: Silver Chloride

Silver chloride, AgCl, is a classic sparingly soluble salt. Its dissolution:

AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)

The $K_{sp}$ expression:

K_{sp} = [Ag^+][Cl^-]

At 25°C, $K_{sp} = 1.8 \times 10^{-10}$ . This tiny number means that in a saturated solution, the product of the two ion concentrations is only $1.8 \times 10^{-10}$ .

If you know the solubility of AgCl is $s$ mol/L, then $[Ag^+] = s$ and $[Cl^-] = s$ , so: …

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For a sparingly soluble salt $A_x^{p+} B_y^{q-}$ with molar solubility $S$ , the solubility product constant $K_{sp}$ is given by $K_{sp} = x^x y^y S^{x+y}$ .

When a salt is described as "sparingly soluble," it means that only a very small amount of it dissolves in a solvent, typically water, to form a solution. Despite this small amount, an equilibrium is established between the undissolved solid salt and its constituent ions in the saturated solution. This equilibrium is crucial for understanding the behavior of such salts.

The solubility product constant ( $K_{sp}$ ) is an equilibrium constant that quantifies this equilibrium. It represents the product of the concentrations of the ions in a saturated solution, each raised to the power of its stoichiometric coefficient in the balanced dissolution equation. Because it's an equilibrium constant, $K_{sp}$ has a constant value at a given temperature for a specific salt.

The molar solubility ( $S$ ) of a salt, on the other hand, is a measure of how much of the salt dissolves. Specifically, it is defined as the number of moles of the salt that dissolve per liter of solution to form a saturated solution. The relationship between $K_{sp}$ and $S$ allows us to calculate one if the other is known, providing a quantitative way to describe the solubility of these salts.

Let's derive this relationship step-by-step for a general salt $A_x^{p+} B_y^{q-}$ .

Write the Dissociation Equilibrium: The salt $A_x^{p+} B_y^{q-}$ dissociates in water to form its respective ions. The balanced chemical equation for this dissolution equilibrium is:

A_x B_y (s) \rightleftharpoons x A^{p+} (aq) + y B^{q-} (aq)

Here, $A_x B_y (s)$ represents the solid salt, and $A^{p+} (aq)$ and $B^{q-} (aq)$ represent the aqueous ions. The coefficients $x$ and $y$ are the stoichiometric coefficients for the cation and anion, respectively.

2. Define Molar Solubility ( $S$ ) and Relate it to Ion Concentrations:

Let $S$ be the molar solubility of the salt $A_x B_y$ . This means that in a saturated solution, $S$ moles of $A_x B_y$ dissolve per liter of solution.

From the balanced dissociation equation, for every 1 mole of $A_x B_y$ that dissolves, $x$ moles of $A^{p+}$ ions and $y$ moles of $B^{q-}$ ions are produced.

Therefore, if $S$ moles of $A_x B_y$ dissolve:

* The concentration of the cation $A^{p+}$ will be $[A^{p+}] = xS$

* The concentration of the anion $B^{q-}$ will be $[B^{q-}] = yS$

> [!WARNING]
> A common mistake is to forget the stoichiometric coefficients when determining the ion concentrations from the molar solubility $S$. For example, if the salt were $CaF_2$, then $S$ moles of $CaF_2$ would yield $S$ moles of $Ca^{2+}$ and $2S$ moles of $F^-$. …

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Q.A sparingly soluble salt having general formula A_x^{p+} B_y^{q-} and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.

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