Q.The molar solubility of AgCl in pure water at 298 K is 1.3×10−5 mol L−1. Calculate its solubility product, Ksp.
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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.
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, Ksp.
The Precise Statement
For a sparingly soluble salt that dissociates in water as:
AmBn(s)⇌mAn+(aq)+nBm−(aq)
the solubility product constant is defined as:
Ksp=[An+]m⋅[Bm−]n
where the square brackets denote molar concentrations (mol/L) at saturation (equilibrium with the solid).
The solid AmBn does not appear in the expression. Its concentration is constant (pure solid) and is absorbed into Ksp. Never write [AmBn] in the Ksp expression.
What Ksp Tells You
- Small Ksp (e.g., 10−30): The salt is very insoluble. Only a tiny amount dissolves.
- Large Ksp (e.g., 10−2): The salt is relatively soluble.
- Ksp is temperature-dependent — always quote the temperature (usually 25°C).
Ksp 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<Ksp) or supersaturated (Q>Ksp), but Ksp itself doesn't change.
A Concrete Example: Silver Chloride
Silver chloride, AgCl, is a classic sparingly soluble salt. Its dissolution:
AgCl(s)⇌Ag+(aq)+Cl−(aq)
The Ksp expression:
Ksp=[Ag+][Cl−]
At 25°C, Ksp=1.8×10−10. This tiny number means that in a saturated solution, the product of the two ion concentrations is only 1.8×10−10.
If you know the solubility of AgCl is s mol/L, then [Ag+]=s and [Cl−]=s, so: …
[!TLDR] For a 1:1 salt, Ksp=s2 where s is the molar solubility. [!ANSWER] $ …
AgCl(s)⇌Ag+(aq)+Cl−(aq); dissolving s mol/L of AgCl produces s mol/L each of Ag+ and Cl−, so Ksp=[Ag+][Cl−]=s×s=s2. With $s = 1.3\times10^{-5}\ \text{mol L}^{- …
For a salt with a different stoichiometry (e.g. Ag2CrO4), the exponents on s must match each ion's own coefficient — sq …
- CBSE 2025Set ANNUAL1 markMCQQ.If the solubility of A2X3 is γ mol dm-3 then its solubility product is(a) 6γ^4(b) 64γ^4(c) 36γ^5(d) 108γ^5
›Reveal solutionSolution
For A2X3 with solubility γ mol/dm3, Ksp = 108γ^5.
A2X3 dissociates as: A2X3 ⇌ 2A^3+ + 3X^2-.
If solubility is γ mol/dm3, then [A3+] = 2γ and [X2-] = 3γ.
…
- CBSE 2018Set ANNUAL1 markMCQQ.Match the columns — Column A:(i) Viscosity,(ii) Sorensen,(iii) Fire works,(iv) Deficient compound,(v) Black precipitate. Column B:(a) B,(b) Ba,(c) Na,(d) Nsm^2,(e) pH,(f) S. Which Column-B entry matches(v) Black precipitate?(a) B(b) Ba(c) Na(d) Nsm^2(e) pH(f) S
›Reveal solutionSolution
'Black precipitate' is matched to Sulfur (S), because passing H2S gas into solutions of many heavy-metal ions precipitates their black metal sulfides.
In classical qualitative inorganic analysis (linked to solubility-product/common-ion-effect equilibria), H2S gas is passed through metal-ion solutions to separate cations into groups. Several common metal sulfides formed this way are black: PbS (lead sulfide), CuS (copper sulfide), HgS (mercuric sulfide), and Ag2S (silver sulf …
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