Solubility and the Solubility Product Constant (Ksp)

Table salt dissolves readily in water, but silver chloride barely dissolves at all, even though both are simple ionic compounds made of a metal and a nonmetal. The difference isn't random; it's governed by a precise equilibrium between a solid and its dissolved ions, and chemists quantify exactly how far that equilibrium favors dissolving with a single number: the solubility product constant, Ksp.

Saturated, Unsaturated, and Supersaturated Solutions

Before discussing Ksp specifically, it helps to be clear on three related terms describing how much solute a solution currently holds:

• Unsaturated: the solution contains less dissolved solute than it's capable of holding at that temperature; more solute could still dissolve.
• Saturated: the solution holds exactly the maximum amount of solute it can dissolve at that temperature; it's in dynamic equilibrium with any undissolved excess solute present.
• Supersaturated: the solution somehow holds more dissolved solute than its normal maximum, an unstable state usually created by carefully cooling a hot saturated solution; the excess solute will crystallize out rapidly if disturbed (for example, by a seed crystal or a scratch on the container).

Ksp specifically describes the equilibrium condition of a saturated solution of a sparingly soluble ionic compound.

What the Solubility Product Constant Actually Represents

For a sparingly soluble ionic compound in equilibrium with its dissolved ions in a saturated solution:

AxBy(s) ⇌ xA(aq) + yB(aq)

The solubility product constant is defined as:

Ksp = [A]^x [B]^y

Notice that, unlike a typical equilibrium constant expression, the solid itself doesn't appear in the expression, since the concentration of a pure solid doesn't meaningfully change and is treated as a constant folded into Ksp itself.

A small Ksp value means very little of the compound dissolves before reaching equilibrium (low solubility), while a larger Ksp value means more of the compound can dissolve before the solution becomes saturated.

A Worked Example: Calculating Molar Solubility From Ksp

Question: Silver chloride, AgCl, has a Ksp of 1.8 × 10⁻¹⁰ at 25°C. What is its molar solubility (the maximum moles of AgCl that can dissolve per liter of water)?

Step 1: Write the dissolution equation and Ksp expression.

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Ksp = [Ag⁺][Cl⁻]

Step 2: Define the unknown.

Let "s" represent the molar solubility, the moles of AgCl that dissolve per liter. Since each formula unit of AgCl produces exactly one Ag⁺ and one Cl⁻ ion:

[Ag⁺] = s     [Cl⁻] = s

Step 3: Substitute into the Ksp expression and solve.

Ksp = [Ag⁺][Cl⁻] = (s)(s) = s²
1.8 × 10⁻¹⁰ = s²
s = √(1.8 × 10⁻¹⁰)
s = 1.34 × 10⁻⁵ mol/L

Answer: At most, about 1.34 × 10⁻⁵ moles of AgCl can dissolve per liter of water, an extremely small amount, which is exactly why silver chloride is classified as "insoluble" in practical terms, even though a tiny amount genuinely does dissolve.

Comparing Ksp Values Across Compounds

Compound	Ksp (approx., 25°C)	Relative Solubility
NaCl	very large (not typically tabulated as Ksp)	Highly soluble
CaSO₄	4.9 × 10⁻⁵	Slightly soluble
AgCl	1.8 × 10⁻¹⁰	Very slightly soluble
CaCO₃	3.3 × 10⁻⁹	Very slightly soluble
AgI	8.5 × 10⁻¹⁷	Extremely slightly soluble

A note of caution when comparing Ksp values directly: this only works cleanly when comparing compounds with the same ion ratio (like two 1:1 compounds). Comparing a 1:1 compound's Ksp directly against a 1:2 compound's Ksp doesn't reliably predict which has higher molar solubility, since the exponents in the Ksp expression differ.

The Common Ion Effect

An important, practical application of Ksp is the common ion effect: adding a soluble compound that shares an ion with a sparingly soluble compound decreases the sparingly soluble compound's solubility, a direct consequence of Le Chatelier's Principle.

Example: Adding sodium chloride (NaCl, highly soluble) to a saturated silver chloride solution increases the concentration of Cl⁻ ions already present. Since Ksp for AgCl must remain constant, the system responds by shifting the AgCl dissolution equilibrium backward, causing some dissolved AgCl to precipitate back out as a solid, reducing the overall amount of AgCl that stays dissolved.

This effect has real practical uses, including selectively precipitating one dissolved ion out of a mixture by deliberately adding a common ion, a technique used in both industrial purification and analytical chemistry.

Predicting Whether a Precipitate Will Form

Ksp also lets you predict whether mixing two solutions will produce a solid precipitate, by comparing the reaction quotient (Q), calculated the same way as Ksp but using the actual, current ion concentrations, against the known Ksp value:

• Q less than Ksp: the solution is unsaturated; no precipitate forms (or existing solid will continue dissolving).
• Q equals Ksp: the solution is exactly saturated; the system is at equilibrium.
• Q greater than Ksp: the solution is supersaturated relative to that equilibrium; a precipitate will form until Q decreases back down to Ksp.

FAQ

Conclusion

Ksp turns the vague idea of "solubility" into a precise, calculable equilibrium constant, letting you determine exactly how much of a sparingly soluble compound will dissolve, predict whether mixing two solutions will produce a precipitate, and understand why adding a common ion can push a dissolved compound back out of solution. Like any equilibrium constant, it's most useful once you can set up the expression correctly from the balanced dissolution equation and remember that pure solids never appear in it.

Here are some useful references if you want to go deeper:

• Khan Academy – Solubility Equilibria — free lessons covering Ksp calculations and the common ion effect.
• Chemguide – Solubility Product Calculations — worked examples of Ksp and molar solubility problems.
• LibreTexts Chemistry – Solubility Equilibria — an open textbook resource covering precipitation and the common ion effect.