Solubility Product (Ksp)

9 min read Updated Mar 26, 2026

The solubility product (Ksp) is just the equilibrium constant for an ionic solid dissolving in water. Everything you learned about K applies here - Ksp is simply K applied to the specific case of dissolution.

Two beakers showing the dissolution equilibrium of an ionic solid: the left beaker shows the solid lattice with some ions dissolving into the water, and the right beaker shows more ions in solution with the solid partially dissolved, with a double arrow between them indicating the reversible equilibrium between dissolution and precipitation — Dissolution equilibrium: an ionic solid dissolves (ions leave the lattice) and precipitates (ions rejoin the lattice) simultaneously. At equilibrium, the rate of dissolution equals the rate of precipitation, and the ion concentrations remain constant. *Credit: OpenStax Chemistry 2e, CC BY 4.0*

Writing the Ksp Expression

For the dissolution of a generic ionic solid:

$A_{x}$ $B_{y}$ (s) ⇌ xA^(y+)(aq) + yB^(x-)(aq)

Examples

Compound	Dissolution reaction	Ksp expression
AgCl	AgCl(s) ⇌ Ag+(aq) + Cl-(aq)	Ksp = [Ag+][Cl-]
CaF₂	CaF₂(s) ⇌ Ca²+(aq) + 2F-(aq)	Ksp = [Ca²+][F-]²
Fe(OH)₃	Fe(OH)₃(s) ⇌ Fe³+(aq) + 3OH-(aq)	Ksp = [Fe³+][OH-]³

Notice how the coefficient (2 for F-, 3 for OH-) becomes the exponent.

Ksp and Molar Solubility

Molar solubility (s) is the number of moles of solid that dissolve per liter of solution to form a saturated solution. You can calculate s from Ksp:

Example: Find the molar solubility of AgCl ( $K_{sp} = 1.8 \times 10^{-10}$ ).

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

If s moles dissolve, then [Ag+] = s and [Cl-] = s.

$K_{sp} = (s)(s) = s^2$

$s = \sqrt{1.8 \times 10^{-10}} \approx 1.3 \times 10^{-5}$ M

Example: Find the molar solubility of CaF₂ ( $K_{sp} = 3.9 \times 10^{-11}$ ).

CaF₂(s) ⇌ Ca²+(aq) + 2F-(aq)

If s moles dissolve, then [Ca²+] = s and [F-] = 2s.

$K_{sp} = (s)(2s)^2 = 4s^3$

$s = (K_{sp}/4)^{1/3} = (3.9 \times 10^{-11} / 4)^{1/3} \approx (9.75 \times 10^{-12})^{1/3} \approx 2.1 \times 10^{-4}$ M

Predicting Precipitation: Qsp vs. Ksp

Just as Q vs. K predicts reaction direction, Qsp vs. Ksp predicts whether a precipitate will form:

Comparison	Solution is…	What happens
Qsp < Ksp	Unsaturated	More solid can dissolve; no precipitate
Qsp = Ksp	Saturated	At equilibrium; no net change
Qsp > Ksp	Supersaturated	Precipitation occurs until Qsp = Ksp

A solution contains

[\text{Ag}^+] = 1.0 \times 10^{-3}

M and

[\text{Cl}^-] = 1.0 \times 10^{-5}

M. The

K_{sp}

of AgCl is

1.8 \times 10^{-10}

. Will a precipitate form?

Click to reveal answer

Yes, a precipitate will form.

Q_{sp} = [\text{Ag}^+][\text{Cl}^-] = (1.0 \times 10^{-3})(1.0 \times 10^{-5}) = 1.0 \times 10^{-8}

. Since

Q_{sp}

(

10^{-8}

) >

K_{sp}

(

1.8 \times 10^{-10}

), the solution is supersaturated and AgCl will precipitate.

Can you compare the molar solubility of two salts just by looking at their Ksp values?

Click to reveal answer

Only if they produce the same number and ratio of ions. You can directly compare Ksp for AgCl vs. AgBr (both 1:1 salts). But you cannot directly compare Ksp for AgCl (1:1) vs. CaF₂ (1:2) without calculating molar solubility, because the Ksp-to-solubility relationship depends on stoichiometry.