AP Chemistry Ksp and Molar Solubility

Calculating molar solubility from Ksp, and the common ion effect, three full worked examples.

Ksp problems ask you to find how much of a sparingly soluble salt actually dissolves, either in pure water or in a solution that already contains one of the salt's own ions. The setup is always an ICE table, the only thing that changes is how many "s" terms a salt's stoichiometry contributes.

The Ksp Equilibrium Setup

For a salt dissolving as A_xB_y(s) ⇌ xA(aq) + yB(aq), let s equal the molar solubility. At equilibrium, [A] = xs and [B] = ys, so Ksp = (xs)^x(ys)^y. Solving that expression for s gives the molar solubility.

Worked Example 1: A 1:1 Salt (AgCl)

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), K_sp = 1.8 × 10⁻¹⁰. Find the molar solubility in pure water.

Let s = molar solubility. [Ag⁺] = s, [Cl⁻] = s.
K_sp = s × s = s² = 1.8 × 10⁻¹⁰
s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M

Worked Example 2: A 1:2 Salt (CaF₂)

CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), K_sp = 3.9 × 10⁻¹¹. Find the molar solubility in pure water.

Let s = molar solubility. [Ca²⁺] = s, [F⁻] = 2s (twice the moles of fluoride per formula unit).
K_sp = s(2s)² = 4s³ = 3.9 × 10⁻¹¹
s³ = 9.75 × 10⁻¹²
s = 2.14 × 10⁻⁴ M

Notice CaF₂ has a smaller K_sp than AgCl but a larger molar solubility. K_sp values only compare directly for salts with the same ion ratio, comparing across different stoichiometries can be misleading.

Worked Example 3: The Common Ion Effect

What is the molar solubility of AgCl (K_sp = 1.8 × 10⁻¹⁰) in a 0.10 M NaCl solution, instead of pure water?

NaCl is a strong electrolyte, so the solution already has [Cl⁻] = 0.10 M before any AgCl dissolves. Let s = additional molar solubility of AgCl.
[Ag⁺] = s, [Cl⁻] = 0.10 + s ≈ 0.10 M (s is negligible next to 0.10)
K_sp = (s)(0.10) = 1.8 × 10⁻¹⁰
s = 1.8 × 10⁻⁹ M

That's about 7,450 times smaller than AgCl's solubility in pure water (Example 1). The common Cl⁻ ion shifts the dissolution equilibrium back toward the solid, exactly what Le Chatelier's principle predicts.

Common Ksp Mistakes

Forgetting the coefficient on an ion's ICE-table entry. A 1:2 salt has [anion] = 2s, not s, that factor of 2 also gets raised to the power of 2 in the K_sp expression.
Comparing K_sp values directly across different salt stoichiometries. A smaller K_sp doesn't always mean a smaller molar solubility unless both salts dissociate into the same number of ions.
Ignoring a common ion already in solution. If the solvent isn't pure water, check for an ion the dissolving salt shares with something already dissolved, and include its starting concentration in the ICE table.
Not approximating (0.10 + s) ≈ 0.10 when s is negligibly small. This approximation is what keeps the common-ion-effect algebra linear instead of needing the full cubic/quadratic solution.

Frequently Asked Questions

What is Ksp?

The solubility product constant, an equilibrium constant for a sparingly soluble ionic compound dissolving in water. For a salt AₓBₙ that dissociates into m mol of cation and n mol of anion per formula unit, Ksp = [cation]^m[anion]^n at equilibrium (saturation).

What is molar solubility?

The number of moles of a salt that dissolve per liter of solution to form a saturated solution. It's a single number (mol/L) you calculate from Ksp using an ICE table, not something you look up directly.

Do I need to memorize Ksp values?

No. Any Ksp value needed for a calculation is given in the question. The skill being tested is setting up and solving the ICE table correctly, not recalling a specific compound's constant.

What is the common ion effect?

When a solution already contains one of a salt's own ions before the salt dissolves, that existing concentration shifts the dissolution equilibrium back toward the solid (Le Chatelier's principle), so the salt's molar solubility in that solution is lower than its molar solubility in pure water.

Why does a 1:2 salt like CaF2 need a cubed term instead of squared?

Because Ksp = [cation][anion]² for a salt releasing one cation and two anions per formula unit, and [anion] = 2s (twice the molar solubility) rather than just s. Substituting gives Ksp = s(2s)² = 4s³, so solving for s requires a cube root, not a square root.

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