Molar Solubility from Ksp

Introduction

A Ksp value is compact, but on its own it does not tell you how many grams of a salt dissolve in a litre of water. To answer that, we translate the equilibrium constant into a molar solubility, s, using the dissolution stoichiometry. The translation depends on the formula type, and it explains a surprising fact: a salt with a smaller Ksp can be more soluble than one with a larger Ksp.

Core explanation

Defining s. If s mol of solid dissolves per dm³ of solution, the ion concentrations follow from the formula. For MₓAᵧ in pure water, [Mᵐ⁺] = xs and [Aⁿ⁻] = ys. Substituting into Ksp:

Ksp = (xs)ˣ (ys)ʸ = xˣ yʸ s⁽ˣ⁺ʸ⁾

AB salts (1:1). AgCl ⇌ Ag⁺ + Cl⁻. [Ag⁺] = s and [Cl⁻] = s, so Ksp = s² and s = √Ksp. For AgCl, s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol dm⁻³.

AB₂ and A₂B salts (1:2 or 2:1). CaF₂ ⇌ Ca²⁺ + 2F⁻. [Ca²⁺] = s and [F⁻] = 2s, so Ksp = s(2s)² = 4s³ and s = ∛(Ksp/4). Silver chromate, Ag₂CrO₄, gives [Ag⁺] = 2s and [CrO₄²⁻] = s, so again Ksp = (2s)²s = 4s³.

Other types. For AB₃ (such as Al(OH)₃) Ksp = 27s⁴; for A₃B₂ (such as Ca₃(PO₄)₂) Ksp = 108s⁵. The pattern always comes from the general formula above.

Comparing across formula types.

Salt Type Ksp s / mol dm⁻³ --- --- --- --- AgCl AB 1.8 × 10⁻¹⁰ 1.3 × 10⁻⁵ Ag₂CrO₄ A₂B 1.1 × 10⁻¹² 6.5 × 10⁻⁵ CaF₂ AB₂ 3.9 × 10⁻¹¹ 2.1 × 10⁻⁴ PbI₂ AB₂ 9.8 × 10⁻⁹ 1.3 × 10⁻³

Silver chromate has a Ksp about 160 times smaller than silver chloride, yet its molar solubility is about five times larger. The cube root in the AB₂ formula shrinks the small number far less than a square root does.

Mass solubility. Multiply s by the molar mass. For AgCl (M = 143.3 g mol⁻¹): 1.34 × 10⁻⁵ × 143.3 = 1.9 × 10⁻³ g dm⁻³, just under 2 mg per litre.

Assumptions. These formulas assume the dissolved ions do not react further, such as by hydrolysis, ion pairing or complex formation, and that activities equal concentrations. For salts with basic anions like carbonate or sulfide, or strongly hydrolysing cations, the real solubility is larger than these simple estimates.

Formulae

AB: Ksp = s², s = √Ksp. AB₂ or A₂B: Ksp = 4s³, s = ∛(Ksp/4). AB₃ or A₃B: Ksp = 27s⁴. A₃B₂ or A₂B₃: Ksp = 108s⁵. Mass solubility = s × M.

Step-by-step reasoning

1. Write the dissolution equation. 2. Express each ion concentration as a multiple of s. 3. Substitute into the Ksp expression, including powers. 4. Collect the numerical factor and the power of s. 5. Solve for s, then convert to g dm⁻³ if required. 6. For the reverse problem, measure s, build the ion concentrations and calculate Ksp.

Visual explanation

Draw one formula unit of CaF₂ splitting into one Ca²⁺ circle and two F⁻ circles. Label the Ca²⁺ "s" and each F⁻ "s", giving 2s in total. Squaring the fluoride term and multiplying shows where the 4 in 4s³ comes from.

Real-world analogy

Ksp is like a recipe total, while s is the number of whole batches made. If each batch uses one egg and two cups of flour, the product "eggs × cups²" grows much faster than the number of batches, so comparing products between different recipes tells you little about how many batches each allows.

Real-world example

Tooth enamel and fluoride treatments involve calcium fluoride and fluorapatite. The low but finite solubility of CaF₂, around 2 × 10⁻⁴ mol dm⁻³, allows it to act as a slow-release reservoir of fluoride ions in the mouth after a dental treatment.

Why?

Why does [F⁻] equal 2s rather than s? Every formula unit of CaF₂ that dissolves releases two fluoride ions. If s mol of CaF₂ dissolves, 2s mol of F⁻ enters solution.

Common misconception

"The smaller the Ksp, the less soluble the salt." This holds only among salts of the same formula type. Silver chromate has a smaller Ksp than silver chloride but a greater molar solubility.

Worked example

Question: Calculate the molar solubility of PbI₂ given Ksp = 9.8 × 10⁻⁹, and hence the iodide concentration in a saturated solution.

Reasoning: Ksp = s(2s)² = 4s³. s³ = 9.8 × 10⁻⁹ ÷ 4 = 2.45 × 10⁻⁹. s = ∛(2.45 × 10⁻⁹) = 1.35 × 10⁻³ mol dm⁻³. [I⁻] = 2s = 2.7 × 10⁻³ mol dm⁻³.

Answer: s = 1.35 × 10⁻³ mol dm⁻³ and [I⁻] = 2.7 × 10⁻³ mol dm⁻³.

Quick check

1. The molar solubility of BaSO₄ is 1.05 × 10⁻⁵ mol dm⁻³. What is its Ksp? Answer: Ksp = s² = (1.05 × 10⁻⁵)² = 1.1 × 10⁻¹⁰.

Exam focus

Show the s and 2s substitution clearly; most lost marks come from writing Ksp = s × 2s instead of s × (2s)². Keep extra significant figures until the final cube root, and state units of s.

Advanced insight

Measured solubilities are often higher than those predicted from Ksp. Ion pairs such as CaF⁺ and neutral dissolved species like AgCl(aq) add to the total dissolved amount without appearing in the Ksp expression. For silver chloride, the neutral dissolved species makes a small but measurable contribution, which careful analytical work must include.

Summary

Molar solubility s is found by expressing ion concentrations as multiples of s and substituting into Ksp. For AB salts Ksp = s²; for AB₂ or A₂B salts Ksp = 4s³. Mass solubility is s multiplied by molar mass. Because the powers differ, Ksp values rank solubility only within one formula type.

Practice questions

1. Calculate the molar solubility of AgBr, Ksp = 5.0 × 10⁻¹³. Answer: s = √(5.0 × 10⁻¹³) = 7.1 × 10⁻⁷ mol dm⁻³. 2. Calculate the molar solubility of CaF₂, Ksp = 3.9 × 10⁻¹¹. Answer: s = ∛(9.75 × 10⁻¹²) = 2.1 × 10⁻⁴ mol dm⁻³. 3. A saturated solution of Mg(OH)₂ has [Mg²⁺] = 1.1 × 10⁻⁴ mol dm⁻³. Calculate Ksp. Answer: [OH⁻] = 2.2 × 10⁻⁴ mol dm⁻³; Ksp = 1.1 × 10⁻⁴ × (2.2 × 10⁻⁴)² = 5.3 × 10⁻¹². 4. Convert the molar solubility of AgCl, 1.34 × 10⁻⁵ mol dm⁻³, to mg dm⁻³ (M = 143.3 g mol⁻¹). Answer: 1.34 × 10⁻⁵ × 143.3 = 1.92 × 10⁻³ g dm⁻³ = 1.92 mg dm⁻³.