Ksp is simply Keq applied to the specific equilibrium of a saturated solution — the solid dissolving into its aqueous ions. The rules are identical to any other Keq expression.

For any sparingly soluble ionic compound M_aX_b:

$$\text{M}_a\text{X}_b(s) \rightleftharpoons a\text{M}^{b+}(aq) + b\text{X}^{a-}(aq)$$

$$K_{sp} = [\text{M}^{b+}]^a \times [\text{X}^{a-}]^b$$

The solid is excluded (pure solid, activity = 1). There is no denominator because the only reactant (the solid) is excluded.

Ksp is Keq for MX(s) ⇌ ions(aq) — solid excluded, no denominator. Formula-type s-formulae: AB (1:1) → Ksp = s²; AB₂ or A₂B (1:2 or 2:1) → Ksp = 4s³; AB₃ or A₃B → Ksp = 27s⁴. Stoichiometric coefficients become powers in Ksp AND appear inside brackets when expressing [ion] in terms of s (e.g. [F⁻] = 2s for CaF₂).

Copy the four Ksp formula types with their s-formulae into your notes before the check below.

We just saw the Ksp expressions for each formula type and the general s-formula for each. That raises a question: how do you actually work from a given Ksp value to a numerical molar solubility — what are the steps and where do students lose marks? This card answers it → by working through the AgCl (1:1) and CaF₂ (1:2) ICE-table calculations with explicit attention to the stoichiometric coefficient inside the bracket.

Molar solubility (s) is the number of moles of the compound that dissolves per litre to reach saturation — calculated from Ksp using an ICE table where initial ion concentrations are zero.

For AgCl (1:1 type): $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$

$K_{sp} = s^2 = 1.8 \times 10^{-10}$  →  $s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}$ mol/L

For CaF₂ (1:2 type): $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$

[Ca²⁺] = s;  [F⁻] = 2s (stoichiometric coefficient is 2)

$K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s(2s)^2 = s \cdot 4s^2 = 4s^3 = 3.9 \times 10^{-11}$

$s^3 = 9.75 \times 10^{-12}$  →  $s = \sqrt[3]{9.75 \times 10^{-12}} = 2.14 \times 10^{-4}$ mol/L

Molar solubility from Ksp: write ICE table with all ions initially zero; set [cation] = s and [anion] = (stoichiometric coefficient × s); substitute into Ksp expression. AgCl: Ksp = s² → s = √Ksp. CaF₂: [F⁻] = 2s → Ksp = s × (2s)² = 4s³ → s = ∛(Ksp/4). Critical: the coefficient must be INSIDE the bracket before squaring.

Write the AgCl and CaF₂ worked calculations into your notes before the check below.

We just saw how to go from a given Ksp to a molar solubility — using an ICE table and the stoichiometric coefficient inside the bracket. That raises a question: what if you measure solubility experimentally — how do you work in the reverse direction to calculate Ksp from that measured s? This card answers it → by reversing the ICE procedure for Mg(OH)₂ and PbCl₂, emphasising the same coefficient-inside-bracket rule.

If you can measure molar solubility experimentally, you can calculate Ksp — the reverse of the Ksp-to-s calculation.

Procedure: write dissolution equation → express all ion concentrations in terms of s using stoichiometric ratios → substitute and calculate.

Example — Mg(OH)₂: s = 1.54 × 10⁻⁴ mol/L at 25°C

$\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq)$

[Mg²⁺] = s = 1.54 × 10⁻⁴;  [OH⁻] = 2s = 3.08 × 10⁻⁴

$K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 = (1.54 \times 10^{-4})(3.08 \times 10^{-4})^2$

$= (1.54 \times 10^{-4})(9.4864 \times 10^{-8}) = 1.46 \times 10^{-11}$

Example — PbCl₂: s = 3.9 × 10⁻² mol/L

[Pb²⁺] = s = 3.9 × 10⁻²;  [Cl⁻] = 2s = 7.8 × 10⁻²

$K_{sp} = (3.9 \times 10^{-2})(7.8 \times 10^{-2})^2 = (3.9 \times 10^{-2})(6.084 \times 10^{-3}) = 2.37 \times 10^{-4}$

Ksp from molar solubility: (1) write dissolution equation; (2) express all ion concentrations using stoichiometric ratios — [cation] = s, [anion] = ns; (3) substitute into Ksp expression. Mg(OH)₂: [Mg²⁺] = s = 1.54×10⁻⁴, [OH⁻] = 2s = 3.08×10⁻⁴ → Ksp = (1.54×10⁻⁴)(3.08×10⁻⁴)² = 1.46×10⁻¹¹. Always apply coefficient INSIDE the bracket first.

Write the Mg(OH)₂ reverse calculation with explicit [OH⁻] = 2s step into your notes before the check below.

We just saw the reverse calculation — converting molar solubility to Ksp by applying stoichiometric ratios before substitution. That raises a question: if two compounds have different Ksp values, can you always say the one with a smaller Ksp is less soluble? This card answers it → by explaining the formula-type comparison rule and showing why CaF₂ (smaller Ksp) is actually MORE soluble than AgCl (larger Ksp).

Ksp values can only be directly compared for compounds with the same ionic formula type — the same total number of ions and stoichiometric ratios.

Same formula type → direct Ksp comparison valid. Both have the same Ksp-to-s algebraic form → larger Ksp → more soluble.

Example (both 1:1): AgCl (Ksp = 1.8 × 10⁻¹⁰) vs AgI (Ksp = 8.5 × 10⁻¹⁷) → AgCl more soluble. The student in the Think First was correct for this comparison.

Different formula types → direct Ksp comparison INVALID. Must calculate s for each.

CaF₂ has a smaller Ksp than AgCl but is more soluble — because the 1:2 formula type produces more ions per formula unit. The student in the Think First was wrong about the AgCl vs CaF₂ comparison.

Ksp comparison rule: same formula type (both 1:1 or both 1:2, etc.) → larger Ksp = more soluble, valid direct comparison. Different formula types → MUST calculate s for each and compare s values directly. Classic HSC trap: CaF₂ (Ksp = 3.9×10⁻¹¹, s = 2.14×10⁻⁴) is MORE soluble than AgCl (Ksp = 1.8×10⁻¹⁰, s = 1.34×10⁻⁵) despite having a smaller Ksp.

Write the comparison rule and the AgCl vs CaF₂ example into your notes before the check below.

We just saw that comparing Ksp across different formula types requires calculating molar solubility — and that a smaller Ksp does not always mean less soluble. That raises a question: how does the Ksp concept apply to a real biological system — and how does fluoride's Ksp relationship directly explain its effectiveness in protecting teeth? This card answers it → by comparing Ksp(hydroxyapatite) vs Ksp(fluorapatite) and using LCP to explain acid dissolution and fluoride's protective mechanism.

Tooth enamel is not a permanent solid — it is in dynamic dissolution equilibrium with saliva, and the balance between dissolution and remineralisation is governed by Ksp.

Tooth enamel is primarily hydroxyapatite: Ca₁₀(PO₄)₆(OH)₂.

Dissolution: $\text{Ca}_{10}(\text{PO}_4)_6(\text{OH})_2(s) \rightleftharpoons 10\text{Ca}^{2+}(aq) + 6\text{PO}_4^{3-}(aq) + 2\text{OH}^-(aq)$

Ksp(hydroxyapatite) ≈ 2.3 × 10⁻⁵⁹ — small, but not zero.

After eating sugar: bacteria produce lactic acid → lowers saliva pH → H⁺ reacts with OH⁻ and PO₄³⁻ (products of dissolution) → LCP: removing products shifts equilibrium right → enamel dissolves.

Fluoride protection: F⁻ replaces OH⁻ in the hydroxyapatite structure, forming fluorapatite Ca₁₀(PO₄)₆F₂.

Ksp(fluorapatite) ≈ 1 × 10⁻¹²¹ — many orders of magnitude smaller than hydroxyapatite.

Fluorapatite is far less soluble than hydroxyapatite — it resists acid dissolution even at lower pH.

Tooth enamel = hydroxyapatite Ca₁₀(PO₄)₆(OH)₂, Ksp ≈ 2.3×10⁻⁵⁹. Acid dissolution: H⁺ reacts with OH⁻ and PO₄³⁻ (products) → LCP shifts equilibrium right → enamel dissolves. Fluoride replaces OH⁻ → forms fluorapatite Ca₁₀(PO₄)₆F₂, Ksp ≈ 1×10⁻¹²¹ — far less soluble → far more resistant to acid. Fluoride protection works because Ksp(fluorapatite) << Ksp(hydroxyapatite).

Write the acid dissolution mechanism and the fluorapatite Ksp comparison into your notes before the check below.