The Solubility Product, Ksp

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 showing that the Ksp-to-solubility relationship has a different algebraic form for each formula type, which is why CaF₂ (smaller Ksp) turns out to be 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 and yet is more soluble. Be careful with the reason. It is not that "more ions makes it dissolve more", the ion count does not push anything into solution. What changes is the algebra connecting Ksp to s. For a 1:1 salt, Ksp = s², so s = √Ksp. For a 1:2 salt, Ksp = 4s³, so s = ∛(Ksp/4). Those are different functions, so the same numerical Ksp maps to a different solubility depending on the formula type, and a smaller Ksp can sit above a larger one once you convert. Comparing the Ksp values directly compares two quantities that are not measuring the same thing. 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. For Ksp below, use the half-formula unit Ca₅(PO₄)₃OH; the Ca₁₀ formula is twice this stoichiometric unit.

Dissolution: $\text{Ca}_{5}(\text{PO}_4)_3\text{OH}(s) \rightleftharpoons 5\text{Ca}^{2+}(aq) + 3\text{PO}_4^{3-}(aq) + \text{OH}^-(aq)$. Ksp = [Ca²⁺]⁵[PO₄³⁻]³[OH⁻].

At 25 °C, a NIST study recommends Ksp ≈ (4.7 ± 2.0) × 10⁻⁵⁹ for hydroxyapatite. The authors note that preparation, impurities and equilibration affect reported values. (McDowell, Gregory & Brown, 1977)

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⁻ can replace OH⁻ in the hydroxyapatite structure, forming fluorapatite Ca₁₀(PO₄)₆F₂. Use its half-formula unit, Ca₅(PO₄)₃F, for the comparison below.

Dissolution: $\text{Ca}_{5}(\text{PO}_4)_3\text{F}(s) \rightleftharpoons 5\text{Ca}^{2+}(aq) + 3\text{PO}_4^{3-}(aq) + \text{F}^-(aq)$. Ksp = [Ca²⁺]⁵[PO₄³⁻]³[F⁻]. At 25 °C, McCann reports fluorapatite pKsp ≈ 59.6, equivalent to Ksp ≈ 2.5 × 10⁻⁶⁰ for this Ca₅ formula unit. This and the hydroxyapatite estimate are from separate studies; treat the comparison as approximate, not a universal constant. (McCann, 1968)

Both Ksp values use the same half-formula stoichiometry. The lower reported fluorapatite Ksp indicates lower solubility, helping explain its greater resistance to dissolution. A Ksp ratio is not itself a molar-solubility ratio, and behavior in saliva also depends on pH and solution composition.

Tooth enamel is primarily hydroxyapatite Ca₁₀(PO₄)₆(OH)₂. For Ksp, use half-formula units: Ca₅(PO₄)₃OH and Ca₅(PO₄)₃F. Reported values include hydroxyapatite Ksp ≈ (4.7 ± 2.0)×10⁻⁵⁹ at 25 °C and fluorapatite Ksp ≈ 2.5×10⁻⁶⁰ at 25 °C (from a reported pKsp ≈ 59.6); these are approximate values from separate studies. H⁺ removes OH⁻ and phosphate ions, shifting dissolution right. With matching stoichiometry, the lower reported fluorapatite Ksp indicates lower solubility and helps explain greater resistance to dissolution; the Ksp ratio is not the solubility ratio.

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