Tooth enamel dissolves and reforms constantly in your mouth — the equilibrium between hydroxyapatite and saliva is governed by Ksp, and fluoride treatment works by replacing hydroxyapatite with a compound that has a lower Ksp, making teeth more resistant to dissolution.
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A student is told that AgCl has Ksp = 1.8 × 10⁻¹⁰ and AgI has Ksp = 8.5 × 10⁻¹⁷ at 25°C. The student concludes: "AgI is less soluble than AgCl because its Ksp is smaller."
Is this conclusion correct? Is this comparison valid? Would the same logic apply to comparing AgCl (Ksp = 1.8 × 10⁻¹⁰) with CaF₂ (Ksp = 3.9 × 10⁻¹¹)? Write your analysis of both comparisons.
Wrong: All ionic compounds dissolve in water because water is polar.
Right: Water dissolves many ionic compounds through ion-dipole interactions, but not all. Solubility depends on the balance between lattice energy and hydration energy. Compounds with very high lattice energy (e.g., AgCl, BaSO₄) are insoluble despite water's polarity.
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 MaXb:
$$\text{M}_a\text{X}_b(s) \rightleftharpoons a\text{M}^{b+}(aq) + b\text{X}^{a-}(aq)$$
$$K_{sp} = [\text{M}^{b+}]^a \times [\text{X}^{a-}]^b$$
The solid is excluded (pure solid, activity = 1). There is no denominator because the only reactant (the solid) is excluded.
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): $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$
| Ag⁺ | Cl⁻ | |
|---|---|---|
| Initial | 0 | 0 |
| Change | +s | +s |
| Equilibrium | s | s |
$K_{sp} = s^2 = 1.8 \times 10^{-10}$ → $s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}$ 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
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 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.
| Compound | Ksp | Formula type | Ksp → s formula | Molar solubility (s) |
|---|---|---|---|---|
| AgCl | 1.8 × 10⁻¹⁰ | 1:1 | s = √Ksp | 1.34 × 10⁻⁵ mol/L |
| AgI | 8.5 × 10⁻¹⁷ | 1:1 | s = √Ksp | 9.22 × 10⁻⁹ mol/L |
| CaF₂ | 3.9 × 10⁻¹¹ | 1:2 | s = ∛(Ksp/4) | 2.14 × 10⁻⁴ mol/L |
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.
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.
The molar solubility of silver sulfate (Ag₂SO₄) is 7.60 × 10⁻³ mol/L at 25°C. (a) Write the dissolution equation. (b) Express ion concentrations in terms of s. (c) Calculate Ksp.
$\text{Ag}_2\text{SO}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{SO}_4^{2-}(aq)$
2:1 type — each formula unit produces 2 Ag⁺ and 1 SO₄²⁻.
[Ag⁺] = 2s = 2(7.60 × 10⁻³) = 1.520 × 10⁻² mol/L
[SO₄²⁻] = s = 7.60 × 10⁻³ mol/L
$K_{sp} = [\text{Ag}^+]^2[\text{SO}_4^{2-}] = (1.520 \times 10^{-2})^2(7.60 \times 10^{-3})$
$(1.520 \times 10^{-2})^2 = 2.3104 \times 10^{-4}$
$K_{sp} = (2.3104 \times 10^{-4})(7.60 \times 10^{-3}) = 1.76 \times 10^{-6}$
(a) Calculate the molar solubility of PbSO₄ (Ksp = 1.6 × 10⁻⁸) and AgBr (Ksp = 5.0 × 10⁻¹³). (b) Can these Ksp values be compared directly? (c) Which compound is less soluble?
Both PbSO₄ and AgBr are 1:1 type compounds. For 1:1: s = √Ksp.
s(PbSO₄) = $\sqrt{1.6 \times 10^{-8}} = 1.27 \times 10^{-4}$ mol/L
s(AgBr) = $\sqrt{5.0 \times 10^{-13}} = 7.07 \times 10^{-7}$ mol/L
Yes — both PbSO₄ and AgBr are 1:1 formula type. Same formula type → direct Ksp comparison is valid. Larger Ksp = more soluble. PbSO₄ (Ksp = 1.6 × 10⁻⁸) > AgBr (Ksp = 5.0 × 10⁻¹³) → PbSO₄ is more soluble than AgBr.
AgBr (s = 7.07 × 10⁻⁷ mol/L) is less soluble than PbSO₄ (s = 1.27 × 10⁻⁴ mol/L) — consistent with the smaller Ksp for the same formula type.
Ksp is the equilibrium constant for the dissolution of a sparingly soluble salt. For AgCl, Ksp = [Ag⁺][Cl⁻]. A smaller Ksp means lower solubility only when comparing salts with the same stoichiometry (e.g. 1:1). To compare AgCl and AgI (both 1:1), the smaller Ksp of AgI (8.5 × 10⁻¹⁷) does mean lower solubility. For salts with different stoichiometries (e.g. AgCl vs Ag₂CrO₄), you must calculate molar solubility s from the Ksp expression before comparing.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Q1. What is the correct Ksp expression for calcium phosphate, Ca₃(PO₄)₂?
Q1. Identify the correct Ksp expression for calcium phosphate, Ca₃(PO₄)₂?
Q2. The Ksp of BaF₂ is 1.0 × 10⁻⁶ at 25°C. What is the molar solubility of BaF₂?
Q2. The Ksp of BaF₂ is 1.0 × 10⁻⁶ at 25°C. Identify the molar solubility of BaF₂?
Q3. A student compares Ksp(CaCO₃) = 3.4 × 10⁻⁹ with Ksp(Ag₂CO₃) = 8.5 × 10⁻¹². They conclude CaCO₃ is less soluble because its Ksp is smaller. Is this valid?
Put your knowledge of Solubility Product Ksp to the test. Answer correctly to deal damage — get it wrong and the boss hits back. Pool: lessons 1–17.
Lesson 17 complete — Solubility Product Ksp