Inorganic

Ksp: from rules to numbers

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The solubility rules tell you whether a salt dissolves. That is a yes/no answer. For anything past a rough gravimetric prediction — buffering a wastewater stream, avoiding a nucleation event in a crystallization, tuning selective precipitation for analysis — you need the how much, and that is Ksp. This guide walks the move from the qualitative rules to a real molar solubility, including the two adjustments (common-ion, pH) that undergraduate problems live and die on.

What Ksp actually is

For a sparingly soluble salt of general form MaXb(s) ⇌ a M+(aq) + b X−(aq), the solubility product is

Ksp = [M+]^a [X-]^b

evaluated at equilibrium in a saturated solution. Two things are baked into the definition:

  • The solid is not in the expression. Pure solids and pure liquids have activity = 1 by convention (IUPAC Gold Book).
  • Concentrations are those at saturation. In an unsaturated solution the ion product Q < Ksp; in a supersaturated one Q > Ksp and the salt eventually precipitates until Q returns to Ksp.

Ksp is temperature-dependent (Le Chatelier — for endothermic dissolutions, warmer water increases Ksp, which is why sugar dissolves better in tea than in iced coffee, and why most sulfates behave the opposite way — CaSO₄ actually becomes less soluble as you heat it). Values in reference tables are almost universally quoted at 25 °C.

From Ksp to molar solubility

The move from Ksp to molar solubility s is stoichiometric. For a 1:1 salt like AgCl,

AgCl(s) ⇌ Ag+ + Cl- Ksp = [Ag+][Cl-] = s · s = s² s = √Ksp

Ksp(AgCl) at 25 °C is 1.77 × 10⁻¹⁰ (CRC 97e §4-105). So s = √(1.77 × 10⁻¹⁰) = 1.33 × 10⁻⁵ M — call it 13 micromolar, or 1.9 mg AgCl per litre of water. That is the “insoluble” side of the solubility rules speaking numerically.

For a 1:2 salt like PbI₂ (Ksp = 9.8 × 10⁻⁹, CRC 97e),

PbI₂(s) ⇌ Pb²⁺ + 2 I⁻ Ksp = [Pb²⁺][I⁻]² = s · (2s)² = 4s³ s = ∛(Ksp / 4)

Plugging in gives s = ∛(9.8 × 10⁻⁹ / 4) = ∛(2.45 × 10⁻⁹) = 1.35 × 10⁻³ M PbI₂. Because two iodides come off per formula unit, [I⁻] in that saturated solution is 2s = 2.7 × 10⁻³ M — the anion is twice as concentrated as the “solubility.”

Key lesson: the exponent on s in the Ksp expression depends on the stoichiometry of the salt. Miss the ² in [Cl⁻]² or the 4 in front of s³, and your answer is wrong by an order of magnitude.

The common-ion effect

If the water already contains one of the ions of your salt, the solubility drops — sometimes dramatically. This is Le Chatelier: raise the concentration of a product, and the equilibrium shifts back toward the reactant (undissolved solid).

Concrete: how much AgCl dissolves in 0.10 M NaCl rather than pure water?

Set up the ICE table with [Cl⁻] starting at 0.10 M (from NaCl, fully dissociated).

Ksp = [Ag+][Cl-] = s · (0.10 + s)

Because Ksp is so small, s ≪ 0.10, so 0.10 + s ≈ 0.10. That gives

s = 1.77 × 10⁻¹⁰ / 0.10 = 1.77 × 10⁻⁹ M

That is 7,500 times less than the pure-water solubility. This is exactly why silver nitrate titrations of chloride (Mohr, Fajans, Volhard methods — Harris Ch. 7) work: at the endpoint, adding more chloride pushes essentially all the silver into the AgCl precipitate.

Warning: the approximation s ≪ [common ion]₀ only works when Ksp is small relative to the common-ion concentration. If Ksp ≈ 10⁻⁵ and [common ion]₀ ≈ 10⁻³ M, the approximation fails. Solve the full quadratic or the full cubic instead of dropping the small term.

The pH effect on solubility of salts of weak acids

The other big surprise: if the anion of the salt is the conjugate base of a weak acid, the solubility depends on pH. Two classes of salt to watch:

  • Carbonates, phosphates, sulfides, oxalates, fluorides — anions of weak acids. Lower pH (more H₃O⁺) protonates the anion, removes it from the ion product, and increases solubility.
  • Hydroxides (Mg(OH)₂, Fe(OH)₃, Al(OH)₃) — solubility increases at low pH and decreases at high pH, in the extreme case dissolving entirely as complex hydroxides.

Textbook example: CaCO₃ (Ksp = 3.36 × 10⁻⁹, CRC 97e). In neutral water, s = √Ksp ≈ 5.8 × 10⁻⁵ M. In slightly acidic rainwater at pH 5, carbonate protonates:

CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻ (Ka2 of H₂CO₃ = 4.7 × 10⁻¹¹)

At pH 5, [OH⁻] = 10⁻⁹ M, and the equilibrium pushes carbonate far to the right (bicarbonate). The effective free [CO₃²⁻] is a small fraction of the total dissolved carbonate, and the ion product cannot reach Ksp until much more CaCO₃ has dissolved. This is exactly the mechanism that dissolves limestone caves — millimeters per year of dissolution driven by mildly acidic groundwater.

For the Ksp calculator, the pH correction is a separate module; the default assumes the anion is not appreciably protonated (chlorides, iodides, sulfates in mildly acidic conditions).

Selective precipitation

When a solution contains two cations that both form insoluble salts with a given anion, you can often precipitate one selectively by choosing conditions where its Q > Ksp but the other’s Q < Ksp. This is the operating principle of qualitative analysis (Housecroft §7.13) and of some purification workups.

Example: a solution 0.010 M in Pb²⁺ and 0.010 M in Ag⁺. Slowly add Cl⁻. Which precipitates first?

Compare each salt’s threshold:

  • AgCl (Ksp = 1.77 × 10⁻¹⁰): starts precipitating when [Cl⁻] > 1.77 × 10⁻¹⁰ / 0.010 = 1.8 × 10⁻⁸ M.
  • PbCl₂ (Ksp = 1.7 × 10⁻⁵, CRC 97e): starts precipitating when [Cl⁻]² > 1.7 × 10⁻⁵ / 0.010, i.e. [Cl⁻] > 0.041 M.

So AgCl starts precipitating first, at a Cl⁻ concentration 2 million times smaller than PbCl₂ needs. You can bring [Cl⁻] up to just below 0.041 M and have essentially all the Ag precipitated with Pb still in solution — that’s the selectivity.

A canonical worked example

Worked example — molar solubility of Ag₂CrO₄
Salt
Ag₂CrO₄(s) ⇌ 2 Ag⁺(aq) + CrO₄²⁻(aq)
Ksp at 25 °C
1.1 × 10⁻¹² (CRC 97e §4-105)
M(Ag₂CrO₄)
331.7 g/mol (IUPAC 2021)
  1. Formula
  2. Substitute
  3. Note

    Let s be the molar solubility (mol Ag₂CrO₄ dissolved per litre). Each formula unit contributes 2 Ag⁺ and 1 CrO₄²⁻, so [Ag⁺] = 2s and [CrO₄²⁻] = s. Substitute into the Ksp expression to get 4s³, then take the cube root: s = ∛(2.75 × 10⁻¹³) = 6.5 × 10⁻⁵ M. As a mass concentration that is 6.5 × 10⁻⁵ × 331.7 = 22 mg/L. Compare to AgCl at 1.9 mg/L — silver chromate is roughly ten times more soluble even though its Ksp is a hundred times smaller, because the 2:1 stoichiometry means one mole of dissolved salt puts three moles of ions into solution. Ksp values from different-stoichiometry salts are not directly comparable; convert to s first.

  4. Result

Compare that to AgCl (1.9 mg/L). Silver chromate is about ten times more soluble than silver chloride even though its Ksp is a hundred times smaller — because the salt’s 2:1 stoichiometry means one mole of dissolved salt puts three moles of ions into solution rather than two. Ksp values from different-stoichiometry salts are not directly comparable; you have to convert to s to compare.

What Ksp does not tell you

Two important limits:

  • Kinetics is not included. Ksp says how much would dissolve at equilibrium; it says nothing about how fast. Barium sulfate has Ksp ≈ 1.1 × 10⁻¹⁰ (thermodynamically very insoluble), which is why it can be safely used as a radiographic contrast agent — the kinetically slow dissolution keeps free Ba²⁺ low enough not to cause acute cardiotoxicity even though Ba²⁺ is toxic in solution.
  • Activities, not concentrations, are the strictly correct quantity. At ionic strengths above ~0.01 M, the concentration-based Ksp starts to disagree with the activity-based one by a factor that grows with ionic strength (Davies equation, or Debye-Hückel in the dilute limit — Harris §7.2). Textbook problems ignore this. Industrial calculations do not.

Practice problems

Problem 1 Easy Ksp(CaF₂) = 3.9 × 10⁻¹¹ at 25 °C. What is the molar solubility in pure water?

Answer: CaF₂ ⇌ Ca²⁺ + 2 F⁻. Ksp = s(2s)² = 4s³. s = ∛(3.9 × 10⁻¹¹ / 4) = ∛(9.75 × 10⁻¹²) = 2.14 × 10⁻⁴ M. Fluoride concentration in the saturated solution is 2s, not s.

Problem 2 Intermediate What is the solubility of PbSO₄ (Ksp = 2.53 × 10⁻⁸) in 0.100 M Na₂SO₄?

Answer: [SO₄²⁻]₀ = 0.100 M from Na₂SO₄. Assume s ≪ 0.100. Ksp ≈ s × 0.100, so s = 2.53 × 10⁻⁸ / 0.100 = 2.53 × 10⁻⁷ M. (Check: s ≪ 0.100 ✓.) Common ion drops solubility 100×.

Problem 3 Hard You have a solution 0.0010 M each in Ba²⁺ and Ca²⁺. You slowly add Na₂SO₄. Which precipitates first, and at what [SO₄²⁻]? Ksp(BaSO₄) = 1.08 × 10⁻¹⁰, Ksp(CaSO₄) = 4.93 × 10⁻⁵.

Answer: BaSO₄ precipitates first at [SO₄²⁻] > 1.08 × 10⁻¹⁰ / 0.0010 = 1.08 × 10⁻⁷ M. CaSO₄ waits until [SO₄²⁻] > 4.93 × 10⁻⁵ / 0.0010 = 0.0493 M. Factor-of-500,000 window for selective barium precipitation.

Where to go next

  • The Ksp / molar solubility calculator runs the s-vs-Ksp conversion for any MaXb stoichiometry, and has a common-ion mode.
  • The solubility-rules cheatsheet covers the qualitative side — worth pinning while you work.
  • If your dissolution is pH-sensitive (carbonates, phosphates, sulfides, hydroxides), pair Ksp with the pH calculator to see where the anion protonates.

Solubility rules are the first check. Ksp is the second. Once you can compute molar solubility, adjust for the common-ion effect, and know when pH matters, you can predict precipitation quantitatively.

Sources

  • Harris, Quantitative Chemical Analysis, 10th ed. (2020), Ch. 6 — Chemical equilibrium
  • Miessler, Fischer & Tarr, Inorganic Chemistry, 5th ed. (2014), §6.6 — Solubility equilibria
  • Housecroft & Sharpe, Inorganic Chemistry, 5th ed. (2018), Ch. 7
  • CRC Handbook of Chemistry and Physics, 97th ed. (2016), §4-105 — Solubility products at 25 °C
  • IUPAC NIST Solubility Data Series