# What is the Ksp?

Finds the solubility product Ksp of a slightly soluble salt from its molar solubility (or its solubility in g/L and molar mass), or the molar solubility from Ksp, for a salt that dissolves into p cations and q anions.

- Page: https://www.acalculator.org/chemistry/ksp-calculator
- JSON spec: https://www.acalculator.org/chemistry/ksp-calculator.json
- Version: a9c52f731f03

## Default answer

Example with the default inputs (Find Ksp, Cations per formula unit (p) 1, Anions per formula unit (q) 2, Solubility in mol/L, Solubility 0.000215): Ksp = 3.97535 × 10⁻¹¹ for an MX₂ salt, where Ksp = s × (2s)² = 4s³.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| find | Find | The solubility product from the solubility, or the molar solubility from Ksp. |
| p | Cations per formula unit (p) | How many cations one formula unit gives: 1 for CaF₂, 2 for Ag₂CrO₄. |
| q | Anions per formula unit (q) | How many anions one formula unit gives: 2 for CaF₂, 1 for Ag₂CrO₄. |
| u | Solubility in | Moles per liter, or grams per liter with the molar mass. |
| s | Solubility | How much salt dissolves in a liter of saturated solution, in the unit chosen above. |
| ksp | Ksp | The solubility product, such as 3.9e-11 for CaF₂. |
| mm | Molar mass (g/mol) | The molar mass of the salt, to convert between grams and moles. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | Ksp or the molar solubility, to 6 significant figures. |
| kspValue | Ksp | The solubility product. |
| molar | Molar solubility (mol/L) | Moles of salt that dissolve per liter of saturated solution. |
| grams | Solubility (g/L) | Grams of salt that dissolve per liter, from the molar mass. |
| cation | Cation concentration (mol/L) | The cation concentration at saturation, p × s. |
| anion | Anion concentration (mol/L) | The anion concentration at saturation, q × s. |
| expression | Ksp in terms of s | The product of the ion concentrations written with the molar solubility s. |
| salt | Salt type | The salt written as MpXq, such as MX₂. |

## Method

Ksp = [M]^p × [X]^q = (p × s)^p × (q × s)^q = p^p × q^q × s^(p + q); s = (Ksp ÷ (p^p × q^q))^(1 ÷ (p + q)); s in mol/L = solubility in g/L ÷ molar mass.

## Assumptions

- The salt dissolves into p cations and q anions in pure water, with no common ion, and the ions do not react further (no hydrolysis or complex ions).
- Concentrations stand for activities, as in introductory chemistry.

## Worked examples

1. find = ksp, p = 1, q = 2, u = mol, s = 0.000215 gives kspValue = 0, anion = 0.00043, result = Ksp = 3.97535 × 10⁻¹¹. Source: OpenStax, Chemistry 2e, §15.1 Precipitation and Dissolution (Ksp from molar solubility and the reverse), https://openstax.org/books/chemistry-2e/pages/15-1-precipitation-and-dissolution, Example 15.2 Calculation of Ksp from Equilibrium Concentrations: Ksp = 3.98 × 10⁻¹¹.
2. find = ksp, p = 1, q = 2, u = mol, s = 0.000131 gives kspValue = 0. Source: OpenStax, Chemistry 2e, §15.1 Precipitation and Dissolution (Ksp from molar solubility and the reverse), https://openstax.org/books/chemistry-2e/pages/15-1-precipitation-and-dissolution, Check Your Learning after Example 15.2: Ksp = 8.99 × 10⁻¹².
3. find = s, p = 1, q = 1, u = mol, ksp = 0 gives molar = 0.000079. Source: OpenStax, Chemistry 2e, §15.1 Precipitation and Dissolution (Ksp from molar solubility and the reverse), https://openstax.org/books/chemistry-2e/pages/15-1-precipitation-and-dissolution, Example 15.3 Determination of Molar Solubility from Ksp: 7.9 × 10⁻⁵ M.
4. find = s, p = 1, q = 2, u = mol, ksp = 0.000001 gives molar = 0.006875. Source: OpenStax, Chemistry 2e, §15.1 Precipitation and Dissolution (Ksp from molar solubility and the reverse), https://openstax.org/books/chemistry-2e/pages/15-1-precipitation-and-dissolution, Example 15.4 Determination of Molar Solubility from Ksp: 6.9 × 10⁻³ M.
5. find = s, p = 1, q = 2, u = mol, ksp = 0 gives molar = 0.000001. Source: OpenStax, Chemistry 2e, §15.1 Precipitation and Dissolution (Ksp from molar solubility and the reverse), https://openstax.org/books/chemistry-2e/pages/15-1-precipitation-and-dissolution, Example 15.6 Determination of Molar Solubility from Ksp: 6.5 × 10⁻⁷ M.
6. find = ksp, p = 1, q = 1, u = g, s = 0.000005, mm = 323.19 gives kspValue = 0. Source: OpenStax, Chemistry 2e, §15.1 Precipitation and Dissolution (Ksp from molar solubility and the reverse), https://openstax.org/books/chemistry-2e/pages/15-1-precipitation-and-dissolution, Example 15.5 Calculation of Ksp from Solubility: 2.0 × 10⁻¹⁶.

## FAQ

### How do I calculate Ksp from molar solubility?

Write the dissolution, find each ion’s concentration from the molar solubility s, and multiply them, each raised to its coefficient. For CaF₂ → Ca²⁺ + 2F⁻ with s = 2.15 × 10⁻⁴ M: [Ca²⁺] = s and [F⁻] = 2s, so Ksp = s × (2s)² = 4s³ = 3.98 × 10⁻¹¹.

### How do I find the molar solubility from Ksp?

Solve Ksp = p^p × q^q × s^(p+q) for s. For a 1:1 salt such as CuBr, s = √Ksp = √(6.3 × 10⁻⁹) = 7.9 × 10⁻⁵ M. For a 1:2 salt such as Ca(OH)₂, s = ∛(Ksp ÷ 4) = 6.9 × 10⁻³ M.

### What are p and q?

The number of cations and of anions one formula unit gives when it dissolves. CaF₂ gives 1 Ca²⁺ and 2 F⁻ (p = 1, q = 2); Ag₂CrO₄ gives 2 Ag⁺ and 1 CrO₄²⁻ (p = 2, q = 1). Mercury(I) chloride, Hg₂Cl₂, gives one Hg₂²⁺ ion and 2 Cl⁻, so p = 1 and q = 2.

### How do I use a solubility in grams per liter?

Divide by the molar mass to get the molar solubility. Lead(II) chromate dissolves to 4.6 × 10⁻⁶ g/L; with a molar mass of 323.19 g/mol that is 1.42 × 10⁻⁸ M, and Ksp = s² = 2.0 × 10⁻¹⁶. Choose g/L and type the molar mass, and the page converts both ways.

### Does a smaller Ksp always mean a less soluble salt?

Only for salts with the same p and q. CuBr (Ksp 6.3 × 10⁻⁹) dissolves to 7.9 × 10⁻⁵ M, but MgF₂, with almost the same Ksp (6.4 × 10⁻⁹), dissolves to 1.2 × 10⁻³ M, about 15 times more, because its Ksp is 4s³ instead of s².

### Does this work in a solution that already has one of the ions?

No. The page assumes the salt dissolves in pure water. A common ion lowers the solubility: in 0.010 M CdBr₂, CdS dissolves to only 1.0 × 10⁻²⁶ M. Ions that react further (hydrolysis, complex ions) also change the answer.

## Sources

- OpenStax, Chemistry 2e, §15.1 Precipitation and Dissolution: the dissolution MpXq(s) ⇌ pM(aq) + qX(aq) and its Ksp, Example 15.2 (CaF₂: Ksp = 3.98 × 10⁻¹¹) and its Check Your Learning (Mg(OH)₂: 8.99 × 10⁻¹²), Example 15.3 (CuBr: 7.9 × 10⁻⁵ M), Example 15.4 (Ca(OH)₂: 6.9 × 10⁻³ M), Example 15.5 (PbCrO₄: 2.0 × 10⁻¹⁶), Example 15.6 (Hg₂Cl₂: 6.5 × 10⁻⁷ M) and its Check Your Learning (MgF₂: 1.2 × 10⁻³ M), Example 15.13 (CdS with a common ion). https://openstax.org/books/chemistry-2e/pages/15-1-precipitation-and-dissolution (retrieved 2026-10-19)