What is the Ksp?
Enter how many cations and anions one formula unit of the salt gives, then its solubility or its Ksp. The calculator gives the solubility product or the molar solubility, the ion concentrations, and Ksp written in terms of s.
- Answer
- Ksp = 3.97535 × 10⁻¹¹
Ksp = 3.97535 × 10⁻¹¹ for an MX₂ salt, where Ksp = s × (2s)² = 4s³.
- Ksp
- 3.97535 × 10⁻¹¹
- Molar solubility (mol/L)
- 0.000215
- Cation concentration (mol/L)
- 0.000215
- Anion concentration (mol/L)
- 0.00043
- Ksp in terms of s
- Ksp = s × (2s)² = 4s³
- Salt type
- MX₂
Answer: Ksp = 3.97535 × 10⁻¹¹. Ksp = 3.97535 × 10⁻¹¹ for an MX₂ salt, where Ksp = s × (2s)² = 4s³.
How to calculate
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.
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.
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.
- 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
Each example is checked against the calculator on every build.
- Find Ksp, Cations per formula unit (p) 1, Anions per formula unit (q) 2, Solubility in mol/L, Solubility 0.000215 gives Ksp 0, Anion concentration (mol/L) 0.00043, Answer 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
- Find Ksp, Cations per formula unit (p) 1, Anions per formula unit (q) 2, Solubility in mol/L, Solubility 0.000131 gives Ksp 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
- Find Molar solubility, Cations per formula unit (p) 1, Anions per formula unit (q) 1, Solubility in mol/L, Ksp 0 gives Molar solubility (mol/L) 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
- Find Molar solubility, Cations per formula unit (p) 1, Anions per formula unit (q) 2, Solubility in mol/L, Ksp 0.000001 gives Molar solubility (mol/L) 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
- Find Molar solubility, Cations per formula unit (p) 1, Anions per formula unit (q) 2, Solubility in mol/L, Ksp 0 gives Molar solubility (mol/L) 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
- Find Ksp, Cations per formula unit (p) 1, Anions per formula unit (q) 1, Solubility in g/L, Solubility 0.000005, Molar mass (g/mol) 323.19 gives Ksp 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
How it works
A salt MpXq dissolves into p cations and q anions:
MpXq(s) ⇌ p M(aq) + q X(aq)
If s mol/L of the salt dissolves (its molar solubility), the ion concentrations are [M] = p × s and [X] = q × s, so
Ksp = [M]^p × [X]^q = (p × s)^p × (q × s)^q = p^p × q^q × s^(p+q)
- Ksp from s: the formula above, worked exactly from the typed decimal.
- s from Ksp: s = (Ksp ÷ (p^p × q^q))^(1 ÷ (p + q)), a root, worked in floating point.
- Grams per liter: s (mol/L) = solubility (g/L) ÷ molar mass (g/mol), and back. Choose g/L and give the molar mass; with mol/L chosen the molar mass is not used.
p and q are whole numbers from 1 to 5. Results are shown to 6 significant figures with trailing zeros dropped; a value below 0.000001 is written with a power of ten (Ksp = 3.97535 × 10⁻¹¹). The page also gives the ion concentrations p × s and q × s and Ksp written in terms of s (for CaF₂: Ksp = s × (2s)² = 4s³).
Rules the page checks. The solubility is from 10⁻²⁰ to 10,000 in its unit, Ksp from 10⁻¹⁰⁰ to 10⁶, and the molar mass from 0.001 to 100,000 g/mol. These bounds keep every result a finite number above 0 (Ksp from about 10⁻²⁴³ to 10⁷⁷).
Assumptions. The salt dissolves in pure water (no common ion), its ions do not react further, and concentrations stand for activities.
Worked examples by hand
CaF₂ from s = 2.15 × 10⁻⁴ M (OpenStax Example 15.2). p = 1, q = 2: Ksp = 1 × 2² × s³ = 4 × (2.15 × 10⁻⁴)³ = 4 × 9.938375 × 10⁻¹² = 3.97535 × 10⁻¹¹ (OpenStax: 3.98 × 10⁻¹¹). [F⁻] = 2s = 4.30 × 10⁻⁴ M.
Mg(OH)₂ from s = 1.31 × 10⁻⁴ M. Ksp = 4 × (1.31 × 10⁻⁴)³ = 8.99236 × 10⁻¹².
CuBr from Ksp = 6.3 × 10⁻⁹ (Example 15.3). p = q = 1: s = √(6.3 × 10⁻⁹) = 7.93725 × 10⁻⁵ M.
Ca(OH)₂ from Ksp = 1.3 × 10⁻⁶ (Example 15.4). s = ∛(1.3 × 10⁻⁶ ÷ 4) = ∛(3.25 × 10⁻⁷) = 6.87534 × 10⁻³ M.
Hg₂Cl₂ from Ksp = 1.1 × 10⁻¹⁸ (Example 15.6). p = 1 (Hg₂²⁺), q = 2: s = ∛(1.1 × 10⁻¹⁸ ÷ 4) = 6.50296 × 10⁻⁷ M.
PbCrO₄ from 4.6 × 10⁻⁶ g/L (Example 15.5). s = 4.6 × 10⁻⁶ ÷ 323.19 = 1.42331 × 10⁻⁸ M; Ksp = s² = 2.02582 × 10⁻¹⁶ (OpenStax rounds s to 1.4 × 10⁻⁸ and gives 2.0 × 10⁻¹⁶).
Other questions people ask
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.