Solubility Product Calculator

Work out solubility product instantly with clear inputs, formula shown and shareable results.

About Solubility Product Calculator

For a salt AxBy dissolving to give x cations and y anions, each ion's concentration is a multiple of the molar solubility, and Ksp is the product of those concentrations each raised to its own coefficient: Ksp = (xs)^x (ys)^y. The exponents are what make Ksp values for 1:2 salts look so much smaller than for 1:1 salts at similar solubility.

Formulas

Solubility product

Ksp = (x.s)^x x (y.s)^y

Logarithmic form

pKsp = -log10(Ksp)

Worked example

With the default values below, Solubility Product Calculator returns Solubility Product Ksp of 8.7880 × 10⁻¹⁵. Change any input above and every figure here is recalculated in your browser.

Inputs used

Molar solubility sminimum 0 mol/L: 0.000013 mol/L
Cations per formula unit (x)minimum 1: 1
Anions per formula unit (y)minimum 1: 2

Results

Solubility Product Ksp: 8.7880 × 10⁻¹⁵
pKsp: 14.056
Cation Concentration: 1.3000 × 10⁻⁵
Anion Concentration: 2.6000 × 10⁻⁵

Step by step

1Values used
2Solubility product
3Logarithmic form
4Solubility Product Ksp= 0.0000
5pKsp= 14.056
6Cation Concentration= 0.0000
7Anion Concentration= 0.0000

Assumptions and limits

The result is derived only from the 3 inputs listed above. Anything not among them is not part of the calculation.
In particular, this calculator does not take inflation, tax, fees or charges or compounding frequency into account.
Figures are rounded for display to 4 decimal places, so totals built from several rounded parts can differ slightly from the unrounded arithmetic.

How these calculators are built and tested is described on the Methodology page. If a figure looks wrong, please report it.

Frequently Asked Questions

Why is Ksp not simply s squared?

Only for a 1:1 salt. A 1:2 salt produces twice as many anions, so the anion term is (2s) and it is squared, giving Ksp = 4s^3.

Does Ksp change with concentration?

No, only with temperature. That is what makes it useful for predicting whether a precipitate forms.