What this calculator does
The Molar Solubility Calculator converts a salt's solubility product constant (Ksp) into its molar solubility (s), the number of moles of the compound that dissolve per litre of solution at equilibrium. It works for any sparingly soluble ionic salt of the general form MxAy, where x and y are the stoichiometric coefficients of the cation and anion.
How to use it
Enter the Ksp value (look it up in a data table), then enter the coefficients x and y from the dissolution equation MxAy → x M + y A. For example, AgCl gives x = 1, y = 1; PbI₂ and Ca(OH)₂ give x = 1, y = 2; Ag₂CrO₄ gives x = 2, y = 1. The tool returns the molar solubility plus the resulting cation and anion concentrations.
The formula explained
At equilibrium, if s mol/L dissolves, then \([\text{M}] = xs\) and \([\text{A}] = ys\). Substituting into \(K_{sp} = [\text{M}]^{x}[\text{A}]^{y}\) gives \(K_{sp} = (xs)^{x}(ys)^{y} = x^{x}y^{y}\cdot s^{(x+y)}\). Solving for \(s\):
$$S = \sqrt[\,x+y\,]{\dfrac{\text{Ksp}}{\text{x}^{\,\text{x}}\cdot\text{y}^{\,\text{y}}}}$$
Worked example
Calcium hydroxide, Ca(OH)₂, has \(K_{sp} = 5.5 \times 10^{-6}\) with x = 1, y = 2. The factor \(x^{x}y^{y} = 1\cdot 4 = 4\). So $$s = \left(\frac{5.5\times 10^{-6}}{4}\right)^{1/3} = (1.375\times 10^{-6})^{1/3} \approx 0.0111\ \text{mol/L}.$$ The hydroxide concentration is \(2s \approx 0.0222\) mol/L.
FAQ
Does this account for the common-ion effect? This version computes solubility in pure water. The common-ion effect lowers solubility when a shared ion is already present; the same equilibrium expression applies but with the extra ion added to its concentration term.
What units does s have? Moles per litre (mol/L), assuming Ksp is the dimensionless standard-state value at 25 °C.
Why must coefficients be at least 1? Every dissolving salt produces at least one cation and one anion, so x and y are whole numbers ≥ 1.
