How this calculation works
Dissolving a sparingly soluble salt adds ions to concentrations already present in solution. A common ion can therefore suppress additional dissolution even when the amount of undissolved solid has not changed. This page solves the entered concentration-form solubility product for additional moles of salt dissolved per litre. It keeps the reaction coefficients distinct from the final ion concentrations.
Ksp = (A₀+a s)^a (B₀+b s)^b for salt → a A + b BA worked example
For a one-to-one salt with Ksp one times ten to the minus eight and no initial ions, both equilibrium ion concentrations equal s. Their product gives s equal to one times ten to the minus four molar. Adding an initial 0.01 molar concentration of ion A changes the equation to 0.01 plus s, multiplied by s. The additional salt solubility falls sharply, while the already present common ion remains part of the final concentration.
Good to know
The calculation assumes ideal concentrations and enough solid available to reach saturation. It does not know whether the entered stock contains sufficient undissolved material or whether the chosen ions balance electrical charge. Complexation, protonation, activity coefficients and competing precipitates are omitted. A real analytical solubility problem may need these coupled equilibria rather than a single dissolution equation.
Reference: OpenStax — precipitation and dissolution (opens in a new tab) ↗
What this calculation answers
Dissolving a sparingly soluble salt adds ions to concentrations already present in solution. A common ion can therefore suppress additional dissolution even when the amount of undissolved solid has not changed.
This page solves the entered concentration-form solubility product for additional moles of salt dissolved per litre. It keeps the reaction coefficients distinct from the final ion concentrations.
The Ionic strength and Davies activity coefficient calculator estimates activity effects from a complete ion list. This solubility model deliberately uses ideal concentrations without coupling those activity coefficients into Ksp.
Use the right input conventions
Coefficients a and b are whole-number stoichiometric amounts in salt becoming a units of ion A and b units of ion B. Enter initial common-ion concentrations before additional salt dissolves.
Ksp must correspond to that same written dissolution reaction. Changing coefficients changes both exponents and added ion amounts; it is not a cosmetic relabeling of the same constant.
Interpret the result
The principal result is additional salt molarity s. Equilibrium ion A is its initial concentration plus a times s, and ion B follows the analogous b relationship. The reported ion product checks those final concentrations.
An initially supersaturated mixture is rejected because positive dissolution alone cannot describe its precipitation balance. The page does not subtract an unspecified amount of solid to repair an incompatible initial state.
Work through the example
For a one-to-one salt with Ksp one times ten to the minus eight and no initial ions, both equilibrium ion concentrations equal s. Their product gives s equal to one times ten to the minus four molar.
Adding an initial 0.01 molar concentration of ion A changes the equation to 0.01 plus s, multiplied by s. The additional salt solubility falls sharply, while the already present common ion remains part of the final concentration.
Where the model stops
The calculation assumes ideal concentrations and enough solid available to reach saturation. It does not know whether the entered stock contains sufficient undissolved material or whether the chosen ions balance electrical charge.
Complexation, protonation, activity coefficients and competing precipitates are omitted. A real analytical solubility problem may need these coupled equilibria rather than a single dissolution equation.
A few questions, answered
Can this solve a supersaturated precipitation mixture?
No. It solves nonnegative additional dissolution with fixed initial common-ion concentrations, not a precipitation inventory balance.
What changes in the comparison?
Add a common ion and compare the lower amount of additional salt that can dissolve.