What the common-ion effect calculator does
The common-ion effect is the drop in solubility of a slightly soluble ionic compound when the solution already contains one of its own ions. This calculator takes a salt's solubility product (Ksp), its ion stoichiometry, and the molar concentration of the shared (common) ion already dissolved, and returns the suppressed molar solubility s′ — how much of the salt can still dissolve. It also reports the pure-water solubility for comparison, so you can see exactly how strongly the common ion pushes the equilibrium back toward the undissolved solid.
How to use it
Enter the salt's Ksp (scientific notation such as 1.8e-10 is accepted). For a salt MxAy, set the cation coefficient x and the anion coefficient y — for AgCl both are 1; for CaF2 use x = 1, y = 2. Type the molar concentration of the ion that is already present, then choose whether that common ion is the cation or the anion. The result shows the suppressed solubility and how many times smaller it is than in pure water.
The formula explained
A salt dissolves according to its equilibrium, and the solubility product fixes the ion-concentration product:
$$K_{sp} = [M^{n+}]^x\,[A^{m-}]^y$$
In pure water, with x cations and y anions released per formula unit, the molar solubility is
$$s = \left( \frac{ K_{sp} }{ x^x\,y^y } \right)^{1/(x+y)}$$
When a common ion is already present at concentration C, that ion's concentration is set mainly by C rather than by the dissolving salt (Le Chatelier's principle). Assuming the added ion dominates — that is, C is much larger than the extra amount contributed by dissolving — the suppressed solubility for a common anion is
$$s' = \left( \frac{ K_{sp} }{ x^x\,C^y } \right)^{1/x}$$
and for a common cation it is
$$s' = \left( \frac{ K_{sp} }{ C^x\,y^y } \right)^{1/y}$$
Worked example
Silver chloride (AgCl, x = 1, y = 1) has Ksp = 1.8 × 10−10. In pure water its solubility is s = √(1.8 × 10−10) ≈ 1.34 × 10−5 mol/L. Now dissolve it in 0.10 M sodium chloride, so the common anion Cl− is already at 0.10 M:
$$s' = \frac{ 1.8\times10^{-10} }{ 0.10 } = 1.8\times10^{-9}\ \text{M}$$
The chloride background cuts silver chloride's solubility roughly 7,450-fold — from about 1.3 × 10−5 M down to 1.8 × 10−9 M.
Frequently asked questions
Why does adding a common ion lower solubility? By Le Chatelier's principle, raising the concentration of a product ion shifts the dissolution equilibrium back toward the undissolved solid, so less salt dissolves. The ion-concentration product must still equal Ksp, so if one ion is forced high, the other must fall.
Does this assume the common ion is much more concentrated than the dissolving salt? Yes. It uses the standard approximation that the added common-ion concentration C sets that ion's total concentration, which is accurate whenever C is far larger than the salt's suppressed solubility — the usual case. When C is comparable to the pure-water solubility, treat the answer as a close estimate.
Which value do I enter for the common-ion concentration? Use the molarity of the shared ion supplied by the other, fully soluble source. For AgCl in 0.10 M NaCl the common ion is Cl− at 0.10 M; for PbCl2 in 0.20 M Pb(NO3)2 the common ion is Pb2+ at 0.20 M.
