Common Ion Effect Calculator

Background.

This calculator answers a question the plain solubility product calculator cannot: how much less of a sparingly soluble salt dissolves when one of its own ions is already in the water. That is the common ion effect, and it is Le Châtelier's principle applied to a dissolution equilibrium — flood the solution with one product and the equilibrium retreats, leaving less solid dissolved.

The arithmetic changes shape as well as value. In pure water, Ksp = xˣ·yʸ·s^(x+y) rearranges into a clean root that you can evaluate by hand. With a background concentration C of the shared ion the equation becomes (C_A + x·s)ˣ·(C_B + y·s)ʸ = Ksp, which for anything beyond the simplest case is a cubic or higher and has no closed-form root. This page solves it numerically, by bisection on a function that is strictly increasing in s, so the answer is exact to the working precision rather than relying on the usual 'assume s is negligible' shortcut. That shortcut is checked here rather than assumed: for a 1:1 salt with s far below C, the exact solution converges on the textbook approximation s ≈ Ksp/C to six significant figures.

The worked default is fluorite, CaF₂, using the U.S. Geological Survey's log Ksp of −10.6, in water that already carries 0.0020 mol/L of calcium — that is 80.2 mg/L, a moderately hard water. Solubility falls from 184.49 µmol/L in pure water to 55.28 µmol/L, a suppression factor of 3.34 and a 70.0 % reduction. In terms of the ion people actually measure, dissolved fluoride drops from 7.010 mg/L to 2.100 mg/L. For scale, the US EPA sets both the maximum contaminant level and its goal for fluoride at 4.0 mg/L, so the same mineral gives water above the US limit in a calcium-poor aquifer and below it in a moderately hard one. That is a statement about a regulatory threshold with a named authority and revision date, not a health assessment, and this page makes no health claims of any kind.

Which ion is the common one matters enormously, and this is the part most treatments gloss over. The shared ion enters the Ksp expression raised to its stoichiometric power, so for CaF₂ the fluoride ion appears squared while calcium appears to the first power. At the same 0.010 mol/L background, calcium suppresses fluorite's solubility 7.4-fold; fluoride suppresses it 735-fold. A hundredfold difference, from a single exponent.

Now the limits, which belong here rather than in a collapsed FAQ, because they change how far the number should be trusted. First, the chemistry: this is an ideal dilute-solution calculation with activity coefficients set to one. The very act of adding a common ion raises the ionic strength, which lowers activity coefficients below one, which weakens the real suppression. The number here is therefore a lower bound on the true solubility. Second, the world: the ideal model describes a closed, single-salt system, and real groundwater is neither. A global analysis of fluoride in groundwater (Podgorski and Berg, Nature Communications 13:4232, 2022) finds that fluoride is indeed strongly negatively correlated with calcium up to about 80 mg/L — and then the trend reverses and climbs again to around 400 mg/L, because bicarbonate reacts with fluorite to release fluoride and because evaporative concentration raises both ions together in arid basins. The default here sits at that 80 mg/L inflection deliberately. Push the common-ion concentration much past it and you are computing a number the field data do not support.

Finally, this page assumes the common ion arrives as a spectator from a fully dissociated soluble salt, that no complex forms between the two ions, and that pH is not doing anything to either of them. Where the anion is the conjugate base of a weak acid, as fluoride is, acidic conditions convert it to the undissociated acid and undo the suppression entirely.

What is common ion effect calculator?

The common ion effect is the reduction in the solubility of an ionic solid caused by the presence of a second, soluble compound that shares one of its ions. Silver chloride is less soluble in sodium chloride solution than in pure water; calcium fluoride is less soluble in a calcium-rich water than in a calcium-free one. The cause is Le Châtelier's principle: the dissolution AB(s) ⇌ A⁺ + B⁻ is an equilibrium, and raising the concentration of a product shifts it back toward the solid.

The quantitative statement is simply that Ksp does not change. Ksp is a function of the salt and the temperature only, so if one ion's concentration is forced up, the other's must come down to keep the product constant — and since the second ion comes only from the dissolving solid, less solid dissolves. The bookkeeping is where care is needed: the total concentration of the shared ion is the background C plus whatever the dissolving salt contributes, which is x·s or y·s depending on which ion it is.

The standard textbook treatment assumes s is negligible compared with C, which reduces the problem to a division. For a 1:1 salt that gives s ≈ Ksp/C. The approximation is excellent when the suppression is strong — for barium sulfate in 0.010 mol/L barium it is accurate to better than one part in a hundred thousand — and it degrades as C approaches s₀, where the correct answer must come from solving the full polynomial. This calculator always solves the full equation, so the two agree where the approximation is valid and the calculator is right where it is not.

The practical uses are everywhere. Gravimetric analysis deliberately adds a slight excess of precipitating agent to drive recovery toward completion. Water treatment exploits it in the opposite direction, dosing lime to precipitate fluoride or phosphate. Scale control in pipes and boilers is a common-ion problem in reverse: the ions accumulate, the product exceeds Ksp, and the solid appears. And in hydrogeology the calcium content of an aquifer is one of the primary controls on how much fluoride the water carries.

One thing the common ion effect is not: it is not the same as the salt effect, sometimes called the diverse ion or inert ion effect. Adding a salt with NO ion in common — sodium nitrate to a silver chloride solution, say — raises the ionic strength, lowers the activity coefficients and INCREASES solubility slightly. The two effects run in opposite directions, and in a real solution containing a common ion both are happening at once. This calculator models only the first.

How to use this calculator.

1. 01Write the dissolution equation and read off the subscripts. CaF₂ → Ca²⁺ + 2F⁻ gives x = 1 and y = 2. Ag₂CrO₄ → 2Ag⁺ + CrO₄²⁻ gives x = 2 and y = 1.
2. 02Enter log₁₀ Ksp for the salt at 25 °C. If your source prints Ksp itself, take its base-10 logarithm first.
3. 03Choose which ion is already in the water. Get this right: for a salt with unequal subscripts the two choices give wildly different suppressions.
4. 04Enter the background concentration of that ion in mol/L. This is the analytical concentration supplied by the other, soluble salt — not the total including what dissolves.
5. 05Read the suppression factor and the percent reduction together with the raw solubility. The factor is usually the more memorable number.
6. 06Look at the equilibrium anion (or cation) concentration if you care about a specific ion rather than the salt. For fluorite suppressed by calcium, the free fluoride is what a water analysis would report.
7. 07Sanity-check against the pure-water baseline shown alongside. If the suppression factor is close to 1, the background concentration is too small to matter and the salt is behaving as if it were in pure water.
8. 08Remember which way the neglected effects push. Ignoring activity coefficients makes this calculation predict MORE suppression than reality delivers, so treat the answer as a lower bound on solubility.

The formula.

SET-UP. For AₓB_y(s) ⇌ x Aⁿ⁺ + y Bᵐ⁻ dissolving into water that already contains C_A of the cation and C_B of the anion, the equilibrium concentrations are [A] = C_A + x·s and [B] = C_B + y·s, so

Ksp = (C_A + x·s)ˣ · (C_B + y·s)ʸ

Only one of C_A and C_B is non-zero here; the other is set to zero by the ion you select.

WHY IT IS SOLVED NUMERICALLY. With a background term inside the bracket the equation no longer factors — for fluorite with common calcium it is a cubic in s — so it is solved by bisection. The left-hand side is strictly increasing in s for s ≥ 0, and adding a common ion can never raise solubility, so s must lie in [0, s₀] where s₀ is the pure-water value. That gives a guaranteed bracket with the residual negative at one end and positive at the other, and 200 halvings drive the interval far below the working precision. Because the function is monotonic there is exactly one root and no possibility of converging on a spurious one.

PURE-WATER BASELINE. s₀ = (Ksp / (xˣ · yʸ))^(1/(x+y)), in closed form — the same expression the solubility product calculator uses, so the two pages agree exactly when the background is set to zero.

WORKED. Fluorite, log₁₀Ksp = −10.6 so Ksp = 2.5119 × 10⁻¹¹, with x = 1, y = 2 and 0.0020 mol/L of background Ca²⁺. Solving (0.0020 + s)(2s)² = 2.5119 × 10⁻¹¹ gives s = 5.5276 × 10⁻⁵ mol/L, that is 55.276 µmol/L. The pure-water value is 184.493 µmol/L, so the suppression factor is 184.493/55.276 = 3.3377 and the reduction is 70.04 %. At M = 78.0748 g/mol that is 4.3157 mg/L of dissolved CaF₂; the equilibrium calcium is 2055.28 µmol/L and the equilibrium fluoride is 110.55 µmol/L, which is 2.1003 mg/L of fluoride against 7.0101 mg/L in calcium-free water.

CHECKING THE TEXTBOOK SHORTCUT. When s is far below C the bracket (C + x·s) is essentially C, and for a 1:1 salt the equation collapses to s ≈ Ksp/C. Barium sulfate (log₁₀Ksp = −9.97) in 0.010 mol/L barium: the shortcut gives 0.010715 µmol/L and the full numerical solution gives 0.010715 µmol/L — agreement to six significant figures. The shortcut is not used in the calculation; it is used to verify it.

WHY THE EXPONENT DOMINATES. The shared ion enters raised to its own subscript. For CaF₂ at the same 0.010 mol/L background, common calcium gives a suppression factor of 7.37 while common fluoride gives 734.55 — a hundredfold difference caused only by y = 2.

CONVENTIONS. All quantities at 25 °C, 0.1 MPa and zero ionic strength. Ksp, the suppression factor and the subscripts are dimensionless; solubilities and ion concentrations are in µmol/L; mass solubility is in mg/L; the background concentration is in mol/L. The suppression factor is s₀/s and is at least 1 by construction; the percent reduction lies in [0, 100).

ROUNDING STAGE. Nothing is rounded at an intermediate step, including inside the 200-iteration bisection. Rounding happens only when a result is returned, at ten decimal places, falling back to twelve significant digits where ten decimal places would collapse a heavily suppressed but genuinely non-zero solubility to exactly zero.

INVALID DOMAIN. A background concentration of exactly zero is legal and returns the pure-water answer with a suppression factor of 1. A negative background, a background above 20 mol/L, fractional or out-of-range subscripts, a non-positive molar mass and a log₁₀Ksp outside −60 to +5 each raise an error naming the field. If suppression drives the solubility below the representable range the calculator raises an error rather than reporting an infinite suppression factor.

A worked example.

Frequently asked questions.

References& sources.

1. [1]Ball, J. W.; Nordstrom, D. K. "User's Manual for WATEQ4F, with Revised Thermodynamic Data Base and Test Cases for Calculating Speciation of Major, Trace, and Redox Elements in Natural Waters." U.S. Geological Survey Open-File Report 91-183 (1991). Database entry 62: Fluorite, CaF₂ = Ca²⁺ + 2F⁻, log K = −10.6 ± 0.02, with the analytic form log K = 66.348 − 4298.2/T − 25.271·log₁₀T (−10.5997 at 298.15 K). Entry 144: Barite, BaSO₄ = Ba²⁺ + SO₄²⁻, log K = −9.97 ± 0.02, used as the closed-form test fixture. Open access, USGS-hosted; retrieved 2026-07-29.
2. [2]SECOND, INDEPENDENT AUTHORITY (BUILD-BRIEF §9.1) — empirical rather than thermodynamic. Podgorski, J.; Berg, M. "Global analysis and prediction of fluoride in groundwater." Nature Communications 13, 4232 (2022), doi:10.1038/s41467-022-31940-x. Open access, peer-reviewed. Confirms the mechanism modelled here — "dissolved calcium can bind with fluoride and remove it from dissolution to again form fluorite" — and confirms the direction at low calcium, while reporting that "the prevalence of fluoride drops precipitously for calcium concentrations up to about 80 mg/L and then rises sharply again to about 400 mg/L". That reversal is why the default background concentration on this page is 0.0020 mol/L (80.2 mg/L) rather than the textbook 0.010 mol/L. Also the source of the WHO 1.5 mg/L guideline value quoted below, which is therefore SECOND-HAND; the WHO Guidelines themselves were not retrieved. Retrieved 2026-07-29.
3. [3]phreeqc.dat, the thermodynamic database distributed with PHREEQC version 3 (U.S. Geological Survey; Parkhurst & Appelo, USGS Techniques and Methods 6-A43, 2013). PHASES block: Fluorite, CaF₂ = Ca⁺² + 2 F⁻, log_k −10.61, attributed to Strübel (1965) and Henry (2018) — an independent data selection that agrees with WATEQ4F inside its stated ±0.02. Barite is listed at −9.88 in the same block, a 0.09 spread against WATEQ4F's −9.97. Open access; retrieved 2026-07-29.
4. [4]U.S. Environmental Protection Agency, National Primary Drinking Water Regulations, inorganic chemicals table: fluoride MCL = 4.0 mg/L and MCLG = 4.0 mg/L. Cited only to give the fluorite worked example a regulatory scale, with a named authority and a revision date; no health, dental or toxicological interpretation is offered anywhere on this page. Page last updated 1 December 2025; retrieved first-hand 2026-07-29.
5. [5]Commission on Isotopic Abundances and Atomic Weights (CIAAW), IUPAC — Standard Atomic Weights, 2021 table incorporating the 2024 revisions. Ca = 40.078(4); F = 18.998403162(5); Ba = 137.327(7); S = 32.06; O = 15.999. Gives M(CaF₂) = 78.0748 g/mol and M(BaSO₄) = 233.383 g/mol, and the 80.2 mg/L equivalent of the default 0.0020 mol/L calcium. Open access; retrieved 2026-07-29.