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Common Ion Effect Solubility Solver

Find how much a common ion suppresses molar solubility, with the common-ion concentration entered directly or calculated from a source compound's mass, a full solubility-vs-concentration curve, and complete step-by-step working.

Common ion effect

Pick the sparingly-soluble salt, then the source of the common ion.

Formula in useKsp = (ps)ᵖ(qs+C₀)ᑫ
Result

New molar solubility (with common ion)

1.596 x 10^-3mol/L

90.1% lower than the 1.620 x 10^-2 mol/L it would have in pure water

s₀

1.620 x 10^-2

Solubility with no common ion

↓%

90.1%

Solubility suppression

[M]

1.596 x 10^-3

Cation from the salt

[X]

3.193 x 10^-3

Anion from the salt

Reading this result: Adding 1.000 x 10^-1 mol/L of the anion before the salt dissolves pushes the equilibrium backward (Le Chatelier's principle), so less of the solid salt can dissolve — the common-ion effect in action.

Solubility vs Common-Ion Concentration

More common ion pushes the equilibrium backward — this curve shows exactly how fast solubility drops (log-log scale).

Your point: 1.000 x 10^-11.596 x 10^-3 mol/L
More solubleLess solubleLess common ionMore common ionBoth axes on a log scale — molar solubility (mol/L) vs common-ion concentration (mol/L)

Step-by-Step: Common Ion Effect on Solubility

Here's exactly how this answer was calculated, one step at a time.

Given: p = 1, q = 2, Ksp = 1.7e-5, common anion = 1.000 x 10⁻¹ mol/L

  1. Step 1: Find the intrinsic solubility (no common ion)

    s₀ = [Ksp / (p^p × q^q)]^(1/(p+q)) = 1.620 x 10⁻² mol/L
  2. Step 2: Add the common-ion concentration to the anion side

    [Xᵐ⁻] = (stoichiometric term from s) + 1.000 x 10⁻¹ mol/L
  3. Step 3: Solve Ksp = [Mⁿ⁺]ᵖ[Xᵐ⁻]ᑫ for s numerically

    Solved by bisection, since the extra common-ion term makes this a non-linear equation in s (more accurate than the 's is negligible' shortcut).
  4. Step 4: Result — new molar solubility

    s = 1.596 x 10⁻³ mol/L
  5. Step 5: Compare to the intrinsic solubility

    Solubility dropped by 90.1%, from 1.620 x 10⁻² mol/L down to 1.596 x 10⁻³ mol/L

New molar solubility (s):

1.596 x 10⁻³ mol/L

Common Ion Effect Solubility Solver: See Exactly How Much a Common Ion Suppresses Solubility

This free calculator answers one of the trickiest questions in a general chemistry solubility unit: how much less of a salt dissolves once one of its own ions is already floating around in the water? It's built for high school and college chemistry students, AP and IB Chemistry prep, MCAT review, and chemistry teachers who need a fast way to check the common-ion effect on any sparingly-soluble salt — with the option to type in a common-ion concentration directly, or calculate it straight from the mass of a source compound like table salt.

In simple words, the common ion effect is what happens when a salt is added to a solution that already contains one of its own ions from a different, fully-soluble source. Instead of dissolving as much as it would in pure water, the salt dissolves less — sometimes dramatically less. This calculator does every step of that math automatically: it works out the salt's normal solubility in pure water, then solves for the new, suppressed solubility once the common ion is added, and shows exactly how big the drop is as a percentage.

Beyond the basic number, this tool adds features a textbook page can't: an option to compute the common-ion concentration directly from a real source compound's mass and volume, a full solubility-vs-concentration curve so the suppression trend is visible at a glance, and a complete, exact numerical solution rather than the rougher approximation most textbooks rely on.

What Is the Common Ion Effect?

The common ion effect is a direct consequence of Le Chatelier's principle applied to a solubility equilibrium. When a sparingly-soluble salt MpXq dissolves in water, it sits at equilibrium: MpXq(s) ⇌ p M(ion) + q X(ion). If a completely different, fully-soluble compound is dissolved in that same water first — one that happens to release the same M or X ion — the concentration of that ion is now higher than it would be from the salt alone.

Because the system is at equilibrium, adding more of one of the products (the common ion) pushes the reaction backward, toward the solid, undissolved form. In plain terms: the salt 'sees' that there's already plenty of its own ion around, so less of the solid needs to dissolve to reach the same equilibrium balance. The result is a lower molar solubility than the salt would have in pure water.

A classic textbook example is dissolving silver chloride (AgCl) in a solution that already contains dissolved sodium chloride (NaCl). NaCl is fully soluble, so it immediately supplies extra Cl⁻ ions. When solid AgCl is then added, its own Cl⁻ contribution has to compete with the Cl⁻ already present, so far less AgCl ends up dissolving than if it had been added to pure water.

The Common Ion Effect Formula

Starting from the same Ksp expression used for any solubility problem, Ksp = [M]^p × [X]^q, the common ion effect changes only one thing: whichever ion is shared with the outside source gets an extra term added to it. If the common ion is the anion, the equation becomes Ksp = (p·s)^p × (q·s + C₀)^q, where C₀ is the concentration of the common ion already present before the salt dissolves. If the common ion is the cation instead, the extra term moves to the other side: Ksp = (p·s + C₀)^p × (q·s)^q.

This equation can't usually be rearranged into a clean algebraic formula for s, the way the plain Ksp-to-solubility formula can — the extra C₀ term makes it non-linear. Many textbooks get around this with a shortcut: assuming s is negligible compared to C₀, so the ion term simplifies to just C₀. That shortcut works fine when C₀ is much larger than the intrinsic solubility, but it can give a noticeably wrong answer otherwise. This calculator skips the shortcut entirely and solves the exact equation using a numerical method (bisection), so the answer stays accurate across the full range of common-ion concentrations.

Worked Example: Common Ion Suppression in Action

Take lead(II) chloride, PbCl2, a 1:2 salt with Ksp = 1.7 × 10⁻⁵. In pure water, its intrinsic solubility works out to about 1.62 × 10⁻² mol/L. Now suppose that same PbCl2 is added to a solution that already contains 0.1 mol/L of Cl⁻ from dissolved NaCl.

Because Cl⁻ is the common ion here (the anion side), the equilibrium expression becomes Ksp = (s)(2s + 0.1)². Solving this numerically gives a new solubility of roughly 1.7 × 10⁻³ mol/L — around 90% lower than the 1.62 × 10⁻² mol/L it would have in pure water. That's the common-ion effect in a single number: adding a Cl⁻-supplying compound shrinks how much PbCl2 can dissolve to a small fraction of its normal amount.

This is exactly the kind of calculation this solver automates — pick the salt, tell it which ion is shared, either type in the concentration or calculate it from a source compound's mass, and it returns the suppressed solubility along with the percent drop.

Calculating a Common-Ion Concentration From a Source Compound

In a real lab setting, the common-ion concentration usually isn't handed to you directly — it comes from weighing out a mass of some other, fully-soluble compound and dissolving it into a known volume. This calculator's 'from a source compound' mode handles that conversion automatically: enter the mass dissolved, the compound's molar mass (or pick a preset like NaCl, CaCl2, or AgNO3), how many of the target ion each formula unit releases, and the total solution volume, and the resulting concentration feeds straight into the common-ion equation above.

This step trips a lot of students up specifically because it's easy to forget that some source compounds release more than one of the common ion per formula unit — CaCl2, for example, releases two Cl⁻ ions per formula unit that dissolves, not one. This calculator's source-compound presets already account for that, so the ion count is never left out by mistake.

Reading the Suppression Curve

The solubility-vs-concentration curve on this page plots exactly how molar solubility changes as more and more of the common ion is added, on a log-log scale so both very small and very large concentrations stay visible on the same chart. At the far left, where almost no common ion is present, solubility sits close to its intrinsic, pure-water value. Moving right along the curve, solubility steadily falls as the common-ion concentration climbs — sharply at first, then more gradually, since the relationship between C₀ and s is a root, not a straight line.

This kind of visual is genuinely useful beyond just checking a single homework answer: it shows at a glance whether a particular common-ion concentration is in the steep, high-impact part of the curve, or the flatter region where adding even more of the ion barely changes solubility any further.

Common Mistakes When Working With the Common Ion Effect

The most frequent mistake is applying the 's is negligible compared to C₀' shortcut without checking whether it actually holds. That approximation is reasonable when C₀ is at least 100 times larger than the intrinsic solubility s₀, but for more soluble salts or smaller common-ion concentrations, it can give a meaningfully wrong answer. This calculator solves the exact non-linear equation instead, so this shortcut error never shows up in the result.

A second common mistake is forgetting to multiply by the stoichiometric coefficient before adding the common-ion term — for a 1:2 salt like PbCl2, the equation needs (2s + C₀)², not (s + C₀)². Skipping that factor of 2 changes the answer significantly once the equation is solved.

A third mistake, specific to the 'source compound' scenario, is forgetting that some source compounds release more than one common ion per formula unit (like CaCl2 releasing two Cl⁻ ions), which under-counts the actual common-ion concentration if missed.

Real-World Uses of the Common Ion Effect

The common ion effect isn't just an exam topic — it's used deliberately in real chemistry and industry. In qualitative analysis, chemists add a reagent that supplies a common ion on purpose, to selectively force one dissolved metal ion to precipitate out of a mixture while leaving others in solution, based on which compound has a small enough Ksp to be pushed past saturation first.

In industrial and municipal water treatment, the common-ion effect is used to intentionally precipitate unwanted dissolved metals like lead or mercury out of contaminated water by adding a source of a matching ion. In geology, the same principle governs how the presence of dissolved salts in seawater or brine affects how much calcium carbonate or gypsum can dissolve or precipitate in natural mineral deposits.

Common Ion Effect Solver: Quick Reference Summary

For a salt MpXq with a common ion already present at concentration C₀, the modified solubility equation is Ksp = (p·s)^p × (q·s + C₀)^q if the common ion is the anion, or Ksp = (p·s + C₀)^p × (q·s)^q if it's the cation. This calculator solves that equation exactly by bisection rather than relying on the 's is negligible' shortcut.

The common-ion concentration itself can either be entered directly, or calculated from a source compound's mass using moles = mass / molar mass, followed by concentration = (moles × ions released per formula unit) / volume.

This free calculator is intended to support learning, homework checking, and everyday solubility-equilibrium questions. Ksp values vary slightly between textbooks and reference sources and change with temperature and ionic strength, so for lab reports, research, or any safety-critical application, always confirm the specific Ksp value with your course materials or a peer-reviewed reference source.

Frequently Asked Questions

What is the common ion effect?

The common ion effect is the decrease in a salt's solubility that happens when one of its own ions is already present in solution from a different, fully-soluble source. It follows from Le Chatelier's principle, since the extra ion pushes the dissolution equilibrium back toward the undissolved solid.

What is the formula for the common ion effect?

Starting from Ksp = [M]^p × [X]^q, add the existing common-ion concentration C₀ to whichever side it belongs to: Ksp = (p·s)^p × (q·s + C₀)^q for a common anion, or Ksp = (p·s + C₀)^p × (q·s)^q for a common cation, then solve for s.

Why does a common ion decrease solubility?

Because the dissolution reaction is at equilibrium, adding more of one of the dissolved products (the common ion) shifts the equilibrium backward toward the solid form, following Le Chatelier's principle — so less of the salt needs to dissolve to reach the same balance.

How do you calculate a common-ion concentration from a source compound?

Convert the source compound's mass to moles using moles = mass / molar mass, then multiply by how many of the common ion each formula unit releases, and divide by the solution volume: concentration = (moles × ions per formula unit) / volume.

Is the 's is negligible' shortcut for the common ion effect always accurate?

No. It works well when the common-ion concentration is much larger (roughly 100 times or more) than the salt's intrinsic solubility, but it can give a noticeably wrong answer otherwise. Solving the exact non-linear equation, as this calculator does, avoids that error.

Can the common ion effect completely stop a salt from dissolving?

No, not entirely — a small amount will always dissolve to satisfy the equilibrium, since Ksp is still greater than zero. But the common ion effect can suppress solubility by well over 90% for a large enough common-ion concentration.

Is the common ion effect the same as the reaction quotient (Q vs Ksp) precipitation check?

They're related but different. The common ion effect describes a new equilibrium solubility once an ion is already present before dissolution starts. The Q vs Ksp check instead predicts whether mixing two solutions right now will cause a precipitate to form.

Does the common ion effect apply to any salt?

It applies specifically to sparingly-soluble ionic compounds governed by a solubility product constant (Ksp) — it doesn't apply to fully soluble salts like NaCl or KNO3, which don't have a meaningful solubility limit under normal conditions.