The common ion effect happens when an ion already present in a solution is added from another source, causing an equilibrium to shift. It matters because it controls how much of a salt dissolves, how much a weak acid ionizes, and how buffers resist pH changes. The effect is a direct application of Le Châtelier's principle, where a system responds to added particles by reducing their impact.
Chemists use it in precipitation, qualitative analysis, medicine, and biological pH control.
For a slightly soluble salt such as AgCl, adding Cl- from NaCl makes the equilibrium AgCl(s) ⇌ Ag+(aq) + Cl-(aq) shift left, so less AgCl dissolves. For a weak acid such as acetic acid, adding acetate ion from sodium acetate shifts HC2H3O2 ⇌ H+ + C2H3O2- left, lowering ionization and helping stabilize pH. In buffers, the weak acid and its conjugate base share a common ion pair that can absorb added acid or base.
A worked solubility example shows the size of the effect: if Ksp for AgCl is 1.8 x 10^-10 and [Cl-] = 0.10 M, then [Ag+] = Ksp/[Cl-] = 1.8 x 10^-9 M.
Understanding Chemistry: The Common Ion Effect
A key distinction is between equilibrium and rate. Adding a common ion does not usually make ions stop moving. Dissolving and recrystallizing continue at the particle level.
The visible change comes from unequal rates while the solution adjusts. When extra chloride is present near silver chloride, silver and chloride ions meet more often at the solid surface.
More ions attach to the crystal than leave it until a new balance is reached. This microscopic picture helps explain why a solid may appear after two clear solutions are mixed.
Use the ion product to predict the first change after mixing solutions. Multiply the current dissolved ion concentrations in the same pattern used for the solubility product. If this value is smaller than the solubility product, more solid can dissolve.
If it is larger, a precipitate forms until the value falls back to the solubility product. If the two values are equal, the system is already at equilibrium. This method is especially useful because concentrations change immediately when solutions are diluted or combined, before the equilibrium has time to respond.
The effect is central to selective precipitation in laboratory analysis. Suppose a mixture contains two positive metal ions whose salts have very different solubilities with chloride. Chloride can be added slowly so that the less soluble chloride forms a solid first.
The other metal ion remains mostly dissolved. Chemists can filter off the solid and test each part separately. This process is not perfectly sharp.
Temperature, total volume, competing reactions, and complex ions can change the result. For example, some metal ions bind to ammonia or other ligands. Binding lowers the amount of free metal ion, which can allow more solid to dissolve even when a common ion is present.
Weak acids require similar care. A weak acid in water produces only a small amount of hydrogen ion. Adding the salt of that acid supplies its conjugate base, but the salt may be highly soluble and nearly fully separated into ions.
The added conjugate base combines with hydrogen ions, leaving fewer hydrogen ions free in solution. The acid then ionizes less. In buffer calculations, use concentrations after all mixing and dilution.
The acid and conjugate base should be present in meaningful amounts. A buffer works best near the acid's pKa, where neither partner is nearly used up. Strong acid or strong base can overwhelm it because the common ion effect has a limited capacity.
Students often make errors by including solids in an equilibrium expression or by treating every listed ion as important. Pure solids have constant activity, so they do not appear in the solubility product expression. Spectator ions such as sodium usually do not enter the key equilibrium directly, though they may affect volume or ionic strength.
Keep track of charges, coefficients, and units. For salts that release unequal numbers of ions, the concentrations must be raised according to the balanced equation.
Most importantly, separate the initial mixing calculation from the final equilibrium calculation. That habit makes common ion problems much easier to organize.
Key Facts
- Common ion effect: adding an ion that is already in an equilibrium mixture shifts the equilibrium away from producing more of that ion.
- Le Châtelier's principle predicts the shift: added product ion drives the equilibrium toward reactants.
- For a salt AB(s) ⇌ A+(aq) + B-(aq), Ksp = [A+][B-].
- If AgCl(s) ⇌ Ag+ + Cl- and Ksp = 1.8 x 10^-10, then in 0.10 M NaCl, [Ag+] = 1.8 x 10^-9 M.
- For a weak acid HA ⇌ H+ + A-, Ka = [H+][A-]/[HA], so adding A- reduces acid ionization.
- Buffer equation: pH = pKa + log([A-]/[HA]), showing how a common ion controls pH.
Vocabulary
- Common ion
- A common ion is an ion that appears in two dissolved substances or in both sides of a solution equilibrium.
- Common ion effect
- The common ion effect is the decrease in solubility or ionization caused by adding an ion already present in an equilibrium.
- Solubility product
- The solubility product, Ksp, is the equilibrium constant for a slightly soluble ionic solid dissolving into its ions.
- Weak acid
- A weak acid is an acid that only partially ionizes in water, forming an equilibrium between HA, H+, and A-.
- Buffer
- A buffer is a solution containing a weak acid and its conjugate base that resists large changes in pH.
Common Mistakes to Avoid
- Ignoring the added ion concentration is wrong because the common ion often dominates the equilibrium concentration and greatly reduces solubility or ionization.
- Treating a common ion as a spectator ion is wrong because it participates in the equilibrium expression even if it comes from a soluble salt.
- Using Ksp = molar solubility for every salt is wrong because Ksp depends on ion powers and any ions already present in solution.
- Assuming adding a common ion changes the value of Ksp or Ka is wrong because equilibrium constants stay fixed at a given temperature while concentrations shift.
Practice Questions
- 1 AgCl has Ksp = 1.8 x 10^-10. Calculate the molar solubility of AgCl in 0.050 M NaCl, assuming the chloride from AgCl is small compared with 0.050 M.
- 2 A weak acid HA has Ka = 1.0 x 10^-5. A buffer contains 0.20 M HA and 0.10 M A-. Calculate the pH using pH = pKa + log([A-]/[HA]).
- 3 Explain why adding sodium acetate to acetic acid lowers the percent ionization of acetic acid but helps the solution resist a drop in pH when a small amount of strong acid is added.