The Common-Ion Effect on Solubility
The common-ion effect is the decrease in the solubility of a sparingly soluble salt when a solution already contains one of its ions, as predicted by Le Châtelier’s principle and calculated with Ksp. In brief: AgCl dissolves to 1.3 × 10−5 M in pure water, but only 1.8 × 10−9 M in 0.10 M NaCl, roughly 7400 times less.
If you add chloride ions to a saturated solution of silver chloride, more solid precipitates. This is a direct consequence of the equilibrium AgCl(s) ⇌ Ag+ + Cl−: the extra Cl− shifts it to the left. The effect is used to make precipitations more complete and to reduce losses of product in analytical work. This article explains the effect, shows how to calculate the solubility in a solution with a common ion and gives tables and examples.
What is the common-ion effect?
A common ion is an ion that is present in the solution from another source and is also produced by the dissolving of the salt. In a saturated solution of AgCl, [Ag+][Cl−] = Ksp. If Cl− is added from NaCl, [Cl−] rises, and the product exceeds Ksp (Q > Ksp), so AgCl precipitates until the product has fallen back to Ksp. At the new equilibrium, [Cl−] is higher than before, and [Ag+], which is a measure of the solubility of AgCl, is lower. The Ksp itself does not change; only the solubility does.
How do you calculate the solubility?
- Write the Ksp expression.
- Set up an ICE table. The initial concentration of the common ion is the concentration from the other source.
- Let s be the molar solubility: the ion that comes only from the salt has the concentration s (times its coefficient), while the common ion has the concentration (initial + its contribution).
- Since Ksp is small, s is much smaller than the initial concentration of the common ion, so it can be neglected in the sum.
- Solve for s and check that it is small compared with the initial concentration (the 5 % rule).
Solubility of AgCl in NaCl solutions
| [Cl−] from NaCl (M) | Solubility s = [Ag+] (M) | Compared with pure water |
|---|---|---|
| 0 (pure water) | 1.3 × 10−5 | 1 (reference) |
| 0.0010 | 1.8 × 10−7 | about 74 times lower |
| 0.010 | 1.8 × 10−8 | about 740 times lower |
| 0.10 | 1.8 × 10−9 | about 7400 times lower |
Each tenfold increase in the chloride concentration decreases the solubility by a factor of ten. At the lowest concentration in the table, the approximation is already good: the chloride from the dissolved AgCl (1.8 × 10−7 M) is only 0.02 % of the added chloride.
Worked examples
Example 1: silver chloride in sodium chloride
What is the solubility of AgCl in 0.10 M NaCl? Let s = [Ag+]. Then [Cl−] = 0.10 + s ≈ 0.10. Ksp = s × 0.10 = 1.8 × 10−10, so s = 1.8 × 10−9 M. The assumption is valid, since s is much less than 0.10.
Example 2: lead(II) iodide in potassium iodide
What is the solubility of PbI2 (Ksp = 9.8 × 10−9) in 0.10 M KI? [Pb2+] = s and [I−] = 0.10 + 2s ≈ 0.10. Ksp = s × (0.10)2, so s = 9.8 × 10−9 / 0.010 = 9.8 × 10−7 M. In pure water the solubility is 1.35 × 10−3 M, which is about 1400 times higher.
Example 3: calcium fluoride in sodium fluoride
What is the solubility of CaF2 (Ksp = 3.9 × 10−11) in 0.010 M NaF? [Ca2+] = s and [F−] ≈ 0.010. s = 3.9 × 10−11 / (0.010)2 = 3.9 × 10−7 M, against 2.1 × 10−4 M in pure water.
Example 4: a common cation
What is the solubility of BaSO4 (Ksp = 1.1 × 10−10) in 0.050 M Na2SO4? [SO42−] ≈ 0.050 M, so s = [Ba2+] = 1.1 × 10−10 / 0.050 = 2.2 × 10−9 M. In pure water, s = 1.0 × 10−5 M.
Example 5: mass lost when washing a precipitate
In gravimetric analysis, a precipitate of BaSO4 is washed with a dilute solution of H2SO4 instead of pure water. Washing with 100 mL of pure water can dissolve about 1.05 × 10−5 mol/L × 0.100 L × 233.4 g/mol = 2.4 × 10−4 g, or 0.24 mg. With 0.010 M sulfate, the solubility is s = 1.1 × 10−10 / 0.010 = 1.1 × 10−8 mol/L, so only 1.1 × 10−9 mol × 233.4 g/mol = 2.6 × 10−7 g (0.26 µg) can dissolve. The possible loss is smaller by a factor of about 1000.
When does the effect reverse?
At a very high concentration of the common ion, the solubility may increase again, because complex ions form. For AgCl, a large excess of Cl− gives soluble complexes such as AgCl2−, and the solubility passes through a minimum at about 10−2–10−3 M chloride. Salts of weak acids are also more soluble in acid, which is a different effect from the common ion effect. And at high total ionic strength, a “salt effect” can raise the solubility slightly by lowering the activity coefficients.
Where is the effect used?
- Gravimetric analysis: an excess of the precipitating agent makes the precipitation of the analyte more complete.
- Washing precipitates: a dilute solution of a common ion reduces the loss.
- Salting out: adding NaCl to a solution of soap or protein decreases the solubility.
- Water treatment and industry: excess lime or soda ash drives the removal of Ca2+.
- Buffers and acids: the same logic explains why the ionization of acetic acid is suppressed by sodium acetate.
Who explained the effect?
The reduction of solubility by a common ion was observed by chemists in the 19th century, and the quantitative explanation followed from the solubility product concept of Nernst (1889) and the work of Ostwald. It was one of the first successful applications of the theory of ionic equilibria to analytical chemistry, and the principle is part of the standard scheme of qualitative analysis.
Which mistakes are common?
- Neglecting the coefficient of the common ion: in PbI2, the iodide contributes with its square.
- Adding s to the common ion concentration when it is not negligible: use the 5 % rule to decide.
- Thinking that Ksp changes: only the solubility changes.
- Forgetting to use the total concentration of the common ion: it can come from more than one source.
- Assuming that more common ion is always better: complex formation can raise the solubility again.