17.5 Common Ion Effect and Precipitation
PRE-HEALTH PREP
Common Ion Effect and Predicting Precipitation
When Qsp Exceeds Ksp
General Chemistry 17.5
The common ion effect describes how the presence of a shared ion from a strong electrolyte suppresses the solubility of a sparingly soluble salt, a direct application of Le Chatelier's Principle to dissolution equilibria. Precipitation can be predicted by calculating the reaction quotient Qsp and comparing it to Ksp—when Qsp exceeds Ksp, the solution is supersaturated and a solid forms. Worked examples demonstrate ICE table modifications for common ion problems and the critical role of dilution calculations (M1V1 = M2V2) when solutions are mixed.
What Is the Common Ion Effect?
The common ion effect occurs when a strong electrolyte introduces an ion that already participates in a solubility equilibrium, reducing the molar solubility of a sparingly soluble salt. This is a direct application of Le Chatelier's Principle: adding an excess product ion drives the dissolution equilibrium back toward the undissolved solid, lowering solubility. Understanding the common ion effect is essential for any general chemistry course covering solubility equilibria and Ksp.
Consider silver chloride as an example. AgCl dissociates in water to produce Ag+ and Cl− ions. Its Ksp is quite small, so it dissolves only sparingly. But what happens if you introduce a strong electrolyte—one that dissociates completely—that contributes either silver ions or chloride ions into the solution?
According to Le Chatelier's Principle, adding extra product ions shifts the equilibrium back toward the solid. Less solid dissolves, which means lower molar solubility. That is the common ion effect: having either Ag+ or Cl− already present in solution from another source reduces the solubility of AgCl.
Molar Solubility of AgCl in 0.1 M HCl
What is the molar solubility of AgCl in 0.1 M HCl, given that Ksp = 1.8 × 10−10? This is a classic common ion effect problem solved with an ICE table.
In pure water, before any AgCl dissociates, neither Ag+ nor Cl− is present. You set up the ICE table with both initial concentrations at zero, let each increase by x, and solve:
Pure Water (Review from Previous Lesson)
Ksp = [Ag+][Cl−] = (x)(x) = x2 = 1.8 × 10−10
x = 1.3 × 10−5 M (molar solubility in pure water)
Now here is the critical change. HCl is a strong acid—a strong electrolyte that dissociates completely. When HCl dissociates, it produces H+ ions and Cl− ions. That chloride is the common ion. Since HCl dissociates in a one-to-one ratio, 0.1 M HCl produces 0.1 M Cl−.
Before any AgCl even begins to dissolve, there is already 0.1 M chloride in the solution. The ICE table now looks different: the initial chloride concentration is 0.1 rather than zero. As AgCl dissolves, both [Ag+] and [Cl−] increase by x, giving equilibrium concentrations of x for silver and (0.1 + x) for chloride.
Example: AgCl in 0.1 M HCl (Common Ion Effect ICE Table)
Ksp = [Ag+][Cl−] = (x)(0.1 + x) = 1.8 × 10−10
Because Ksp is extremely small, x will be tiny. Adding a very small number to 0.1 still gives approximately 0.1, so we can simplify:
1.8 × 10−10 = (0.1)(x)
x = 1.8 × 10−9 M
Compare the two results. In pure water, the molar solubility was 1.3 × 10−5 M. With the common ion present, it plummets to 1.8 × 10−9 M—a roughly 10,000-fold reduction in solubility. That is the power of the common ion effect.
Tip: The "small x" approximation works the same way here as it did with Ka and Kb problems. When the equilibrium constant is very small, x will be negligible compared to the initial concentration. Verify by checking that x is less than 5% of the value it was added to or subtracted from.
Common Mistake: In pure-water solubility problems, you may have used a shortcut where the Ksp expression simplifies to a neat power of x. That shortcut only works when no common ion is present. With a common ion, one of your initial concentrations is nonzero, and you must account for it in the ICE table.
Molar Solubility of Ag₂S in 0.001 M AgNO₃
What is the molar solubility of Ag₂S in 0.001 M AgNO3, given that Ksp = 6.0 × 10−51? This example shows how the common ion effect applies when the salt contains more than one cation per formula unit.
First, identify the strong electrolyte and the common ion. Silver nitrate is a soluble ionic compound—recall that all nitrate salts are soluble, making AgNO3 a strong electrolyte. It dissociates completely in a one-to-one ratio, producing 0.001 M Ag+.
Pro Tip: Whenever you recognize a problem as part of solubility equilibria but are unsure where to begin, start by writing the dissolution reaction and the Ksp expression. It is the best possible launching point for nearly any problem in this chapter.
The dissolution of Ag2S produces two silver ions and one sulfide ion:
Ag₂S(s) → 2 Ag+(aq) + S2−(aq)
Ksp = [Ag+]2[S2−]
In the ICE table, the initial silver concentration is 0.001 M (from AgNO3) rather than zero. As Ag2S dissolves by an amount x, silver increases by 2x and sulfide increases by x. At equilibrium, [Ag+] = 0.001 + 2x and [S2−] = x.
Example: Ag₂S in 0.001 M AgNO₃ (Common Ion ICE Table)
Ksp = (0.001 + 2x)2(x) = 6.0 × 10−51
With a Ksp of 10−51, x will be astronomically small. Even doubling it produces a value far smaller than 0.001, so the approximation holds easily:
6.0 × 10−51 = (0.001)2(x) = (1 × 10−6)(x)
x = 6.0 × 10−45 M
This molar solubility is dramatically smaller than the value obtained in pure water, once again confirming how substantially a common ion suppresses solubility.
Note: Without the small-x approximation, this problem produces a cubic equation. You will never need to solve a cubic on a general chemistry exam. The approximation is not a shortcut—it is the expected approach when K is very small.
3 Problem Types: Common Ion Effect Summary
The common ion effect is a direct consequence of Le Chatelier's Principle applied to solubility equilibria. Any strong electrolyte that introduces an ion already present in the dissolution reaction will shift the equilibrium toward the solid, reducing solubility.
Mathematically, the only change from a standard molar solubility calculation is in the ICE table: instead of starting with zero for all ion concentrations, one of them begins at the concentration contributed by the strong electrolyte. Everything else—setting up the Ksp expression, applying the small-x approximation, solving for x—proceeds the same way.
| Problem Type | ICE Table Setup | Key Difference |
|---|---|---|
| Solubility in pure water | All initial ion concentrations = 0 | Standard Ksp calculation |
| Solubility with common ion | One initial concentration ≠ 0 | Common ion from strong electrolyte; lower molar solubility |
| Precipitation (Qsp problems) | No ICE table needed | Plug concentrations directly into Qsp expression |
Predicting Precipitation with Qsp
Predicting precipitation in general chemistry relies on the reaction quotient Qsp compared to the solubility product Ksp. Precipitation is the formation of a solid within a solution—picture a snow globe: you shake it and tiny flakes appear throughout the liquid. In chemistry, you might mix two clear solutions and watch a milky cloudiness develop. If you wait long enough, that cloudiness often settles into a solid pellet at the bottom of the container. That is precipitation—ions in an aqueous solution coming together to form an insoluble solid.
Precipitation is actually the reverse of the dissolution reactions we have been writing. Instead of a solid breaking apart into ions, ions in solution combine to form a solid. The question becomes: under what conditions does this reverse reaction occur? The answer is always found by calculating Qsp and comparing it to Ksp.
Maximum Hydroxide Concentration in a Mg2+ Solution
What is the maximum concentration of OH− that can exist in a solution where [Mg2+] = 0.01 M? The Ksp of magnesium hydroxide is 1.6 × 10−12. This type of problem asks you to use Ksp as an upper limit—exceeding it triggers precipitation.
Start by writing the dissolution reaction and Ksp expression for the salt whose Ksp was provided:
Mg(OH)₂(s) → Mg2+(aq) + 2 OH−(aq)
Ksp = [Mg2+][OH−]2
Introducing Qsp: The Reaction Quotient for Solubility
The expression Qsp (the reaction quotient for solubility) looks identical to Ksp, but with one crucial distinction. Ksp uses ion concentrations measured at equilibrium—the maximum amounts that can coexist in solution. Qsp uses the current concentrations, which may or may not be at equilibrium. Comparing Qsp to Ksp tells you exactly what the system will do next.
No ICE Table Needed for Precipitation Problems
Notice something important: in this problem, there is no solid magnesium hydroxide dissolving. You simply have a solution containing Mg2+ ions, and you want to know the maximum hydroxide concentration before a precipitate forms. There is no solid to set up an ICE table around, no dissolution happening, and therefore no x to solve for.
Key Distinction: ICE tables and the variable x (molar solubility) apply when a solid is dissolving—when the reaction proceeds in the forward direction. Precipitation problems deal with the reverse direction, where ions might combine to form a solid. No ICE table is needed; you simply plug known concentrations into the Ksp or Qsp expression.
The Ksp represents the maximum product of ion concentrations at equilibrium. If you know two of the three variables (Ksp, [Mg2+], [OH−]), you can solve for the third.
Example: Maximum [OH−] Before Precipitation at [Mg2+] = 0.01 M
1.6 × 10−12 = (0.01)[OH−]2
Divide both sides by 0.01 (equivalent to multiplying by 100):
[OH−]2 = 1.6 × 10−10
Take the square root:
[OH−] = 1.3 × 10−5 M
This is the maximum hydroxide concentration that can coexist with 0.01 M Mg2+ without forming a precipitate. Exceed that threshold, and magnesium hydroxide solid begins to form.
3 Outcomes When Comparing Qsp to Ksp
When you have ions in solution and want to know whether a precipitate will form, calculate Qsp and compare it to Ksp. There are exactly 3 possible outcomes when predicting precipitation:
| Condition | Solution Status | What Happens |
|---|---|---|
| Qsp < Ksp | Unsaturated | No precipitate; more solid could still dissolve |
| Qsp = Ksp | Saturated (at equilibrium) | No net change; system is at equilibrium |
| Qsp > Ksp | Supersaturated | Precipitate forms; equilibrium shifts left until Qsp = Ksp |
When Qsp exceeds Ksp, the ion concentrations are higher than the equilibrium allows. The system responds by shifting the equilibrium to the left—ions combine to form a solid precipitate, reducing the ion concentrations until Qsp equals Ksp once again. This is Le Chatelier's Principle in action.
Did a Precipitate Form? (Worked Example)
To a solution of 1.0 × 10−4 M magnesium nitrate, NaOH was added to a final concentration of 2.0 × 10−4 M. Did a precipitate form? (Ksp of Mg(OH)2 = 1.6 × 10−12)
First, identify the relevant ion concentrations. Magnesium nitrate is a soluble ionic compound (all nitrate salts are soluble), so it dissociates completely. In a one-to-one ratio, 1.0 × 10−4 M Mg(NO3)2 yields 1.0 × 10−4 M Mg2+. The nitrate ions are spectators—they play no role in the equilibrium.
Sodium hydroxide is a strong base, another strong electrolyte. It dissociates completely in a one-to-one ratio, giving 2.0 × 10−4 M OH−. The sodium ions are irrelevant.
Example: Calculating Qsp to Predict Precipitation
Qsp = [Mg2+][OH−]2
Qsp = (1.0 × 10−4)(2.0 × 10−4)2
Qsp = (1.0 × 10−4)(4.0 × 10−8)
Qsp = 4.0 × 10−12
Since Qsp (4.0 × 10−12) > Ksp (1.6 × 10−12), a precipitate forms.
The Dilution Twist: Mixing Equal Volumes
Professors love to add one sneaky twist to precipitation problems. Consider this version: 50 mL of 1.0 × 10−4 M magnesium nitrate is mixed with 50 mL of 2.0 × 10−4 M sodium hydroxide. Did a precipitate form? This problem requires a dilution calculation before you can compare Qsp to Ksp.
You might think this is identical to the previous problem—but it is not. Think of it this way: if you mix 50 mL of orange juice with 50 mL of water, the concentration of the orange juice drops. The same principle applies here. When two solutions are combined, they dilute each other.
Common Mistake: Using the pre-mixing concentrations in Qsp without accounting for dilution. The concentrations given are before mixing. After combining the solutions, the total volume increases, and every concentration decreases.
Applying M1V1 = M2V2 Before Calculating Qsp
Use the dilution equation to find each concentration after mixing:
Example: Dilution of Mg(NO3)2
M1V1 = M2V2
(1.0 × 10−4)(50 mL) = M2(100 mL)
M2 = 5.0 × 10−5 M Mg(NO3)2
Therefore [Mg2+] = 5.0 × 10−5 M after mixing.
Example: Dilution of NaOH
(2.0 × 10−4)(50 mL) = M2(100 mL)
M2 = 1.0 × 10−4 M NaOH
Therefore [OH−] = 1.0 × 10−4 M after mixing.
Quick Rule: When you mix equal volumes of two solutions, every concentration is cut in half. This is a handy shortcut that saves time on exams.
Recalculating Qsp with Diluted Concentrations
Now recalculate Qsp using the post-mixing concentrations:
Example: Qsp After Dilution Shows No Precipitation
Qsp = [Mg2+][OH−]2
Qsp = (5.0 × 10−5)(1.0 × 10−4)2
Qsp = (5.0 × 10−5)(1.0 × 10−8)
Qsp = 5.0 × 10−13
Since Qsp (5.0 × 10−13) < Ksp (1.6 × 10−12), no precipitate forms. The solution is unsaturated.
The same concentrations that produced a precipitate in the previous problem no longer do so after dilution. This is a favorite exam trick—always check whether the problem gives you concentrations before or after mixing, and apply M1V1 = M2V2 when volumes are combined.
7 Key Takeaways
- The common ion effect reduces molar solubility when a strong electrolyte contributes an ion already involved in the dissolution equilibrium, shifting the reaction toward the solid via Le Chatelier's Principle.
- In common ion problems, the ICE table changes only in that one initial ion concentration is nonzero—set it to the concentration provided by the strong electrolyte, then solve as usual with the small-x approximation.
- Qsp (the reaction quotient for solubility) uses the same expression as Ksp but with current rather than equilibrium concentrations. Comparing Qsp to Ksp determines whether a precipitate forms.
- When Qsp > Ksp, a precipitate forms. When Qsp < Ksp, the solution is unsaturated and no precipitate appears. When Qsp = Ksp, the solution is exactly saturated.
- Precipitation problems require no ICE table and no variable x—plug the known ion concentrations directly into the Qsp or Ksp expression.
- When solutions are mixed, their concentrations change due to dilution. Use M1V1 = M2V2 to find post-mixing concentrations before calculating Qsp.
- Mixing equal volumes cuts all concentrations in half—a detail that can flip a precipitation prediction from "yes" to "no."