E. Chemical Systems and Equilibrium · Acid-Base and Solubility EquilibriaPredicting Precipitation and the Common Ion Effect
E2.4 E2.1 E3.4 E2.2About 75 minutes
Learning goals- I can calculate the ion product Q and compare it with Ksp to predict whether a precipitate forms.
- I can account for dilution when two solutions are mixed.
- I can explain the common ion effect with Le Châtelier's principle and calculate solubility in a solution with a common ion.
- I can use Ksp to plan a selective precipitation and to explain why some solids dissolve in acid.
Words to know
ion product, Q
The reaction quotient for a dissolving equilibrium, calculated with the actual ion concentrations at any moment (for AgCl, Q = [Ag⁺][Cl⁻]).
common ion effect
The decrease in the solubility of a salt when a solution already contains one of its ions.
selective precipitation
Separating ions by adding a reagent that precipitates one ion while leaving others in solution.
supersaturated
Containing more dissolved solute than the equilibrium amount; Q > Ksp. Precipitation is expected.
Will it go cloudy?
Water-treatment operators, mining engineers and lab analysts constantly ask the same question: if I mix these two solutions, will a solid form? The Grade 11 solubility table answers "probably" for whole compounds, but not for real concentrations. A tiny amount of Ag⁺ in a dilute chloride solution might stay dissolved; a larger amount will not. In this lesson you will make that prediction with numbers, using the reaction quotient you met in The Reaction Quotient Q: Which Way Will It Shift? (unit Chemical Equilibrium).
Q versus Ksp
For any slightly soluble salt, calculate the ion product Q the same way as Ksp, but with the concentrations that are actually present at that moment:
- Q < Ksp: the solution is unsaturated. No precipitate forms (and more solid would dissolve).
- Q = Ksp: the solution is saturated, at equilibrium.
- Q > Ksp: the solution is supersaturated. A precipitate forms until Q falls to Ksp.
When two solutions are mixed, dilute first. Each ion is spread through the combined volume: c(after) = c(before) × V(own) ÷ V(total). Forgetting this is the most common error.
Worked exampleExample 1: A precipitate forms
50.0 mL of 0.0010 mol/L AgNO₃ is mixed with 50.0 mL of 0.0020 mol/L NaCl. Will AgCl (Ksp = 1.8 × 10⁻¹⁰) precipitate?
- Total volume = 100.0 mL. Each solution is diluted by a factor of 2.
- [Ag⁺] = 0.0010 × 50.0 ÷ 100.0 = 5.0 × 10⁻⁴ mol/L; [Cl⁻] = 0.0020 × 50.0 ÷ 100.0 = 1.0 × 10⁻³ mol/L.
- Q = [Ag⁺][Cl⁻] = (5.0 × 10⁻⁴)(1.0 × 10⁻³) = 5.0 × 10⁻⁷.
- Q (5.0 × 10⁻⁷) > Ksp (1.8 × 10⁻¹⁰).
Yes, AgCl precipitates. Q is more than 2000 times Ksp.
Worked exampleExample 2: No precipitate
100.0 mL of 1.0 × 10⁻³ mol/L Pb(NO₃)₂ is mixed with 100.0 mL of 2.0 × 10⁻³ mol/L NaCl. Will PbCl₂ (Ksp = 1.7 × 10⁻⁵) precipitate?
- After mixing (200.0 mL): [Pb²⁺] = 5.0 × 10⁻⁴ mol/L; [Cl⁻] = 1.0 × 10⁻³ mol/L.
- Q = [Pb²⁺][Cl⁻]² = (5.0 × 10⁻⁴)(1.0 × 10⁻³)² = 5.0 × 10⁻¹⁰.
- Q (5.0 × 10⁻¹⁰) < Ksp (1.7 × 10⁻⁵).
No precipitate. The solubility table lists PbCl₂ as having low solubility, but at these low concentrations it stays dissolved.
Worked exampleExample 3: What pH starts a precipitate?
A solution contains 0.050 mol/L Mg²⁺. Above what pH will Mg(OH)₂ (Ksp = 5.6 × 10⁻¹²) begin to precipitate?
- Precipitation begins when Q = Ksp: [Mg²⁺][OH⁻]² = 5.6 × 10⁻¹².
- [OH⁻] = √(5.6 × 10⁻¹² ÷ 0.050) = 1.06 × 10⁻⁵ mol/L.
- pOH = 4.98, so pH = 9.02.
Mg(OH)₂ begins to precipitate above pH 9.02. Raising pH is how water-treatment plants remove many metal ions.
The common ion effect
What happens to the solubility of AgCl if the water already contains chloride ions, for example from NaCl?
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
By Le Châtelier's principle, the extra Cl⁻ (the common ion) shifts the equilibrium to the left. Less AgCl can dissolve. This is the common ion effect: a salt is less soluble in a solution that already contains one of its ions. (You met the same effect in buffers: added A⁻ suppresses the ionization of HA.)
To calculate the new solubility, put the common ion's concentration into the ICE table. For AgCl in 0.10 mol/L NaCl:
- [Ag⁺] = s, [Cl⁻] = 0.10 + s ≈ 0.10 (s is tiny)
- Ksp = s(0.10) = 1.8 × 10⁻¹⁰, so s = 1.8 × 10⁻⁹ mol/L
Compare: in pure water s = 1.3 × 10⁻⁵ mol/L. The common ion lowered the solubility by a factor of about 7000.
Worked exampleExample 4: Common ion with an AB₂ salt
Find the molar solubility of CaF₂ (Ksp = 3.5 × 10⁻¹¹) in 0.010 mol/L NaF, and compare it with its solubility in pure water (2.1 × 10⁻⁴ mol/L).
- [Ca²⁺] = s; [F⁻] = 0.010 + 2s ≈ 0.010.
- Ksp = s(0.010)² = 3.5 × 10⁻¹¹, so s = 3.5 × 10⁻¹¹ ÷ 1.0 × 10⁻⁴ = 3.5 × 10⁻⁷ mol/L.
- Check: 2s = 7 × 10⁻⁷, which really is negligible compared with 0.010.
- Ratio: 2.1 × 10⁻⁴ ÷ 3.5 × 10⁻⁷ ≈ 600.
s = 3.5 × 10⁻⁷ mol/L, about 600 times less than in pure water.
Selective precipitation
If a solution contains several ions that form precipitates with the same reagent, the one needing the lowest concentration of reagent precipitates first. For a solution 0.010 mol/L in both Cl⁻ and I⁻, slowly adding Ag⁺:
- AgI starts to form when [Ag⁺] = 8.5 × 10⁻¹⁷ ÷ 0.010 = 8.5 × 10⁻¹⁵ mol/L.
- AgCl starts to form when [Ag⁺] = 1.8 × 10⁻¹⁰ ÷ 0.010 = 1.8 × 10⁻⁸ mol/L.
So AgI precipitates first. By the time AgCl begins to form, [I⁻] has dropped to 8.5 × 10⁻¹⁷ ÷ 1.8 × 10⁻⁸ = 4.7 × 10⁻⁹ mol/L: essentially all the iodide is out of solution. The two ions have been separated.
Acids dissolve some "insoluble" solids
If a salt's anion is a weak base (CO₃²⁻, OH⁻, F⁻, PO₄³⁻), adding acid removes that anion: for example, CO₃²⁻ + H₃O⁺ → HCO₃⁻ + H₂O. By Le Châtelier's principle, more solid dissolves to replace it. That is why limestone and marble (CaCO₃) dissolve in acid rain, why Mg(OH)₂ antacids dissolve in stomach acid, and why tooth enamel weakens in acidic conditions. Salts of strong-acid anions, such as AgCl, are not made more soluble by acid.
Just for funChemistry teachers have a favourite motto for this lesson: "If you're not part of the solution, you're part of the precipitate."
Hands-on activityPrecipitate, or not? A dilution test
You need: calcium chloride ice melt (the kind sold for driveways in Canadian winters; check the label says calcium chloride), baking soda, 5 clear glasses, a teaspoon, a tablespoon, warm water, a dark sheet of paper, rubber gloves and eye protection.
Grown-up note: calcium chloride irritates skin and eyes and gets hot when it dissolves. An adult should handle it, with gloves and eye protection; keep it away from mouths and pets, and rinse any spill with water.
- Stock solutions. In one glass, dissolve 1 teaspoon of calcium chloride in ½ cup of water (solution A). In another, dissolve 1 teaspoon of baking soda in ½ cup of water (solution B).
- Full strength. In glass 3, mix 1 tablespoon of A with 1 tablespoon of B. Hold it in front of the dark paper. Watch for cloudiness (CaCO₃) and bubbles (CO₂).
- Diluted. Make A and B ten times more dilute (1 tablespoon of each with 9 tablespoons of water) and repeat the mixing in glass 4.
- Very dilute. Dilute those again tenfold and mix in glass 5.
- Record which mixtures turned cloudy.
What to look for: the full-strength mixture turns cloudy white at once, with some fizzing: Ca²⁺(aq) + 2 HCO₃⁻(aq) → CaCO₃(s) + H₂O(l) + CO₂(g). As you dilute, the cloudiness becomes fainter, and in the most dilute mixture you probably won't see any. Each tenfold dilution of both solutions lowers the ion product a lot, so eventually Q is no longer large enough for a visible precipitate. The exact cutoff is hard to calculate here, because only part of the HCO₃⁻ supplies CO₃²⁻, but the trend is exactly what Q versus Ksp predicts.
Your turn
1Will a precipitate form if [Ba²⁺] = 1.0 × 10⁻⁴ mol/L and [SO₄²⁻] = 9.0 × 10⁻⁵ mol/L? (Ksp BaSO₄ = 1.1 × 10⁻¹⁰)
Q = (1.0 × 10⁻⁴)(9.0 × 10⁻⁵) = 9.0 × 10⁻⁹ > 1.1 × 10⁻¹⁰, so yes.
225.0 mL of 1.0 × 10⁻³ mol/L Pb(NO₃)₂ is mixed with 75.0 mL of 2.0 × 10⁻³ mol/L KI. Will PbI₂ (Ksp = 9.8 × 10⁻⁹) precipitate?
[Pb²⁺] = 1.0 × 10⁻³ × 25.0 ÷ 100.0 = 2.5 × 10⁻⁴; [I⁻] = 2.0 × 10⁻³ × 75.0 ÷ 100.0 = 1.5 × 10⁻³. Q = (2.5 × 10⁻⁴)(1.5 × 10⁻³)² = 5.6 × 10⁻¹⁰ < 9.8 × 10⁻⁹: no precipitate.
3Find the molar solubility of AgCl in 0.010 mol/L NaCl.
s = Ksp ÷ [Cl⁻] = 1.8 × 10⁻¹⁰ ÷ 0.010 = 1.8 × 10⁻⁸ mol/L.
4Find the molar solubility of BaSO₄ in 0.010 mol/L Na₂SO₄.
s = 1.1 × 10⁻¹⁰ ÷ 0.010 = 1.1 × 10⁻⁸ mol/L (compared with 1.0 × 10⁻⁵ in pure water).
5Use Le Châtelier's principle to explain why adding NaF lowers the solubility of CaF₂.
CaF₂(s) ⇌ Ca²⁺ + 2 F⁻. Adding F⁻ (a product) shifts the equilibrium left, toward the solid, so less CaF₂ can dissolve.
6Which dissolves more in hydrochloric acid than in water: CaCO₃ or AgCl? Explain.
CaCO₃. H₃O⁺ reacts with CO₃²⁻ (a weak base), removing it and shifting CaCO₃(s) ⇌ Ca²⁺ + CO₃²⁻ to the right. Cl⁻ is not a base, so acid has no such effect on AgCl (the extra Cl⁻ from HCl even lowers its solubility slightly).
More practice: warm-up
1If Q > Ksp, what happens?
A precipitate forms until Q = Ksp.
2If Q < Ksp, is the solution saturated?
4Does a common ion increase or decrease solubility?
5When 50 mL of one solution is mixed with 50 mL of another, by what factor is each concentration diluted?
By a factor of 2 (each is halved).
More practice: on your own
1100.0 mL of 0.020 mol/L CaCl₂ is mixed with 100.0 mL of 0.010 mol/L Na₂SO₄. Will CaSO₄ (Ksp = 4.9 × 10⁻⁵) precipitate?
[Ca²⁺] = 0.010 mol/L; [SO₄²⁻] = 0.0050 mol/L. Q = 5.0 × 10⁻⁵, just greater than 4.9 × 10⁻⁵, so yes, barely: a small amount of precipitate may form.
2A solution contains 0.010 mol/L Cu²⁺. Above what pH does Cu(OH)₂ (Ksp = 2.2 × 10⁻²⁰) begin to precipitate?
[OH⁻] = √(2.2 × 10⁻²⁰ ÷ 0.010) = 1.5 × 10⁻⁹ mol/L; pOH = 8.83; pH = 5.17.
3Find the molar solubility of PbI₂ (Ksp = 9.8 × 10⁻⁹) in 0.10 mol/L KI.
[I⁻] ≈ 0.10, so s(0.10)² = 9.8 × 10⁻⁹ and s = 9.8 × 10⁻⁷ mol/L (versus 1.3 × 10⁻³ in pure water).
4A solution is 0.010 mol/L in both Ba²⁺ and Ca²⁺. Na₂SO₄ is added slowly. Which sulfate precipitates first? (Ksp BaSO₄ = 1.1 × 10⁻¹⁰, CaSO₄ = 4.9 × 10⁻⁵)
BaSO₄ starts at [SO₄²⁻] = 1.1 × 10⁻¹⁰ ÷ 0.010 = 1.1 × 10⁻⁸ mol/L; CaSO₄ at 4.9 × 10⁻⁵ ÷ 0.010 = 4.9 × 10⁻³ mol/L. BaSO₄ precipitates first.
5For the previous question, what is [Ba²⁺] when CaSO₄ just begins to precipitate? What percentage of the barium remains in solution?
[Ba²⁺] = 1.1 × 10⁻¹⁰ ÷ 4.9 × 10⁻³ = 2.2 × 10⁻⁸ mol/L, which is 2.2 × 10⁻⁸ ÷ 0.010 × 100% = 0.00022%: the separation is essentially complete.
Challenge
1A lab worker wants to remove Mg²⁺ from a solution (0.050 mol/L) by raising the pH to 11.00. What concentration of Mg²⁺ remains, and what percentage has been removed?
At pH 11.00, [OH⁻] = 1.0 × 10⁻³ mol/L. [Mg²⁺] = 5.6 × 10⁻¹² ÷ (1.0 × 10⁻³)² = 5.6 × 10⁻⁶ mol/L. Removed: (0.050 − 0.0000056) ÷ 0.050 × 100% ≈ 99.99%.
2Mohr's method finds chloride by titrating with Ag⁺ using a little CrO₄²⁻ as the indicator. Red-brown Ag₂CrO₄ (Ksp 1.1 × 10⁻¹²) forms only after almost all Cl⁻ has precipitated as white AgCl. If [CrO₄²⁻] = 5.0 × 10⁻³ mol/L, find [Ag⁺] when Ag₂CrO₄ starts to form, and [Cl⁻] at that point.
[Ag⁺]² = 1.1 × 10⁻¹² ÷ 5.0 × 10⁻³ = 2.2 × 10⁻¹⁰, so [Ag⁺] = 1.5 × 10⁻⁵ mol/L. Then [Cl⁻] = 1.8 × 10⁻¹⁰ ÷ 1.5 × 10⁻⁵ = 1.2 × 10⁻⁵ mol/L, a tiny amount, so the colour appears very close to the equivalence point.
3Why does the approximation [Cl⁻] = 0.10 + s ≈ 0.10 work for AgCl in 0.10 mol/L NaCl, but not for PbCl₂ (Ksp 1.7 × 10⁻⁵) in 0.0010 mol/L NaCl?
For AgCl, s = 1.8 × 10⁻⁹, which is negligible next to 0.10. For PbCl₂ in 0.0010 mol/L NaCl, the approximation would give s = 1.7 × 10⁻⁵ ÷ (0.0010)² = 17 mol/L, which is impossible and far larger than 0.0010, so the extra Cl⁻ from PbCl₂ (2s) dominates. The common ion is too dilute to matter; s is close to the pure-water value (about 0.016 mol/L) and must be found by solving s(0.0010 + 2s)² = Ksp.
Quiz
1. Mixing two solutions gives [Ag⁺] = 1.0 × 10⁻⁶ mol/L and [Cl⁻] = 1.0 × 10⁻⁵ mol/L. Will AgCl (Ksp 1.8 × 10⁻¹⁰) precipitate?
2. What is the first step when predicting a precipitate after mixing two solutions?
3. The solubility of AgBr in 0.10 mol/L NaBr is
4. What is the molar solubility of CaF₂ (Ksp 3.5 × 10⁻¹¹) in 0.10 mol/L NaF?
5. A solution contains equal concentrations of Cl⁻ and I⁻. When Ag⁺ is added slowly, what happens first?
6. Why does CaCO₃ dissolve in acid rain?
Big idea (E2.4, E3.4): compare the ion product Q with Ksp: a precipitate forms only when Q > Ksp. Mixing solutions dilutes them, so recalculate concentrations first. A common ion lowers solubility (Le Châtelier), and acid raises the solubility of salts whose anions are weak bases.
How to help: Ask your teen to say the three cases aloud (Q less than, equal to, greater than Ksp) and what each means. When checking work, look first for the dilution step after mixing. The ice-melt activity needs an adult with gloves and eye protection; it is a good chance to talk about why water treatment and winter road salt involve chemistry.
Common difficulties:
- Forgetting to dilute both solutions when they are mixed.
- Forgetting exponents in Q (for PbI₂, [I⁻] is squared).
- Adding the common ion to only one side, or including s when it is negligible (and not checking that it is).
- Thinking acid makes every salt more soluble; it only helps when the anion is a weak base.