Chemistry 2e · Equilibria of Other Reaction Classes
Precipitation and Dissolution
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In 30 seconds
When an ionic compound is placed in water, two opposing processes begin immediately: ions leave the solid surface and enter solution (dissolution), while ions already in solution collide with the solid and rejoin it (precipitation). In many cases these processes reach a dynamic equilibrium in which a saturated solution coexists with undissolved solid. This is a heterogeneous equilibrium Equilibrium between species in different phases (here, solid and aqueous ions) Full entry →, because it involves both a solid phase and dissolved ions.
The quantitative language for this equilibrium is the solubility product constant, Ksp, which is just an equilibrium constant written for a dissolution process. Because the solid's concentration is fixed, it does not appear in the Ksp expression. From Ksp we can compute molar solubility (how much solid dissolves per liter), predict whether a precipitate will form when solutions are mixed, and control solubility with the common ion effect Reduced solubility caused by an ion already present from another source Full entry →.
Why this matters
Precipitation and dissolution govern chemistry everywhere that water meets a sparingly soluble salt. Kidney stones are calcium oxalate and calcium phosphate solids; understanding Ksp explains why urine chemistry and hydration affect their formation. Drinking-water plants remove ions like Ca²⁺, Mg²⁺, and Fe³⁺ by deliberately precipitating them as insoluble hydroxides or carbonates. Silver halide precipitates were the basis of traditional photography, and gravimetric analysis — determining an ion's amount by weighing a precipitate — is still used in environmental and clinical labs. Predicting when a solid forms (or dissolves) is also central to geochemistry: cave formation, scale buildup in pipes, and the release of lead from aging plumbing all obey these same equilibria.
The college version
Core Concepts
The dissolution equilibrium and Ksp
For silver chloride, the equilibrium is
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
and the solubility product is
Ksp = [Ag+][Cl-]
The solid is omitted because its "concentration" is constant. Ksp is small for sparingly soluble salts (e.g., 1.8 × 10-10 for AgCl at 25 °C) and larger for more soluble ones. Like every equilibrium constant, it depends on temperature.
Molar solubility connects Ksp to a concentration
Molar solubility, s, is the number of moles of solid that dissolve per liter of solution. Stoichiometry links s to the ion concentrations. For a 1:1 salt such as AgCl, [Ag+] = [Cl-] = s, so Ksp = s2. For a 1:2 salt such as CaF₂,
CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)
gives [Ca2+] = s and [F-] = 2s, so Ksp = [Ca2+][F-]2 = (s)(2s)2 = 4s3. Always write the balanced dissolution equation first: the exponents in the Ksp expression come from the coefficients.
The reaction quotient Q predicts precipitation
Before mixing solutions, compute the ion product Q from the actual concentrations. Compare it with Ksp:
- Q < Ksp: unsaturated; no precipitate forms (more solid could dissolve).
- Q = Ksp: exactly saturated; equilibrium holds.
- Q > Ksp: supersaturated; precipitation occurs until Q falls to Ksp.
This is Le Châtelier's principle applied to a dissolution equilibrium: adding ions drives the equilibrium toward solid.
The common ion effect
If the solution already contains one of the salt's ions from a different source, dissolution is suppressed. For BaSO₄ in a solution that also contains SO₄²⁻ from Na₂SO₄, the sulfate ion shifts the equilibrium
BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)
to the left. The molar solubility drops dramatically. Chemists exploit this to precipitate ions as completely as possible in gravimetric analysis.
Selective precipitation
When several ions could precipitate with the same reagent, the salt with the smaller Ksp precipitates first, provided its stoichiometry is comparable. Adding Cl⁻ to a solution containing Ag⁺, Pb²⁺, and Hg₂²⁺ precipitates them in stages — the classic qualitative-analysis scheme. The same idea separates ions in industrial and environmental chemistry.
How It Works / Step-by-Step Process
- Write the balanced dissolution equation and its Ksp expression.
- Relate ion concentrations to molar solubility s through the stoichiometric coefficients.
- Substitute into Ksp and solve for s, or invert to find Ksp from measured solubility.
- For mixing problems, compute Q from the final (diluted) concentrations and compare with Ksp.
- Apply the common ion effect by including the extra source of the shared ion in the Ksp expression.
Common Confusions
| Common Confusion | Correct Understanding |
|---|---|
| "Ksp is the same as molar solubility." | Ksp is the equilibrium constant; solubility s is the amount dissolved. They are related by stoichiometry, not equal. |
| "A salt with a smaller Ksp is always less soluble." | Only for identical stoichiometry. CaF₂ (Ksp = 4.0 × 10-11) is more soluble than AgCl (Ksp = 1.8 × 10-10) because of the s3 vs s2 relationship. |
| "The solid's concentration appears in Ksp." | Pure solids are omitted; their activity is constant. |
| "Adding a common ion makes the salt dissolve more." | It does the opposite: the equilibrium shifts toward solid, lowering solubility. |
| "Q > Ksp means nothing happens." | It means precipitation will happen — the system is supersaturated. |
| "Exponents in Ksp come from the ion charges." | They come from the balanced-equation coefficients, not the charges. For CaF₂ it's [F-]2, not [F-]1 from the −1 charge. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine a crowded dance floor. Dancers keep leaving the floor (dissolving), and dancers at the edge keep stepping back on (precipitating). At some point, just as many leave as come back, so the crowd size stops changing — that's equilibrium. The solubility product is the "crowd limit": if you push too many dancers on at once, the extras get pushed back off as solid.
Worked examples
(a) AgCl. With Ksp = 1.8 × 10-10, the relation Ksp = s2 gives
s = Ksp = 1.8 × 10-10 = 1.34 × 10-5
so the molar solubility of AgCl in pure water is 1.3 × 10-5 mol/L.
(b) CaF₂. With Ksp = 4.0 × 10-11, use Ksp = 4s3:
s = 3Ksp4 = 34.0 × 10-114 = 31.0 × 10-11 = 2.2 × 10-4
The molar solubility is 2.2 × 10-4 mol/L. Notice that CaF₂ is roughly ten times more soluble than AgCl even though its Ksp is smaller — the s3 relationship punishes small Ksp values less harshly than s2. Comparing salts of different stoichiometry by Ksp alone is a trap.
Barium sulfate, Ksp = 1.1 × 10-10, is used in medical imaging (barium swallow). In pure water, its molar solubility is
s = 1.1 × 10-10 = 1.05 × 10-5 mol/L
Now find the solubility in 0.010 M Na₂SO₄. The sulfate comes from both the dissolving salt and the added Na₂SO₄, so [SO42-] = 0.010 + s. Because s is tiny compared with 0.010, approximate:
Ksp = [Ba2+][SO42-] = s(0.010 + s) ≈ s(0.010)
s = 1.1 × 10-100.010 = 1.1 × 10-8 mol/L
The added sulfate reduces the solubility by a factor of about 1000. This is why excess precipitating agent is used in gravimetric analysis: the common ion forces nearly all of the target ion out of solution.
A lab technician mixes 50.0 mL of 0.020 M AgNO₃ with 50.0 mL of 0.040 M NaCl. Mixing doubles the total volume, so each concentration is halved: [Ag+] = 0.010 M and [Cl-] = 0.020 M. The ion product is
Q = [Ag+][Cl-] = (0.010)(0.020) = 2.0 × 10-4
Since Q = 2.0 × 10-4 is enormously larger than Ksp = 1.8 × 10-10, the solution is supersaturated and AgCl precipitates until the remaining ion product equals Ksp.
Key takeaways
- Ksp is an equilibrium constant for dissolution; solids never appear in its expression.
- Write the balanced dissolution equation before writing Ksp; coefficients become exponents.
- For AB: Ksp = s2. For AB2 or A2B: Ksp = 4s3.
- Q > Ksp means precipitation; Q < Ksp means no precipitation; Q = Ksp means saturated.
- A common ion lowers molar solubility (Le Châtelier); this is why "like dissolves in like" fails for salts.
- Smaller Ksp generally means a less soluble salt, but compare only salts of the same stoichiometry (e.g., 1:1 vs 1:1).
- Ksp values are tabulated at 25 °C and change with temperature.
- Precipitation can be forced by removing a product ion (e.g., by complexation or acid) — the topic of coupled equilibria.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Write the Ksp expression for PbI₂. What relation does it give between s and Ksp?
Show answer
PbI2(s) ⇌ Pb2+ + 2I-, so Ksp = [Pb2+][I-]2 = s(2s)2 = 4s3.
The molar solubility of AgI is about 9.2 × 10-9 mol/L. Estimate its Ksp.
Show answer
For a 1:1 salt, Ksp = s2 = (9.2 × 10-9)2 = 8.5 × 10-17, matching the tabulated value.
A solution is 0.0010 M in both Ba²⁺ and Ag⁺. If solid Na₂SO₄ is added slowly, which precipitates first, BaSO₄ (Ksp = 1.1 × 10-10) or Ag₂SO₄ (Ksp = 1.2 × 10-5)?
Show answer
BaSO₄ precipitates first: it needs only [SO42-] = Ksp/[Ba2+] = 1.1 × 10-7 M, while Ag₂SO₄ needs [SO42-] = 1.2 × 10-5/(0.0010)2 = 12 M.
Why does BaSO₄ dissolve so much less in 0.010 M Na₂SO₄ than in pure water?
Show answer
The common ion SO₄²⁻ from Na₂SO₄ shifts the dissolution equilibrium toward solid (Le Châtelier), cutting [Ba2+] to roughly Ksp/0.010.
If Q = Ksp, what happens macroscopically?
Show answer
The solution is exactly saturated: solid and solution coexist at equilibrium, and no net change is visible.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- heterogeneous equilibrium
- Equilibrium between species in different phases (here, solid and aqueous ions)
- solubility product (Ksp)
- Equilibrium constant for dissolution; product of ion concentrations, each raised to its coefficient
- molar solubility (s)
- Moles of solid that dissolve per liter of saturated solution
- saturated solution
- Solution holding the maximum dissolved salt at equilibrium with solid
- reaction quotient (Q)
- Ion product computed from real, pre-equilibrium concentrations
- common ion effect
- Reduced solubility caused by an ion already present from another source
- selective (fractional) precipitation
- Separating ions by precipitating them one at a time using Ksp differences
Sources & references
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