AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/7/7-11)
Unit 7 · Topic 7.11
7.11 Introduction to Solubility Equilibria
Dissolving a slightly soluble salt is a reversible process, and its equilibrium constant is the solubility product, Ksp. You can calculate a salt's molar solubility from Ksp, find Ksp from a measured solubility, and use Q versus Ksp to predict whether a precipitate will form.
Key terms
- solubility product (Ksp)
- molar solubility
- saturated solution
- dissolution
- precipitate
The solubility equilibrium
When you add a salt like AgCl to water, a little dissolves into ions. Once the solution is saturated (holding as much as it can), dissolving and precipitating happen at the same rate: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).
The equilibrium constant is the solubility product: Ksp = [Ag⁺][Cl⁻]. The solid is left out, as with any solid. For a salt like CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), the coefficient becomes an exponent: Ksp = [Ca²⁺][F⁻]².
Molar solubility from Ksp
Molar solubility, s, is the number of moles of the salt that dissolve per liter of saturated solution. Write each ion's concentration in terms of s using the formula, then substitute into Ksp.
For CaF₂, each formula unit that dissolves gives 1 Ca²⁺ and 2 F⁻, so [Ca²⁺] = s and [F⁻] = 2s. Then Ksp = (s)(2s)² = 4s³.
To turn molar solubility into grams per liter, multiply by the molar mass.
| Salt type | Example | Ion concentrations | Ksp in terms of s |
|---|---|---|---|
| 1 : 1 | AgCl | s, s | s² |
| 1 : 2 | CaF₂, Mg(OH)₂ | s, 2s | 4s³ |
| 2 : 1 | Ag₂CrO₄ | 2s, s | 4s³ |
| 1 : 3 | Fe(OH)₃ | s, 3s | 27s⁴ |
Comparing solubilities
If two salts produce the same number of ions in the same ratio (both 1 : 1, for example), the one with the smaller Ksp is less soluble. If the ratios differ, you can't compare Ksp values directly; calculate s for each.
The solubility rules from topic 4.7 fit this picture. Salts that are called soluble, such as any salt of Na⁺, K⁺, NH₄⁺ or NO₃⁻, have Ksp values greater than 1. Salts called insoluble have very small Ksp values.
Will a precipitate form?
When two solutions are mixed, calculate Q (the ion product, written just like Ksp but with the current concentrations after mixing). If Q > Ksp, the solution holds more ions than it can at equilibrium, so solid precipitates until Q = Ksp. If Q < Ksp, no precipitate forms. If Q = Ksp, the solution is exactly saturated.
Example: mixing 50.0 mL of 0.0010 M AgNO₃ with 50.0 mL of 0.0010 M NaCl doubles the volume, so [Ag⁺] = [Cl⁻] = 0.00050 M. Q = (0.00050)² = 2.5 × 10⁻⁷, which is far larger than 1.8 × 10⁻¹⁰, so AgCl precipitates.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Molar solubility of a 1 : 1 salt
Ksp of AgCl is 1.8 × 10⁻¹⁰ at 25 °C. What is its molar solubility in pure water?
Show the solutionHide the solution
- Step 1: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). If s mol/L dissolves, [Ag⁺] = s and [Cl⁻] = s.
- Step 2: Ksp = s² = 1.8 × 10⁻¹⁰.
- Step 3: s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M.
Answer: s = 1.3 × 10⁻⁵ M
- Example 2Calculator allowed
Molar solubility of a 1 : 2 salt
Ksp of CaF₂ is 3.9 × 10⁻¹¹ at 25 °C. Find its molar solubility and the equilibrium [F⁻] in a saturated solution.
Show the solutionHide the solution
- Step 1: CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq). [Ca²⁺] = s and [F⁻] = 2s.
- Step 2: Ksp = (s)(2s)² = 4s³ = 3.9 × 10⁻¹¹.
- Step 3: s³ = 9.75 × 10⁻¹², so s = 2.1 × 10⁻⁴ M.
- Step 4: [F⁻] = 2s = 4.3 × 10⁻⁴ M. Forgetting to square the 2 (writing Ksp = 2s³) is a frequent error.
Answer: s = 2.1 × 10⁻⁴ M; [F⁻] = 4.3 × 10⁻⁴ M
- Example 3Calculator allowed
Trap: Ksp from solubility, and comparing salts
The molar solubility of Ag₂CrO₄ in water is 6.5 × 10⁻⁵ M. (a) Calculate Ksp. (b) AgCl has Ksp = 1.8 × 10⁻¹⁰. Which salt has the higher molar solubility?
Show the solutionHide the solution
- Step 1: (a) Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq), so [Ag⁺] = 2s = 1.3 × 10⁻⁴ M and [CrO₄²⁻] = s = 6.5 × 10⁻⁵ M.
- Step 2: Ksp = [Ag⁺]²[CrO₄²⁻] = (1.3 × 10⁻⁴)²(6.5 × 10⁻⁵) = 4s³ = 1.1 × 10⁻¹².
- Step 3: (b) The trap: Ag₂CrO₄'s Ksp (1.1 × 10⁻¹²) is smaller than AgCl's (1.8 × 10⁻¹⁰), so it's tempting to call Ag₂CrO₄ less soluble.
- Step 4: But the salts have different ion ratios, so compare molar solubilities: Ag₂CrO₄ is 6.5 × 10⁻⁵ M and AgCl is 1.3 × 10⁻⁵ M. Ag₂CrO₄ is about five times more soluble.
Answer: (a) Ksp = 1.1 × 10⁻¹²; (b) Ag₂CrO₄ has the higher molar solubility despite its smaller Ksp.
Common mistakes
- Including the solid in the Ksp expression.
- Writing [F⁻] = s instead of 2s for CaF₂, or forgetting to square 2s.
- Comparing Ksp values directly for salts with different ion ratios.
- Forgetting that mixing two solutions dilutes both before you calculate Q.
On the exam
- Expect to write a Ksp expression, then calculate molar solubility or Ksp. Write the dissolving equation first so the coefficients guide your ion concentrations.
- Ranking questions often hide a trap with salts of different formulas. Calculate s when the ion ratios differ.
Connected topics
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Check yourself
4 questions on 7.11 Introduction to Solubility Equilibria. Pick an answer to see if you got it, and why.
The molar solubility of CaF₂ in water at 25 °C is 2.1 × 10⁻⁴ M. What is Ksp for CaF₂ at this temperature?
| Salt | Ksp at 25 °C |
|---|---|
| AgCl | 1.8 × 10⁻¹⁰ |
| Ag₂CrO₄ | 1.1 × 10⁻¹² |
Solubility-product constants
Which salt has the greater molar solubility in pure water at 25 °C, and why?
Which of the following is the Ksp expression for calcium phosphate, Ca₃(PO₄)₂?
Ksp for PbI₂ is 9.8 × 10⁻⁹ at 25 °C. What is the molar solubility of PbI₂ in pure water at 25 °C?
0 of 4 answered