Introduction to Solubility Equilibria
Revisit saturated dissolution and free ions before interpreting standard dissolution free energy.
AP Chemistry Topic 7.11 · CED printed p. 139 / PDF p. 146. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Dissociation stoichiometry
- Q versus K
- Dilution
A sparingly soluble salt still dissolves to some extent. In a saturated solution with undissolved solid at fixed temperature, dissolution and precipitation balance dynamically. Write the balanced dissolution equation and omit the pure solid from Ksp. A very small solubility product explains why a salt described by solubility rules as insoluble can still have measurable dissolved ions. In the course’s standard-state convention, Ksp greater than 1 corresponds to soluble behavior; exact concentration predictions still need the full expression and suitable activity model.
Molar solubility s is moles of salt formula units dissolved per liter, not automatically the concentration of every ion. For MX, Ksp = s2 in pure water. For MX2, the concentrations are s and 2s, so Ksp = 4s3. Work from the equation each time. You can calculate Ksp from measured saturated ion concentrations, or solve for s from Ksp. Comparing Ksp values gives a direct relative-solubility ranking only when the dissolution stoichiometry and conditions are comparable; otherwise calculate s first. If you need mass solubility, multiply s by the salt’s molar mass.
The ion product Qsp uses the same expression before equilibrium. Qsp > Ksp indicates net precipitation; Qsp < Ksp indicates that more solid can dissolve if present. At Qsp = Ksp the solution is saturated in this model. An unsaturated solution with no solid present does not somehow create ions to force equality. Use concentrations after mixing, account for dilution, and state the assumptions: fixed temperature, a pure specified solid phase, dilute activities approximated by molarity, and no important complexation or acid–base side reactions.
Let's work through it
- For M2X(s) ⇌ 2M+ + X2−, a saturated solution in pure water has s = 1.50 × 10−4 M.
- Ion concentrations are [M+] = 3.00 × 10−4 M and [X2−] = 1.50 × 10−4 M.
- Ksp = [M+]2[X2−] = 1.35 × 10−11. The solid contributes no variable term.
- Check that Ksp = 4s3 gives the same result. Another solution must be compared using its actual Qsp, not this s unless it is saturated under the same conditions.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
Is s equal to [X−] for MX2?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
Is s equal to [X−] for MX2?
Worked answer and reason
No, [X−] = 2s in pure water with no other source.
One dissolved formula unit releases two X ions.
Retrieval 2
Qsp is below Ksp with no solid in the container. Must precipitation occur?
Worked answer and reason
No. The solution is unsaturated; precipitation is not favored.
Equality is a saturated equilibrium condition, not a demand on every solution.
Catch the tempting mistake
- The salt with the larger Ksp is always more soluble.
- Different formulas relate Ksp to different powers of s; calculate comparable molar solubilities.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 7 problem 12 · ap synthesis · symbolic
- Unit 7 problem 13 · ap application · data
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For the teacher
- A complete answer must state the solid phase and mixing volume, distinguish salt solubility from ion concentration, and check possible side reactions.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 7.11.A
- Calculate the solubility of a salt based on the value of K_sp for the salt.
- 7.11.A.1
- The dissolution of a salt is a reversible process whose extent can be described by K_sp, the solubility-product constant.
- 7.11.A.2
- The solubility of a substance can be calculated from the K_sp for the dissolution process. This relationship can also be used to predict the relative solubility of different substances.
- 7.11.A.3
- The solubility rules (see 4.7.A.5) can be quantitatively related to K_sp, in which K_sp values >1 correspond to soluble salts.
- 7.11.A.4
- The molar solubility of one or more species in a saturated solution can be used to calculate the K_sp of a substance.
Science references for this teaching
- AP Chemistry Course and Exam Description, effective Fall 2024 — Required topic-numbered learning objectives and essential knowledge, all continuations and AP exclusions; supplied PDF SHA256 b5dfe8677ef3d88c613865d2e2a3e8d6125d652e2b24c71ef1e8ce4e011094f0.
- OpenStax Chemistry 2e §15.1, Precipitation and Dissolution — Stoichiometric relation between molar solubility and Ksp; common-ion and precipitation reasoning.
Glossary and deeper reading · 7 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Stoichiometry
Stoichiometry uses a balanced chemical equation to relate amounts of reactants and products. Equation coefficients provide mole ratios rather than direct mass ratios. For example, forming water from hydrogen and oxygen consumes two moles of hydrogen molecules per mole of oxygen molecules.
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Solubility
Solubility describes how much of a substance can dissolve in a specified solvent under stated conditions. It depends on the solute, solvent, temperature, and other relevant conditions and is distinct from how fast dissolution occurs. For example, two salts can have different equilibrium solubilities in water at the same temperature.
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Saturated solution
A saturated solution has the dissolved concentration corresponding to equilibrium with the undissolved phase under the stated conditions. Microscopic dissolution and precipitation can continue at equal rates when the solid is present. For example, additional solid may remain undissolved after a saturated salt solution is reached at fixed temperature.
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Mole
A mole is an amount of substance containing exactly 6.02214076 × 10²³ specified entities. The entities can be atoms, molecules, ions, or formula units, so the counted entity must be named. For example, one mole of oxygen molecules contains two moles of oxygen atoms.
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Formula unit
A formula unit specifies the simplest whole-number ratio of ions in an ionic compound. It is a counting unit for an extended solid rather than an isolated molecule. For example, one calcium chloride formula unit represents one calcium ion and two chloride ions.
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Ion
An ion is an atom or group of atoms with a net electric charge. Ordinary ion formation changes electron count without changing the element's proton count. For example, a sodium atom losing one electron becomes a sodium ion with charge +1.
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Chemical equilibrium
Chemical equilibrium is a dynamic state in which forward and reverse reaction rates are equal. Macroscopic composition remains constant in a closed system under fixed conditions, but reactant and product concentrations need not be equal. For example, a reversible reaction can reach equilibrium with much more reactant than product.