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AP Chemistry · Unit 7 of 9
Equilibrium
7–9% of the exam6 lessons · 82 min17 terms
What this unit covers
The topics below follow the published Chemistry course framework for Unit 7. This unit is worth 7–9% of the exam, so budget your time against that rather than against how long the unit takes to teach.
Keq & QLe ChâtelierSolubilityCommon ion effect
Lessons in this unit
- Dynamic Equilibrium & the Equilibrium Constant13 min · 3 objectivesDescribe chemical equilibrium as a dynamic balance of forward and reverse rates · Write the equilibrium-constant expression (Kc and Kp) for a reaction · Interpret the size of K and use the reaction quotient Q to predict shift direction
- ICE Tables & Calculating K14 min · 3 objectivesSet up an ICE (Initial, Change, Equilibrium) table for a reversible reaction · Calculate K from initial amounts and a single equilibrium measurement · Use a known K to solve for equilibrium concentrations
- Le Châtelier's Principle12 min · 3 objectivesState Le Châtelier’s principle and predict shifts from concentration changes · Predict the effect of pressure and volume changes on gaseous equilibria · Predict how temperature changes shift equilibrium and change the value of K
- Solubility Equilibria (Ksp)13 min · 3 objectivesWrite the solubility-product expression Ksp for a slightly soluble salt · Convert between Ksp and molar solubility for different stoichiometries · Predict and explain the common-ion effect on solubility
- ICE Tables & K — Advanced15 min · 3 objectivesDecide when the small-x approximation is valid and apply it to solve an ICE problem · Solve an equilibrium problem that genuinely requires the quadratic formula · Convert between Kp and Kc using Kp = Kc(RT)^Δn
- Solubility Equilibria — Ksp, Common Ion & Precipitation15 min · 3 objectivesConvert between Ksp and molar solubility for salts with 1:2 stoichiometry (Ksp = 4s³) · Calculate the reduced solubility produced by the common-ion effect · Predict whether a precipitate forms by comparing the ion product Q to Ksp
Formulas in Unit 7
Equilibrium-constant expression
For aA + bB ⇌ cC + dD: Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)
Products on top, reactants on the bottom, each raised to its coefficient. Omit any pure solid (s) or pure liquid (l).
Kp for gaseous equilibria
Kp = Kc (RT)^Δn, where Δn = (moles gas product) − (moles gas reactant)
Kp uses partial pressures instead of concentrations. If the number of gas moles is unchanged (Δn = 0), then Kp = Kc.
Reaction quotient Q vs. K
Q < K → shift right (forward); Q = K → at equilibrium; Q > K → shift left (reverse)
Q has the exact same form as K but uses the current (not necessarily equilibrium) concentrations. Comparing Q to K predicts the direction of change.
ICE relationship
Equilibrium concentration = Initial + Change (Change = ± coefficient × x)
Reactants are consumed (−), products are formed (+). The coefficient in front of x must match the balanced equation.
Temperature and K (heat-as-reagent rule)
Exothermic (ΔH < 0): heat is a product → raising T shifts left, K decreases. Endothermic (ΔH > 0): heat is a reactant → raising T shifts right, K increases.
Only a temperature change alters the numerical value of K. Concentration, volume, pressure, and catalyst changes never do.
Solubility product
For AₘBₙ(s) ⇌ m Aⁿ⁺(aq) + n Bᵐ⁻(aq): Ksp = [Aⁿ⁺]ᵐ [Bᵐ⁻]ⁿ
Only the dissolved ions appear, each raised to its coefficient. A larger Ksp (for the same ion-count stoichiometry) means a more soluble salt.
Ksp ↔ molar solubility
MX type: Ksp = s². MX₂ type: Ksp = 4s³. M₂X type: Ksp = 4s³.
Always rebuild the relationship from the balanced dissolution equation; do not assume Ksp = s² for every salt.
Quadratic formula (for the exact ICE solve)
ax² + bx + c = 0 → x = (−b ± √(b² − 4ac)) / 2a
Use when the small-x approximation fails the 5% test. Discard the root that gives a negative concentration or a value exceeding the initial amount.
Kp from Kc
Kp = Kc (RT)^Δn, Δn = (moles of gaseous product) − (moles of gaseous reactant)
R = 0.0821 L·atm·mol⁻¹·K⁻¹ and T is in kelvin. If Δn = 0 the factor is 1, so Kp = Kc. Only gas-phase species count toward Δn.
Solubility product and molar solubility
AₘBₙ(s) ⇌ m Aⁿ⁺ + n Bᵐ⁻: Ksp = [Aⁿ⁺]ᵐ[Bᵐ⁻]ⁿ. MX: Ksp = s². MX₂ or M₂X: Ksp = 4s³. MX₃: Ksp = 27s⁴.
The numerical prefix (4, 27, …) is (coefficient)^(coefficient) summed over the ions. To get s from Ksp for a 1:2 salt, take the cube root of Ksp/4.
Precipitation criterion
Q > Ksp → precipitate forms; Q = Ksp → just saturated; Q < Ksp → no precipitate
Q uses concentrations in the combined solution (after mixing dilutes them). Same Q-vs-K logic as reaction equilibria, applied to dissolving.
Every term in Unit 7
All 17 terms we publish for Equilibrium, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.
- Dynamic equilibrium
- Forward and reverse rates are equal, so concentrations stay constant while both reactions continue.
- Equilibrium constant K
- Products over reactants, each raised to its coefficient. Pure solids and liquids are omitted because their concentrations do not change.
- Kc and Kp relationship
- Kp = Kc(RT)^Δn, where Δn is moles of gaseous products minus gaseous reactants. They are equal when Δn = 0.
- Reaction quotient Q
- Same expression as K but at any moment. Q < K shifts forward, Q > K shifts reverse, Q = K means equilibrium.
- Le Châtelier's principle
- A system at equilibrium shifts to partially counteract an imposed change in concentration, pressure or temperature.
- Effect of temperature on K
- Temperature is the only change that alters K itself. Raising it shifts an endothermic reaction forward and increases K.
- Effect of a catalyst on equilibrium
- None. A catalyst speeds both directions equally, so equilibrium is reached sooner at the same position.
- ICE table
- Initial, change, equilibrium. The change row must respect the stoichiometric ratios, which is where most errors occur.
- Small-x approximation
- Valid when K is small relative to the initial concentration; check that x is under 5% of the initial value, or solve the quadratic.
- Solubility product Ksp
- The equilibrium constant for a solid dissolving. Molar solubility is derived from Ksp using the ion stoichiometry.
- Common ion effect
- Adding an ion already present shifts the dissolution equilibrium back and lowers solubility — a direct application of Le Châtelier.
- pH effect on solubility
- Salts of weak-acid anions dissolve more in acid, because H⁺ removes the anion and pulls the dissolution equilibrium forward.
- Adding an inert gas at constant volume
- No effect on equilibrium, because the partial pressures of the reacting species are unchanged.
- Effect of volume change
- Reducing volume shifts equilibrium toward the side with fewer moles of gas. No shift occurs when both sides have equal gas moles.
- Magnitude of K
- K much greater than 1 means products dominate at equilibrium; K much less than 1 means reactants do. K says nothing about how fast equilibrium is reached.
- Relating K for reversed and scaled reactions
- Reversing gives 1/K; multiplying coefficients by n raises K to the nth power; adding reactions multiplies their K values.
- Q vs K reasoning on a graph
- A concentration-time graph reaching constant non-zero values for all species indicates equilibrium, not completion.
What examiners penalize here
- On the exam, state the reason with the comparison. "Q < K, so the reaction proceeds forward" earns the point; "it goes forward" alone often does not. Also remember: reversing a reaction inverts K (K becomes 1/K), and multiplying an equation through by n raises K to the nth power.
- To decide whether a precipitate forms, compute the ion product Q for the trial concentrations and compare to Ksp: Q > Ksp means the solution is supersaturated and a precipitate forms; Q < Ksp means it stays dissolved; Q = Ksp is exactly saturated. It is the same Q-versus-K logic from Lesson 1, applied to dissolving.
- On the free-response, the point is often awarded for the justification, not just the number. After a small-x solve, explicitly write the ratio (e.g. "x/[initial] = 1% < 5%, so the approximation is valid"). If it fails, say so and switch to the quadratic — graders reward recognizing the boundary.
- Two exam reflexes for solubility problems: (1) after mixing solutions, recompute every concentration in the combined volume before finding Q — dilution is the most common oversight; (2) always carry the ion coefficient as an exponent, so a common ion or trial ion with coefficient 2 is squared. Both slips move the answer by orders of magnitude.
Practice Chemistry
Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.
Questions about this unit
How much of the AP Chemistry exam is Unit 7?
Unit 7, Equilibrium, is worth 7–9% of the Chemistry multiple-choice section according to the published course framework. Across all 9 units that makes it a middling share, roughly what an even split across units would give.
What topics are covered in Chemistry Unit 7?
Equilibrium covers Keq & Q, Le Châtelier, Solubility and Common ion effect. We publish 17 terms with definitions for this unit, all of them on this page.
How should I study Chemistry Unit 7?
Read the 6 lessons below first — about 80 minutes — then drill the 17 terms in cram mode until you can produce each definition from memory rather than just recognize it. Recognition is what makes a unit feel finished when it is not. Finish with practice questions and read the explanation for every one you get right by elimination as well as the ones you miss.
All 9 units of AP Chemistry
Unit names, topics and exam weights follow the published College Board course framework for AP Chemistry. AP® is a trademark registered by the College Board, which does not endorse this site.