Dynamic equilibrium: physical and chemical
Link to Std 9–10 : You know that some reactions seem to stop before all reactants are used up; now we see that nothing has stopped. Forward and backward changes simply go on at equal speed.
A crowd moves up and down a busy escalator pair. If the same number goes up as comes down, the people on each floor stay constant although everybody keeps moving.
Equilibrium is the state in which the forward and backward processes occur at the same rate in a closed system, so the observable properties (concentration, pressure, colour) stay constant. It is dynamic, not static.
In the figure : The reactant concentration falls and the product concentration rises until the two curves become flat. After the dashed line the rates are equal and the concentrations do not change.
Rule 1: Physical equilibria — Examples are ice ⇌ water at 273 K and 1 atm, water ⇌ water vapour in a closed vessel, and undissolved solid ⇌ dissolved solid in a saturated solution.
- Easy : Ice and water in a closed flask at 0 °C and 1 atm are in equilibrium; the mass of each stays constant.
- Medium : In a closed bottle of water, the vapour pressure at a given temperature stays constant because evaporation and condensation go at the same rate.
- Exam-type : A saturated sugar solution with undissolved crystals is at equilibrium: sugar dissolves and crystallises at the same rate.
Rule 2: Chemical equilibrium — In a reversible reaction such as H₂ + I₂ ⇌ 2HI, the equilibrium composition is the same whether we start from H₂ and I₂ or from pure HI at the same temperature.
- Easy : At equilibrium the rate of forward reaction equals the rate of backward reaction.
- Medium : Starting from 1 mol H₂ and 1 mol I₂, or from 2 mol HI, in the same flask and temperature gives the same final mixture.
- Exam-type : An equilibrium is reached only in a closed system; if CO₂ escapes from CaCO₃ heated in an open vessel, the reaction goes to completion.
Rule 3: Features of equilibrium — It is dynamic; macroscopic properties are constant; it can be approached from either side; a catalyst helps to reach it faster but does not change the position.
- Easy : A catalyst does not change the final concentrations.
- Medium : The equilibrium constant is fixed at a given temperature.
- Exam-type : The mixture at equilibrium has both reactants and products present, so "no further visible change" does not mean "no reaction".
Rule 4: Irreversible and reversible — Some reactions go to completion (burning paper); reversible reactions end in a mixture.
- Easy : Burning of paper is irreversible under normal conditions.
- Medium : Dissociation of N₂O₄ ⇌ 2NO₂ is reversible; the mixture shows a brown colour of fixed depth at a given temperature.
- Exam-type : A reaction that forms a gas or a precipitate and removes it from the vessel runs to completion.
At equilibrium rates are equal, not concentrations.
Saying that the concentrations of reactants and products are equal at equilibrium. They are only constant.
Memory clue: "Equal rates, constant amounts".
Spot the difference
| Point | Static | Dynamic equilibrium |
|---|---|---|
| Particles | Nothing happens | Forward and backward changes go on |
| Rates | Zero | Equal and non-zero |
| Example | A closed book | Ice and water at 0 °C |
Practice questions
- 1Define dynamic equilibrium.
- 2Name one physical equilibrium.
- 3Does a catalyst change the position of equilibrium?
- 4Why must the system be closed for equilibrium?
- 5Does the H₂ + I₂ ⇌ 2HI mixture depend on whether we start from H₂ and I₂ or from HI?
- 6What stays constant and what is equal at equilibrium?
- 7Why does CaCO₃ ⇌ CaO + CO₂ not reach equilibrium in an open vessel?
- 8Explain how a saturated solution with crystals at the bottom shows dynamic equilibrium.
Show answers (try them yourself first)
- 1A state in which forward and backward processes go on at the same rate so that the properties of the system stay constant.
- 2Ice ⇌ water at 0 °C (or water ⇌ vapour in a closed vessel).
- 3No; it only helps to reach equilibrium sooner.
- 4In an open system the products can escape, so the backward change cannot catch up.
- 5No; at the same temperature the final mixture is the same.
- 6Concentrations, pressure and colour stay constant; the forward and backward rates are equal.
- 7CO₂ escapes, so the backward reaction cannot occur and the reaction keeps going to completion.
- 8Crystals keep dissolving and dissolved particles keep crystallising at the same rate, so the concentration and the mass of solid stay constant.
Law of mass action: Kc and Kp
Link to Std 9–10 : You have used balanced equations to count moles; now the coefficients also decide the power of each concentration in the equilibrium expression.
A cricket scoreboard shows a ratio such as runs per over. An equilibrium constant is a ratio of products to reactants that stays fixed for a given temperature.
For aA + bB ⇌ cC + dD the equilibrium constant is Kc = [C]ᶜ[D]ᵈ ÷ ([A]ᵃ[B]ᵇ), with concentrations in mol L⁻¹ at equilibrium. For gases Kp uses partial pressures, and Kp = Kc (RT)^Δn.
Law of mass action: Kc and Kp
In the app you can move this figure with your finger.
Open in the appIn the figure : A large K (above about 10³) means the mixture is mostly products; a small K (below about 10⁻³) means mostly reactants. The calculator below converts Kc to Kp.
Rule 1: Writing K — Products on top, reactants below, each raised to its coefficient. Pure solids and pure liquids are left out because their concentration is constant.
- Easy : For 2SO₂ + O₂ ⇌ 2SO₃, Kc = [SO₃]² ÷ ([SO₂]²[O₂]).
- Medium : For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂] and Kp = p(CO₂).
- Exam-type : For H₂ + I₂ ⇌ 2HI with [HI] = 0.40, [H₂] = 0.20 and [I₂] = 0.10 mol L⁻¹, Kc = 0.40² ÷ (0.20 × 0.10) = 8.0.
Rule 2: Changing the equation — Reverse the reaction: K becomes 1 ÷ K. Multiply the equation by n: K becomes Kⁿ. Add two reactions: multiply their K values.
- Easy : If A ⇌ B has K = 4, then B ⇌ A has K = 0.25.
- Medium : For ½A ⇌ ½B, K = √4 = 2.
- Exam-type : For 2A ⇌ 2B, K = 4² = 16.
Rule 3: Kp and Kc — Kp = Kc (RT)^Δn with R = 0.0831 L bar K⁻¹ mol⁻¹ when pressures are in bar. Δn = moles of gaseous products − moles of gaseous reactants.
- Easy : If Δn = 0 (H₂ + I₂ ⇌ 2HI), then Kp = Kc.
- Medium : For N₂ + 3H₂ ⇌ 2NH₃ with Kc = 60 at 500 K (a sample value), Δn = −2 and RT = 0.0831 × 500 = 41.55, so Kp = 60 × 41.55⁻² ≈ 0.035.
- Exam-type : For PCl₅ ⇌ PCl₃ + Cl₂ with Kc = 0.50 at 400 K (sample value), Δn = +1, RT = 33.24, so Kp = 0.50 × 33.24 ≈ 16.6.
Rule 4: What K tells and does not tell — K shows the extent of reaction, not the speed. It changes only with temperature.
- Easy : A large K means that the reaction goes nearly to completion.
- Medium : A catalyst or a change of concentration does not change K.
- Exam-type : For an exothermic reaction K falls when temperature rises; for an endothermic reaction K rises.
K depends only on temperature, and on how the equation is written.
Including pure solids or liquids in the expression, or using initial instead of equilibrium concentrations.
Memory clue: "Products over reactants, powers = coefficients".
Spot the difference
| Point | Kc | Kp |
|---|---|---|
| Uses | Concentrations | Partial pressures |
| Relation | Kp = Kc (RT)^Δn | Kc = Kp (RT)^−Δn |
| Units | Powers of mol L⁻¹ | Powers of bar |
Practice questions
- 1Write Kc for N₂ + 3H₂ ⇌ 2NH₃.
- 2What is the effect of a catalyst on K?
- 3Why are pure solids left out of K?
- 4Find Kc when [HI] = 0.40, [H₂] = 0.20 and [I₂] = 0.10 mol L⁻¹.
- 5If A ⇌ B has K = 4, find K for 2B ⇌ 2A.
- 6Write Kc and Kp for CaCO₃(s) ⇌ CaO(s) + CO₂(g).
- 7Find Kp for N₂ + 3H₂ ⇌ 2NH₃ at 500 K if Kc = 60 (R = 0.0831 L bar K⁻¹ mol⁻¹).
- 8Find Kp for PCl₅ ⇌ PCl₃ + Cl₂ at 400 K if Kc = 0.50.
Show answers (try them yourself first)
- 1Kc = [NH₃]² ÷ ([N₂][H₂]³).
- 2None; K depends only on temperature.
- 3Their concentration is constant, so it is already included in the value of K.
- 40.40² ÷ (0.20 × 0.10) = 0.16 ÷ 0.020 = 8.0.
- 5Reverse: 0.25. Squared: 0.25² = 0.0625.
- 6Kc = [CO₂]; Kp = p(CO₂).
- 7Δn = −2, Kp = 60 × (0.0831 × 500)⁻² = 60 ÷ 1726 ≈ 0.035.
- 8Δn = +1, Kp = 0.50 × (0.0831 × 400) = 0.50 × 33.24 ≈ 16.6.
Reaction quotient and equilibrium calculations
Link to Std 9–10 : You solve equations in maths; here we set up an unknown x to find the equilibrium mixture.
A seesaw tilts to the heavier side until it levels. The reaction quotient tells which way the reaction will tilt.
The reaction quotient Q has the same form as K but uses the concentrations at any moment. Compare Q with K: if Q < K the reaction goes forward; if Q > K it goes backward; if Q = K it is at equilibrium.
In the figure : Q at the centre means equilibrium; Q above K pushes the mixture backward; Q below K pushes it forward.
Rule 1: Predicting direction — Calculate Q from the given concentrations and compare with K.
- Easy : For H₂ + I₂ ⇌ 2HI with K = 50, a mixture [H₂] = [I₂] = 0.5 and [HI] = 2.0 has Q = 2.0² ÷ (0.5 × 0.5) = 16. Q < K, so the reaction goes forward.
- Medium : With [HI] = 4.0 and the same H₂ and I₂, Q = 16 ÷ 0.25 = 64. Q > K, so the reaction goes backward.
- Exam-type : If Q = K the concentrations do not change, however they are written.
Rule 2: ICE table — Write Initial, Change and Equilibrium concentrations in a table, put the unknown change as x, and put equilibrium values into K.
- Easy : 1.0 mol each of H₂ and I₂ in 1.0 L with K = 50: change is −x for H₂ and I₂ and +2x for HI.
- Medium : K = (2x)² ÷ (1 − x)² = 50, so 2x ÷ (1 − x) = √50 = 7.07 and x = 7.07 ÷ 9.07 = 0.78. Then [HI] = 1.56 mol L⁻¹ and [H₂] = [I₂] = 0.22 mol L⁻¹.
- Exam-type : Check the answer by putting the values back: 1.56² ÷ 0.22² ≈ 50.
Rule 3: Degree of dissociation — The fraction of a reactant that breaks down to give products.
- Easy : For N₂O₄ ⇌ 2NO₂ starting with 0.50 mol L⁻¹ N₂O₄, let x dissociate: [N₂O₄] = 0.50 − x and [NO₂] = 2x.
- Medium : With Kc = 0.10 (sample value): (2x)² ÷ (0.50 − x) = 0.10, which gives 4x² + 0.10x − 0.050 = 0, so x = 0.10.
- Exam-type : So [NO₂] = 0.20, [N₂O₄] = 0.40 mol L⁻¹ and the degree of dissociation is 0.10 ÷ 0.50 = 20%.
Rule 4: Approximations — If K is very small, x is tiny compared with the initial concentration and the "(C − x)" can be written as "C". Check that x is less than about 5% of C.
- Easy : For K = 1 × 10⁻⁶ and C = 0.10, x² ÷ C = 10⁻⁶ gives x = 3.2 × 10⁻⁴, far less than C.
- Medium : The approximation is not allowed when x is more than 5% of C; then solve the quadratic equation.
- Exam-type : In the N₂O₄ example, x = 0.10 is 20% of 0.50, so the approximation would give a wrong answer.
Q ≠ K means the system is not yet at equilibrium and will move towards K.
Forgetting that the "2x" must be squared when the coefficient is 2.
Memory clue: "Q smaller, go forward; Q bigger, go back".
Spot the difference
| Point | Q | K |
|---|---|---|
| Concentrations | At any moment | At equilibrium |
| Value | Changes with time | Fixed at a temperature |
| Use | Gives the direction | Gives the extent |
Practice questions
- 1State what Q > K means.
- 2What does Q = K mean?
- 3What is an ICE table?
- 4For H₂ + I₂ ⇌ 2HI, K = 50. If [H₂] = [I₂] = 0.5 and [HI] = 2.0 mol L⁻¹, in which direction does the reaction go?
- 5Same reaction with [HI] = 4.0 mol L⁻¹: which direction?
- 6For H₂ + I₂ ⇌ 2HI (K = 50) start with 1.0 mol L⁻¹ each of H₂ and I₂. Find [HI] at equilibrium.
- 7For N₂O₄ ⇌ 2NO₂ with Kc = 0.10 and initial [N₂O₄] = 0.50 mol L⁻¹, find the equilibrium concentrations.
- 8Find the percentage dissociation in question 7.
Show answers (try them yourself first)
- 1The mixture has too many products; the reaction goes backward.
- 2The system is at equilibrium.
- 3A table of Initial, Change and Equilibrium concentrations used to solve for x.
- 4Q = 4.0 ÷ 0.25 = 16 < 50, so it goes forward.
- 5Q = 16 ÷ 0.25 = 64 > 50, so it goes backward.
- 6x = 0.78, so [HI] = 2x ≈ 1.56 mol L⁻¹.
- 7x = 0.10, so [NO₂] = 0.20 and [N₂O₄] = 0.40 mol L⁻¹.
- 80.10 ÷ 0.50 × 100 = 20%.
Le Chatelier's principle
Link to Std 9–10 : You know that factories make ammonia and sulphuric acid; Le Chatelier's principle explains which conditions give the best yield.
When a crowded bus gets another door opening, passengers shift to reduce the crowding. A system at equilibrium also shifts to reduce any disturbance.
Le Chatelier's principle: if a system at equilibrium is disturbed by a change of concentration, pressure or temperature, it shifts in the direction that reduces the effect of the change.
Le Chatelier's principle
In the app you can move this figure with your finger.
Open in the appIn the figure : The effect of four changes on N₂ + 3H₂ ⇌ 2NH₃ (exothermic). The live panel below lets you change the conditions and watch the shift.
Rule 1: Change of concentration — Adding a reactant shifts the equilibrium forward; adding a product shifts it backward. Removing a substance shifts towards making it. K does not change; Q changes.
- Easy : Adding more H₂ to the ammonia mixture makes more NH₃.
- Medium : Removing NH₃ as it forms keeps pushing the reaction forward.
- Exam-type : After adding extra reactant, Q becomes smaller than K, so the reaction goes forward until Q = K again.
Rule 2: Change of pressure — Raising the pressure shifts towards the side with fewer moles of gas. No effect if the moles of gas are equal.
- Easy : In N₂ + 3H₂ ⇌ 2NH₃ (4 → 2 moles), high pressure favours NH₃.
- Medium : For PCl₅ ⇌ PCl₃ + Cl₂ (1 → 2 moles), high pressure shifts the reaction backward.
- Exam-type : For H₂ + I₂ ⇌ 2HI (2 → 2 moles), pressure has no effect on the position.
Rule 3: Change of temperature — Heating favours the endothermic direction; cooling favours the exothermic direction. This is the only change that alters K.
- Easy : The Haber reaction is exothermic, so low temperature favours NH₃ but the rate is slow.
- Medium : For N₂O₄ ⇌ 2NO₂ (endothermic), heating makes the mixture browner as more NO₂ forms.
- Exam-type : In industry about 200 atm pressure, about 700 K and an iron catalyst are used for ammonia as a compromise between yield and rate.
Rule 4: Catalyst and inert gas — A catalyst speeds up both directions equally, so it does not shift the equilibrium. Adding an inert gas at constant volume does not change the partial pressures, so there is no shift.
- Easy : Adding iron to the ammonia reactor makes equilibrium come sooner but gives no extra NH₃ at equilibrium.
- Medium : Adding argon at constant volume does not change any concentration, so the equilibrium stays.
- Exam-type : Contact process, 2SO₂ + O₂ ⇌ 2SO₃ (exothermic, 3 → 2 moles): the best yield is at low temperature and high pressure.
Only temperature changes the value of K.
Saying that a catalyst increases the yield. It only saves time.
Memory clue: "Add it, it goes away; remove it, it comes back" — the system acts to undo the change.
Spot the difference
| Change | Shift | K changes? |
|---|---|---|
| Concentration | Yes | No |
| Pressure | Yes (if Δn ≠ 0) | No |
| Temperature | Yes | Yes |
| Catalyst | No | No |
Practice questions
- 1State Le Chatelier's principle.
- 2Which change shifts the equilibrium and also changes K?
- 3Does a catalyst shift the equilibrium?
- 4How does raising the pressure affect N₂ + 3H₂ ⇌ 2NH₃?
- 5How does heating affect an exothermic equilibrium?
- 6What is the effect of adding argon at constant volume?
- 7Predict the effect of high pressure and heating on PCl₅ ⇌ PCl₃ + Cl₂ (endothermic).
- 8Why is a moderately high temperature used for ammonia though the reaction is exothermic?
Show answers (try them yourself first)
- 1A system at equilibrium shifts in the direction that opposes a change imposed on it.
- 2A change of temperature.
- 3No; it only speeds up both directions equally.
- 4The equilibrium shifts forward (4 → 2 moles), giving more NH₃.
- 5It shifts backward, towards the reactants, and K decreases.
- 6Nothing; the partial pressures are the same.
- 7High pressure shifts it backward (fewer moles); heating shifts it forward (endothermic).
- 8A low temperature gives a higher yield but a very slow rate; the moderately high temperature and a catalyst give a good rate.
Acids, bases, pH and weak electrolytes
Link to Std 9–10 : You tested lemon juice and soap with indicators; now we measure acidity on the pH scale and learn why vinegar is milder than hydrochloric acid.
Tomato ketchup, soap and tap water all feel different in the mouth or on the hand. Their pH values tell how acidic or basic they are.
pH = −log[H⁺]. In water at 298 K, Kw = [H⁺][OH⁻] = 10⁻¹⁴, so pH + pOH = 14. A weak acid ionises only partly and its strength is given by Ka.
Acids, bases, pH and weak electrolytes
In the app you can move this figure with your finger.
Open in the appIn the figure : The pH scale from 0 (strongly acidic) through 7 (neutral) to 14 (strongly basic). Each step is a ten-fold change in [H⁺]. The calculator below finds the pH of a weak acid.
Rule 1: Definitions of acids and bases — Arrhenius: acids give H⁺ and bases give OH⁻ in water. Brønsted–Lowry: an acid is a proton donor and a base is a proton acceptor. Lewis: an acid accepts an electron pair and a base donates one.
- Easy : HCl donates a proton to water, so HCl is a Brønsted acid and H₂O is a Brønsted base.
- Medium : In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ the conjugate pairs are NH₄⁺/NH₃ and H₂O/OH⁻.
- Exam-type : BF₃ is a Lewis acid (it accepts a lone pair) although it has no H⁺ to give.
Rule 2: Strong acids and pH — A strong acid ionises completely, so [H⁺] = concentration of the acid.
- Easy : 0.01 M HCl has pH = −log 0.01 = 2.
- Medium : 0.001 M NaOH has pOH = 3, so pH = 14 − 3 = 11.
- Exam-type : Diluting a pH 1 solution 100 times (10 mL made up to 1000 mL) gives [H⁺] = 0.1 ÷ 100 = 10⁻³ mol L⁻¹, pH = 3.
Rule 3: Weak acids, Ka and degree of ionisation — For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻] ÷ [HA]. When ionisation is small, [H⁺] = √(Ka × C) and degree of ionisation α = √(Ka ÷ C).
- Easy : A weak acid with Ka = 2 × 10⁻⁵ and C = 0.05 M: [H⁺] = √(2 × 10⁻⁵ × 0.05) = 10⁻³ M, so pH = 3.
- Medium : For the same acid α = √(2 × 10⁻⁵ ÷ 0.05) = 0.02, which is 2%.
- Exam-type : Ka × Kb (of the conjugate base) = Kw. If Ka = 10⁻⁵, then Kb = 10⁻¹⁴ ÷ 10⁻⁵ = 10⁻⁹.
Rule 4: Salts and polyprotic acids — Salts of a strong acid with a weak base give acidic solutions, the reverse gives basic ones, and strong–strong gives neutral. For a polyprotic acid Ka1 > Ka2 > Ka3.
- Easy : NaCl solution is neutral.
- Medium : NH₄Cl solution is acidic; sodium acetate solution is basic.
- Exam-type : In H₃PO₄ the first proton comes off most easily because it leaves from a neutral molecule; later protons leave from negative ions.
A lower pH means a more acidic solution; a change of one unit means ten times change in [H⁺].
Using pH = −log C for a weak acid. Use [H⁺] = √(Ka C) instead.
Memory clue: "pH plus pOH equals 14", and for weak acids "root of Ka times C".
Spot the difference
| Point | Strong acid | Weak acid |
|---|---|---|
| Ionisation | Complete | Partial |
| [H⁺] | Equals C | √(Ka C) |
| Example | HCl | CH₃COOH |
Practice questions
- 1Define pH.
- 2Find the pH of 0.01 M HCl.
- 3Write the relation between Kw, [H⁺] and [OH⁻].
- 4Find the pH of 0.001 M NaOH.
- 5Find the pH of a 0.05 M weak acid with Ka = 2 × 10⁻⁵.
- 6Find Kb of the conjugate base of an acid whose Ka = 10⁻⁵.
- 7A solution of pH 1 is diluted 100 times. Find its pH.
- 8Why is NH₄Cl solution acidic?
Show answers (try them yourself first)
- 1pH = −log[H⁺].
- 2pH = −log 0.01 = 2.
- 3Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 298 K.
- 4pOH = 3, so pH = 11.
- 5[H⁺] = √(2 × 10⁻⁵ × 0.05) = 10⁻³ M, so pH = 3.
- 6Kb = 10⁻¹⁴ ÷ 10⁻⁵ = 10⁻⁹.
- 7[H⁺] falls from 0.1 to 0.001 mol L⁻¹, so pH = 3.
- 8NH₄⁺ (a weak acid) gives H⁺ to water, while Cl⁻ does not react, so [H⁺] > [OH⁻].
Buffers, common ion effect and solubility product
Link to Std 9–10 : You know that blood and sea water stay at an almost fixed pH; buffers explain how. You also know that some salts dissolve only slightly; Ksp measures how little.
A shock absorber in a scooter takes up bumps. A buffer takes up added acid or base so that the pH hardly moves.
A buffer resists a change in pH on adding small amounts of acid or base; it is a weak acid with its salt (or a weak base with its salt). The solubility product Ksp is the product of the ion concentrations (each to its power) in a saturated solution of a sparingly soluble salt.
Buffers, common ion effect and solubility product
In the app you can move this figure with your finger.
Open in the appBuffers, common ion effect and solubility product
In the app you can move this figure with your finger.
Open in the appIn the figure : Left: a buffer made from a weak acid HA and its salt A⁻. Right: AB(s) ⇌ A⁺ + B⁻, where a precipitate forms when the ionic product is larger than Ksp. The two calculators give the buffer pH and the solubility.
Rule 1: Henderson equation — pH = pKa + log ([salt] ÷ [acid]) for an acidic buffer.
- Easy : Acetic acid pKa = 4.75 with equal acid and salt: pH = 4.75.
- Medium : With [acetate] = 0.40 M and [acid] = 0.10 M: pH = 4.75 + log 4 = 5.35.
- Exam-type : With [acetate] = 0.20 M and [acid] = 0.40 M: pH = 4.75 + log 0.5 = 4.45.
Rule 2: Common ion effect — Adding a common ion to a solution of a weak electrolyte (or sparingly soluble salt) suppresses its ionisation (or solubility).
- Easy : Adding sodium acetate to acetic acid lowers [H⁺] because it adds CH₃COO⁻.
- Medium : AgCl is less soluble in NaCl solution than in pure water.
- Exam-type : In a buffer the common ion keeps the acid mostly un-ionised, which is why the ratio salt : acid fixes the pH.
Rule 3: Solubility product — For AB: Ksp = s² and for AB₂: Ksp = 4s³, where s is the solubility in mol L⁻¹.
- Easy : AB with Ksp = 1.0 × 10⁻¹⁰ has s = √(1.0 × 10⁻¹⁰) = 1.0 × 10⁻⁵ mol L⁻¹.
- Medium : AB₂ with Ksp = 4 × 10⁻¹² has 4s³ = 4 × 10⁻¹², so s³ = 10⁻¹² and s = 1 × 10⁻⁴ mol L⁻¹.
- Exam-type : AB (Ksp = 1.0 × 10⁻¹⁰) in 0.010 M of B⁻ has s = Ksp ÷ [B⁻] = 10⁻¹⁰ ÷ 10⁻² = 1.0 × 10⁻⁸ mol L⁻¹, far lower than in pure water.
Rule 4: Precipitation — Compare the ionic product Q with Ksp. If Q > Ksp, a precipitate forms; if Q < Ksp, the solution is unsaturated.
- Easy : For AB with Ksp = 10⁻¹⁰, mixing to [A⁺] = [B⁻] = 10⁻³ gives Q = 10⁻⁶ > Ksp, so a precipitate forms.
- Medium : With [A⁺] = [B⁻] = 10⁻⁶, Q = 10⁻¹² < Ksp, so there is no precipitate.
- Exam-type : For AB₂ (Ksp = 4 × 10⁻¹²) with [A²⁺] = 10⁻⁵ and [B⁻] = 10⁻⁴, Q = 10⁻⁵ × (10⁻⁴)² = 10⁻¹³ < Ksp, so there is no precipitate.
A buffer works best when the salt : acid ratio is near 1, so pH is near pKa.
Using s² for AB₂ salts. Use 4s³ for AB₂ or A₂B.
Memory clue: "AB gives s squared; AB₂ gives four s cubed".
Spot the difference
| Point | Ionic product Q | Ksp |
|---|---|---|
| Meaning | Product at any time | Product in a saturated solution |
| Use | Predicts a precipitate when Q > Ksp | Fixed reference value |
| Changes with | Concentration | Temperature only |
Practice questions
- 1Define a buffer solution.
- 2What is the pH of an acetic acid and sodium acetate buffer with equal concentrations (pKa = 4.75)?
- 3State the common ion effect.
- 4Find the pH when [acetate] = 0.40 M and [acetic acid] = 0.10 M (pKa = 4.75).
- 5Find the solubility of AB with Ksp = 1.0 × 10⁻¹⁰.
- 6Find the solubility of AB₂ with Ksp = 4 × 10⁻¹².
- 7Find the solubility of AB (Ksp = 1.0 × 10⁻¹⁰) in 0.010 M B⁻.
- 8Will a precipitate form if [A⁺] = [B⁻] = 10⁻³ mol L⁻¹ and Ksp = 10⁻¹⁰?
Show answers (try them yourself first)
- 1A solution that resists a change in pH when small amounts of acid or base are added.
- 2pH = pKa = 4.75.
- 3A common ion lowers the ionisation of a weak electrolyte or the solubility of a sparingly soluble salt.
- 44.75 + log 4 = 4.75 + 0.60 = 5.35.
- 5s = √(1.0 × 10⁻¹⁰) = 1.0 × 10⁻⁵ mol L⁻¹.
- 64s³ = 4 × 10⁻¹², so s = 1 × 10⁻⁴ mol L⁻¹.
- 7s = 10⁻¹⁰ ÷ 10⁻² = 1.0 × 10⁻⁸ mol L⁻¹.
- 8Q = 10⁻⁶ > 10⁻¹⁰, so a precipitate forms.