Equilibrium Class 11 Chemistry: Chemical and Ionic Equilibrium with Solved Numericals

What Equilibrium is about

Weightage: This chapter is Unit VI, Equilibrium, which the syllabus gives 7 of the 70 theory marks in the school annual exam.Source: CBSE Chemistry (043) syllabus 2026-27, Class XI course structure, Unit VI Equilibrium

Key concepts

Equilibrium in physical and chemical processes; dynamic equilibrium

NCERT section 6.1 Equilibrium in Physical Processes; 6.2 Equilibrium in Chemical Processes – Dynamic Equilibrium

Physical equilibria involve changes of state or dissolution:

• Solid ⇌ liquid: ice and water at 273 K and 1 atm; the rate of melting equals the rate of freezing. The temperature at which they coexist at 1 atm is the normal melting point.
• Liquid ⇌ vapour: in a closed container, evaporation and condensation reach equal rates; the pressure of vapour then is the equilibrium vapour pressure. The boiling point is the temperature at which this equals the atmospheric pressure.
• Solid ⇌ vapour: sublimation of iodine or camphor in a closed vessel.
• Solid in liquid: a saturated solution, where dissolution and crystallisation rates are equal.
• Gas in liquid: CO₂ in a soda bottle, governed by Henry's law: mass dissolved is proportional to the pressure of the gas.

General features: equilibrium is reached only in a closed system at constant temperature; it is dynamic (both processes continue at equal rates); measurable properties become constant; and the concentrations at equilibrium relate through a constant.

Chemical equilibrium: for A + B ⇌ C + D, the forward rate falls and the backward rate rises until they are equal. Equilibrium can be reached from either side. Experiments with radioactive isotopes (such as making ammonia with deuterium, D₂) show that reactant and product molecules keep exchanging even at equilibrium, which proves it is dynamic.

Worked example

A small amount of water is placed in a closed flask at 300 K. (a) Describe what happens to the rates of evaporation and condensation with time. (b) What can you say about the vapour pressure once equilibrium is reached? (c) Why would equilibrium never be reached if the flask were open? (3 marks)

Try it first, then show the solutionSolution

1. (a) At first only evaporation occurs; as vapour builds up, condensation starts and speeds up while evaporation stays constant, until the two rates become equal. (1)
2. (b) The vapour pressure then stays constant at the equilibrium vapour pressure of water at 300 K, though molecules keep escaping and returning (dynamic). (1)
3. (c) In an open vessel, vapour escapes into the room, so condensation can never catch up; the water simply evaporates completely. (1)

Answer: Rates become equal (dynamic equilibrium) and vapour pressure becomes constant; an open vessel lets vapour escape, so no equilibrium.

Law of chemical equilibrium and the equilibrium constant

NCERT section 6.3 Law of Chemical Equilibrium and Equilibrium Constant; 6.4 Homogeneous Equilibria (6.4.1, 6.4.2)

For a general reaction $aA+bB\rightleftharpoons cC+dD$, at a fixed temperature,
$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$
where the square brackets are molar concentrations at equilibrium. This is the law of chemical equilibrium (Guldberg and Waage's law of mass action).

Rules for K:

• K depends only on temperature, not on starting amounts.
• For the reverse reaction, $K^{\prime }=\frac{1}{K}$.
• If the equation is multiplied by n, $K^{\prime }=K^n$ (for halving it, $K^{\prime }=\sqrt{K}$).
• If two equations are added, their K values multiply.
• K has units that depend on Δn, but at standard states it is treated as dimensionless.

Gas-phase equilibria: using partial pressures,
$K_p=\frac{p_C^c{p}_D^d}{p_A^a{p}_B^b}$ and $K_p=K_c(RT{)}^{\Delta n}$
where Δn = (c + d) − (a + b), counting gaseous moles only, and R = 0.0831 L bar K⁻¹ mol⁻¹ (when pressure is in bar).

• Δn = 0 (H₂ + I₂ ⇌ 2HI): Kp = Kc.
• Δn ≠ 0: Kp ≠ Kc.

Worked example

For $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$, Kc = 0.50 at 700 K (concentrations in mol L⁻¹). Calculate (a) Kc for $2N{H}_3(g)\rightleftharpoons N_2(g)+3H_2(g)$, (b) Kc for $\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\rightleftharpoons N{H}_3$, and (c) Kp for the original reaction. (R = 0.0831 L bar K⁻¹ mol⁻¹) (3 marks)

Try it first, then show the solutionSolution

1. (a) Reverse: K' = 1/0.50 = 2.0. (1)
2. (b) Equation halved: K'' = √0.50 = 0.71. (1)
3. (c) Δn = 2 − 4 = −2; Kp = Kc(RT)⁻² = 0.50 ÷ (0.0831 × 700)² = 0.50 ÷ 3384 = 1.48 × 10⁻⁴. (1)

Answer: (a) 2.0 (b) 0.71 (c) Kp ≈ 1.5 × 10⁻⁴

Heterogeneous equilibria and applications of K: extent, direction and equilibrium concentrations

NCERT section 6.5 Heterogeneous Equilibria; 6.6 Applications of Equilibrium Constants; 6.7 Relationship between K, Q and G

Heterogeneous equilibrium: substances are in more than one phase, for example $CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$. The concentration (molar density) of a pure solid or pure liquid is constant, so it is left out of K: here $K_p=p_{C{O}_2}$. Pure solids and liquids must still be present for the equilibrium to exist, but their amounts do not matter.

Using K:
1. Extent of reaction: if K > 10³, products dominate (nearly complete); if K < 10⁻³, reactants dominate (hardly any reaction); between these, both are present in appreciable amounts.
2. Direction: the reaction quotient Q has the same form as K but uses concentrations at any moment.

• Q < K: net reaction goes forward (towards products).
• Q > K: net reaction goes backward.
• Q = K: at equilibrium.

3. Equilibrium concentrations: use an ICE table (Initial, Change, Equilibrium) in terms of x, write the K expression, and solve.

K, Q and G: $\Delta G=\Delta G^{\ominus }+RT\ln Q$. At equilibrium ΔG = 0 and Q = K, so $\Delta G^{\ominus }=-RT\ln K$, or $K=e^{-\Delta G^{\ominus }/RT}$. Negative ΔG⦵ gives K > 1.

Worked example

For $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$, Kc = 49 at a certain temperature. 0.50 mol of H₂ and 0.50 mol of I₂ are placed in a 1.0 L flask. Calculate the equilibrium concentrations. (3 marks)

Try it first, then show the solutionSolution

1. ICE: H₂ and I₂ each 0.50 − x; HI = 2x. (1)
2. Kc = (2x)² / (0.50 − x)² = 49, so 2x / (0.50 − x) = 7; 2x = 3.5 − 7x; 9x = 3.5; x = 0.389. (1)
3. [HI] = 0.778 M; [H₂] = [I₂] = 0.50 − 0.389 = 0.111 M. Check: 0.778² ÷ 0.111² = 49 ✓. (1)

Answer: [HI] ≈ 0.78 M; [H₂] = [I₂] ≈ 0.11 M

Factors affecting equilibria: Le Chatelier's principle

NCERT section 6.8 Factors Affecting Equilibria

Le Chatelier's principle: if a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to undo the change.

• Concentration: adding a reactant (or removing a product) shifts the equilibrium forward; adding a product shifts it back. K does not change; the system adjusts until Q = K again. Example: adding SCN⁻ to the red Fe³⁺ + SCN⁻ ⇌ [Fe(SCN)]²⁺ mixture deepens the colour; removing Fe³⁺ with oxalic acid fades it.
• Pressure (by changing volume): increasing pressure shifts the equilibrium towards the side with fewer gas molecules. For N₂ + 3H₂ ⇌ 2NH₃, high pressure favours ammonia. If Δn_g = 0 (H₂ + I₂ ⇌ 2HI), pressure has no effect.
• Inert gas: adding argon at constant volume does not change the partial pressures, so the equilibrium is unaffected.
• Temperature: the only factor that changes K. For an exothermic forward reaction (ΔH < 0), raising T lowers K and shifts the equilibrium back; for an endothermic one, raising T raises K. Example: the pink [Co(H₂O)₆]²⁺ ⇌ blue [CoCl₄]²⁻ equilibrium changes colour with temperature, and N₂O₄ (colourless) ⇌ 2NO₂ (brown) is endothermic, so the gas gets browner when heated.
• Catalyst: speeds up forward and backward reactions equally, so equilibrium is reached faster but its position and K are unchanged.

Industrial use: the Haber process (ammonia, exothermic, fewer gas moles) uses high pressure (about 200 atm), a moderate temperature (about 700 K) as a compromise between yield and rate, and an iron catalyst.

Worked example

For $2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$, ΔH = −197 kJ mol⁻¹. Predict the effect on the amount of SO₃ of (a) increasing pressure, (b) raising temperature, (c) adding a catalyst, (d) adding helium at constant volume. (2 marks)

Try it first, then show the solutionSolution

1. (a) 3 gas moles → 2: higher pressure shifts forward; SO₃ increases. (½)
2. (b) Exothermic forward: higher temperature shifts back; SO₃ decreases (K falls). (½)
3. (c) No change in amount; equilibrium is reached faster. (½)
4. (d) No change: partial pressures of the reacting gases are unchanged. (½)

Answer: (a) increases (b) decreases (c) no change (d) no change

Ionic equilibrium; acids, bases and salts

NCERT section 6.9 Ionic Equilibrium in Solution; 6.10 Acids, Bases and Salts

Substances that conduct electricity in water are electrolytes. Strong electrolytes (NaCl, HCl) ionise almost completely; weak electrolytes (acetic acid, ammonia) ionise only partly, and an equilibrium exists between ions and un-ionised molecules: this is ionic equilibrium.

Three definitions of acids and bases:
1. Arrhenius: acids give H⁺ (as H₃O⁺, the hydronium ion) in water; bases give OH⁻ in water. Limited to aqueous solutions.
2. Brønsted-Lowry: an acid is a proton donor; a base is a proton acceptor. When an acid loses a proton it becomes its conjugate base; when a base gains one it becomes its conjugate acid. In $HCl+H_{2}O\rightleftharpoons H_3{O}^{+}+C{l}^{-}$, Cl⁻ is the conjugate base of HCl, and H₃O⁺ the conjugate acid of water. A strong acid has a very weak conjugate base. Water is amphoteric: it acts as an acid with NH₃ and as a base with HCl.
3. Lewis: an acid is an electron-pair acceptor (BF₃, AlCl₃, H⁺, Co³⁺); a base is an electron-pair donor (NH₃, H₂O, OH⁻, F⁻). This includes substances without protons.

Worked example

Write the conjugate acid and the conjugate base of (a) HS⁻ (b) HPO₄²⁻. Classify Fe³⁺ and CN⁻ as Lewis acids or bases. (2 marks)

Try it first, then show the solutionSolution

1. (a) HS⁻: gaining H⁺ gives the conjugate acid H₂S; losing H⁺ gives the conjugate base S²⁻. (½)
2. (b) HPO₄²⁻: conjugate acid H₂PO₄⁻, conjugate base PO₄³⁻. (½)
3. Fe³⁺ has empty orbitals and accepts electron pairs (for example from water or CN⁻): Lewis acid. CN⁻ donates a lone pair: Lewis base. (1)

Answer: (a) H₂S / S²⁻ (b) H₂PO₄⁻ / PO₄³⁻; Fe³⁺ Lewis acid, CN⁻ Lewis base

Ionisation of water, pH, and weak acids and bases

NCERT section 6.11 Ionization of Acids and Bases (6.11.1 to 6.11.5)

Water ionises slightly: $2H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$. The ionic product of water is
$K_w=[H_3{O}^{+}][O{H}^{-}]=1.0\times 10^{-14}$ at 298 K.
In pure water both concentrations are 10⁻⁷ M. Kw rises with temperature (ionisation is endothermic).

pH = −log[H₃O⁺] (Sørensen). At 298 K, pH + pOH = 14. Acidic: pH < 7; neutral: 7; basic: pH > 7. A change of 1 pH unit means a tenfold change in [H⁺].

Strong acids and bases ionise completely: [H⁺] = concentration × number of H⁺ per formula (for HCl, 0.01 M gives pH 2).

Weak acid HA with concentration c and degree of ionisation α: $HA+H_{2}O\rightleftharpoons H_3{O}^{+}+A^{-}$.
$K_a=\frac{c{\alpha }^2}{1-\alpha }\approx c{\alpha }^2$ (when α is small), so $\alpha =\sqrt{\frac{K_a}{c}}$ and $[H^{+}]=c\alpha =\sqrt{K_{a}c}$. The larger the Ka (smaller pKa), the stronger the acid. pKa = −log Ka.

Weak base BOH: $K_b\approx c{\alpha }^2$ and $[O{H}^{-}]=\sqrt{K_{b}c}$.

Conjugate pairs: $K_a\times K_b=K_w$, or pKa + pKb = 14. Strong acid ↔ weak conjugate base, and vice versa.

Polybasic acids (H₂S, H₂CO₃, H₃PO₄, oxalic acid) ionise in steps: $K_{a1}>K_{a2}>K_{a3}$, because removing a proton from a negative ion is harder.

Acid strength depends on the strength of the H–A bond and the polarity of the bond: down a group bond strength dominates (HF < HCl < HBr < HI); across a period electronegativity dominates (CH₄ < NH₃ < H₂O < HF).

Worked example

Calculate the pH and the degree of ionisation of 0.10 M acetic acid. (Ka = 1.8 × 10⁻⁵; log 1.34 = 0.127) (3 marks)

Try it first, then show the solutionSolution

1. α = √(Ka/c) = √(1.8 × 10⁻⁵ ÷ 0.10) = √(1.8 × 10⁻⁴) = 1.34 × 10⁻² (1.34%). (1)
2. [H⁺] = cα = 0.10 × 1.34 × 10⁻² = 1.34 × 10⁻³ M. (1)
3. pH = −log(1.34 × 10⁻³) = 3 − 0.127 = 2.87. (1)

Answer: pH ≈ 2.87; α ≈ 0.0134 (1.34%)

Common ion effect and hydrolysis of salts

NCERT section 6.11 (6.11.6 Common Ion Effect; 6.11.7 Hydrolysis of Salts and the pH of their Solutions)

Common ion effect: adding an ion already present in a weak electrolyte's equilibrium suppresses its ionisation (Le Chatelier). Adding sodium acetate to acetic acid adds CH₃COO⁻, so CH₃COOH ⇌ CH₃COO⁻ + H⁺ shifts back and [H⁺] falls (pH rises). Similarly NH₄Cl suppresses the ionisation of NH₄OH.

Hydrolysis of salts: ions of salts of weak acids or weak bases react with water and change the pH.
1. Strong acid + strong base (NaCl, KNO₃): no hydrolysis, pH = 7.
2. Weak acid + strong base (CH₃COONa): the anion takes H⁺ from water and releases OH⁻: basic solution.
$pH=7+\frac{1}{2}(p{K}_a+\log c)$
3. Strong acid + weak base (NH₄Cl): the cation gives H⁺ to water: acidic solution.
$pH=7-\frac{1}{2}(p{K}_b+\log c)$
4. Weak acid + weak base (CH₃COONH₄): both ions hydrolyse; the pH is independent of concentration:
$pH=7+\frac{1}{2}(p{K}_a-p{K}_b)$
(close to 7 for ammonium acetate since pKa ≈ pKb).

Worked example

Calculate the pH of 0.10 M sodium acetate solution. Explain why it is basic. (pKa of acetic acid = 4.74) (2 marks)

Try it first, then show the solutionSolution

1. pH = 7 + ½(pKa + log c) = 7 + ½(4.74 + log 0.10) = 7 + ½(4.74 − 1) = 7 + 1.87 = 8.87. (1)
2. CH₃COO⁻ is the conjugate base of a weak acid: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, producing OH⁻, so the solution is basic. (1)

Answer: pH ≈ 8.87; acetate ions hydrolyse to give OH⁻.

Buffer solutions

NCERT section 6.12 Buffer Solutions (6.12.1 Designing Buffer Solution)

A buffer solution resists a change in pH when it is diluted or when a small amount of acid or base is added. Blood (pH about 7.4), many medicines and cosmetics need buffers.

• Acidic buffer: a weak acid and its salt with a strong base (CH₃COOH + CH₃COONa), pH about 4.75.
• Basic buffer: a weak base and its salt with a strong acid (NH₄OH + NH₄Cl), pH about 9.25.

How it works (acetate buffer): added H⁺ is mopped up by the large reserve of CH₃COO⁻ (forming CH₃COOH); added OH⁻ is neutralised by CH₃COOH (forming CH₃COO⁻ and water). The ratio [salt]/[acid] changes only a little, so the pH barely changes.

Designing a buffer (Henderson-Hasselbalch equation):
$pH=p{K}_a+\log \frac{[\text{conjugate base, }{A}^{-}]}{[\text{acid, }HA]}$
for an acidic buffer, and for a basic buffer
$pOH=p{K}_b+\log \frac{[\text{conjugate acid, }B{H}^{+}]}{[\text{base, }B]}$.
When [A⁻] = [HA], pH = pKa: choose a weak acid whose pKa is close to the pH you need, then adjust the ratio. Diluting a buffer does not change the ratio, so the pH stays almost the same.

Worked example

Calculate the pH of a buffer containing 0.10 M acetic acid and 0.20 M sodium acetate. What happens to the pH if the solution is diluted to twice its volume? (pKa = 4.74; log 2 = 0.301) (2 marks)

Try it first, then show the solutionSolution

1. pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.20/0.10) = 4.74 + 0.301 = 5.04. (1)
2. On dilution both concentrations halve, so their ratio and the pH stay the same, 5.04. (1)

Answer: pH ≈ 5.04, unchanged on dilution

Solubility equilibria of sparingly soluble salts

NCERT section 6.13 Solubility Equilibria of Sparingly Soluble Salts (6.13.1, 6.13.2)

Salts can be classed by solubility: soluble (more than 0.1 M), slightly soluble (0.01 to 0.1 M) and sparingly soluble (less than 0.01 M). Solubility depends on the balance between lattice enthalpy (holding ions together) and solvation enthalpy (released when ions are surrounded by solvent).

For a sparingly soluble salt in a saturated solution, there is an equilibrium between the undissolved solid and its ions. For $BaS{O}_4(s)\rightleftharpoons B{a}^{2+}(aq)+S{O}_4^{2-}(aq)$,
$K_{sp}=[B{a}^{2+}][S{O}_4^{2-}]$
The solid is left out (constant). $K_{sp}$ is the solubility product constant.

Relation to molar solubility S (mol L⁻¹), for $M_x{X}_y\rightleftharpoons x{M}^{y+}+y{X}^{x-}$:
$K_{sp}=(xS{)}^x(yS{)}^y=x^x{y}^y{S}^{x+y}$

• AB type (AgCl, BaSO₄): $K_{sp}=S^2$.
• AB₂ or A₂B type (CaF₂, Ag₂CrO₄): $K_{sp}=4S^3$.
• A₂B₃ type (Bi₂S₃): $K_{sp}=108S^5$.

The ionic product Q_sp (same expression with current concentrations) predicts precipitation: if Q_sp > Ksp, a precipitate forms; if Q_sp < Ksp, more solid can dissolve.

Common ion effect on solubility: solubility falls sharply when a common ion is already present (AgCl is far less soluble in NaCl solution than in pure water). This is used to precipitate salts completely, for example to purify sodium chloride by passing HCl gas into its saturated solution.

Worked example

Ksp of AgCl at 298 K is 1.8 × 10⁻¹⁰. Calculate its molar solubility (a) in pure water and (b) in 0.10 M NaCl. Comment. (3 marks)

Try it first, then show the solutionSolution

1. (a) Ksp = S², so S = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol L⁻¹. (1)
2. (b) [Cl⁻] ≈ 0.10 M from NaCl; Ksp = S × 0.10, so S = 1.8 × 10⁻¹⁰ ÷ 0.10 = 1.8 × 10⁻⁹ mol L⁻¹. (1)
3. The solubility drops about 7500 times because of the common ion Cl⁻. (1)

Answer: (a) 1.34 × 10⁻⁵ M (b) 1.8 × 10⁻⁹ M; common ion effect

Chemistry in real life: Equilibrium

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Equilibrium Constant, Reaction Quotient and Le Chatelier's Principle
• Ionic Equilibrium: pH, Weak Electrolytes, Buffers and Solubility Product

Equilibrium formula sheet

Equilibrium constant (concentrations)

$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$

Leave out pure solids and liquids.

Kp and Kc

$K_p=K_c(RT{)}^{\Delta n_g}$

R = 0.0831 L bar K⁻¹ mol⁻¹; Δn_g = gas moles of products − reactants.

Manipulating K

$K_{reverse}=\frac{1}{K},K_{n\times eqn}=K^n,K_{sum}=K_1{K}_2$

Halving an equation gives √K.

K and Gibbs energy

$\Delta G=\Delta G^{\ominus }+RT\ln Q,\Delta G^{\ominus }=-2.303RT\log K$

Q < K: forward; Q > K: backward.

Ionic product of water and pH

$K_w=[H_3{O}^{+}][O{H}^{-}]=10^{-14},pH=-\log [H_3{O}^{+}],pH+pOH=14$

At 298 K.

Weak acid and weak base

$K_a=\frac{c{\alpha }^2}{1-\alpha }\approx c{\alpha }^2,[H^{+}]=\sqrt{K_{a}c},[O{H}^{-}]=\sqrt{K_{b}c}$

Use the approximation when α < 5%.

Conjugate pair

$K_a\times K_b=K_w,p{K}_a+p{K}_b=14$

At 298 K.

pH of salt solutions

$pH=7+\frac{1}{2}(p{K}_a+\log c);pH=7-\frac{1}{2}(p{K}_b+\log c);pH=7+\frac{1}{2}(p{K}_a-p{K}_b)$

Salt of weak acid + strong base; strong acid + weak base; weak acid + weak base.

Henderson-Hasselbalch equation

$pH=p{K}_a+\log \frac{[A^{-}]}{[HA]},pOH=p{K}_b+\log \frac{[B{H}^{+}]}{[B]}$

Acidic and basic buffers.

Solubility product

$K_{sp}=x^x{y}^y{S}^{x+y}$

AB: S²; AB₂: 4S³; A₂B₃: 108S⁵. Precipitation if Q_sp > Ksp.

Equilibrium: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Equilibrium

The question patterns to expect, and the method that works for each.

Tricks to remember Equilibrium

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Q vs K: the see-saw

  Q smaller than K: not enough product yet, so go forward. Q bigger than K: too much product, go back.

  Use it when: Predicting direction.

• Only T changes K

  Concentration, pressure, inert gas and catalyst shift or speed things up; only temperature changes the number K.

  Use it when: Assertion-reason and MCQs.

• Strong acid, weak partner

  The stronger the acid, the weaker its conjugate base: Ka × Kb = Kw keeps them on a see-saw.

  Use it when: Conjugate pair questions.

• Root of K times c

  Weak acid: [H⁺] = √(Ka c); weak base: [OH⁻] = √(Kb c).

  Use it when: Quick pH of weak electrolytes.

• Ksp patterns 1-4-27-108

  AB: S²; AB₂: 4S³; AB₃: 27S⁴; A₂B₃: 108S⁵.

  Use it when: Ksp and solubility conversions.

Equilibrium: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Equilibrium

The slips examiners see most often, and what to write instead.

• ✗Mistake: Including solids or liquids in K

  Students copy every species into the expression.

  Wrong working

  CaCO₃(s) ⇌ CaO(s) + CO₂(g): Kc = [CaO][CO₂]/[CaCO₃]

  Right working

  Kc = [CO₂]; Kp = p(CO₂)

  ✓Do this: Strike out (s) and (l) before writing K.

  Usually costs: 1 mark.

• ✗Mistake: Using total moles instead of gas moles in Δn

  State symbols are ignored.

  ✓Do this: Count only (g) species.

  Usually costs: 1 mark.

• ✗Mistake: Saying a catalyst increases yield

  Faster rate is confused with more product.

  ✓Do this: A catalyst helps reach equilibrium faster; K and yield are unchanged.

  Usually costs: 1 mark.

• ✗Mistake: Treating a weak acid as strong

  Students write [H⁺] = c.

  Wrong working

  0.10 M CH₃COOH: pH = 1

  Right working

  [H⁺] = √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ M; pH = 2.87

  ✓Do this: Check whether a Ka is given or whether the acid is in the weak list.

  Usually costs: 2 marks.

• ✗Mistake: Forgetting the coefficient in Ksp

  Students write Ksp = S² for every salt.

  Wrong working

  Ag₂CrO₄: Ksp = S²

  Right working

  Ag₂CrO₄ → 2Ag⁺ + CrO₄²⁻: Ksp = (2S)²(S) = 4S³

  ✓Do this: Write the dissociation and express each ion in terms of S.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong pH of a strong base

  Students take −log[OH⁻] as pH.

  ✓Do this: Find pOH = −log[OH⁻], then pH = 14 − pOH. Count two OH⁻ for Ba(OH)₂.

  Usually costs: 1 mark.

NCERT exercise guide: Equilibrium

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Equilibrium

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The conjugate base of H₂PO₄⁻ is:
   (a) H₃PO₄ (b) HPO₄²⁻ (c) PO₄³⁻ (d) H₂PO₃⁻

   Hint

   Remove one H⁺.

   Answer

   (b) HPO₄²⁻

2. Q2 · 1 mark · MCQExam level

   Which of the following is a Lewis acid?
   (a) NH₃ (b) H₂O (c) AlCl₃ (d) OH⁻

   Hint

   Electron-pair acceptor.

   Answer

   (c) AlCl₃

3. Q3 · 2 marks · Very short answerExam level

   Write the expressions for Kc and Kp for $3Fe(s)+4H_{2}O(g)\rightleftharpoons F{e}_3{O}_4(s)+4H_2(g)$. (2 marks)

   Hint

   Leave out solids.

   Answer

   Kc = [H₂]⁴/[H₂O]⁴ (1); Kp = p(H₂)⁴/p(H₂O)⁴ (1). Here Δn = 0, so Kp = Kc.

4. Q4 · 2 marks · Very short answerExam level

   What is the pH of 1.0 × 10⁻³ M HNO₃, and what is [OH⁻] in it? (2 marks)

   Hint

   Strong acid.

   Answer

   [H⁺] = 10⁻³ M, pH = 3 (1); [OH⁻] = 10⁻¹⁴ ÷ 10⁻³ = 10⁻¹¹ M (1).

5. Q5 · 3 marks · Short answerHarder exam level

   Suppose 0.20 mol of HI is placed in a 1.0 L flask at a temperature where Kc for $2HI(g)\rightleftharpoons H_2(g)+I_2(g)$ is 0.016. Calculate the equilibrium concentrations. (3 marks)

   Hint

   Let 2x of HI react; take a square root.

   Answer

   HI = 0.20 − 2x; H₂ = I₂ = x (1). x²/(0.20 − 2x)² = 0.016, so x/(0.20 − 2x) = 0.1265; x = 0.0253 − 0.253x; x = 0.0202 M (1). [H₂] = [I₂] ≈ 0.020 M; [HI] ≈ 0.16 M (1).

6. Q6 · 2 marks · Very short answerHarder exam level

   The pH of 0.10 M solution of a monobasic acid is 3.0. Calculate Ka and the degree of ionisation. (2 marks)

   Hint

   [H⁺] = 10⁻³.

   Answer

   α = 10⁻³ ÷ 0.10 = 0.010 (1%) (1). Ka ≈ cα² = 0.10 × (0.010)² = 1.0 × 10⁻⁵ (1).

7. Q7 · 3 marks · Short answerHarder exam level

   Predict whether the solutions of NaNO₃, NH₄Cl and Na₂CO₃ are acidic, basic or neutral, with reasons. (3 marks)

   Hint

   Parent acid and base strengths.

   Answer

   NaNO₃: strong acid + strong base, no hydrolysis, neutral (1). NH₄Cl: NH₄⁺ hydrolyses giving H₃O⁺, acidic (1). Na₂CO₃: CO₃²⁻ hydrolyses giving OH⁻, basic (1).

8. Q8 · 2 marks · Very short answerHarder exam level

   Equal volumes of 0.002 M AgNO₃ and 0.002 M NaCl are mixed. Will AgCl precipitate? (Ksp = 1.8 × 10⁻¹⁰) (2 marks)

   Hint

   Concentrations halve on mixing.

   Answer

   After mixing [Ag⁺] = [Cl⁻] = 0.001 M; ionic product = 1.0 × 10⁻⁶ (1). This exceeds 1.8 × 10⁻¹⁰, so AgCl precipitates (1).

9. Q9 · 2 marks · Very short answerHarder exam level

   Arrange in increasing acid strength and explain: HF, HCl, HBr, HI. (2 marks)

   Hint

   Bond strength down a group.

   Answer

   HF < HCl < HBr < HI (1). Down the group the H–X bond gets longer and weaker, so H⁺ is released more easily; bond strength matters more than electronegativity here (1).

10. Q10 · 3 marks · Short answerHOTS / competency

    Suppose a chemist dissolves 0.10 mol each of acetic acid and sodium acetate in water to make 1.0 L. She then adds 0.010 mol of NaOH. Calculate the pH before and after, and compare with the pH change if the same NaOH were added to 1.0 L of pure water. (pKa = 4.74; log 1.22 = 0.087) (3 marks)

    Hint

    NaOH converts acid to acetate.

    Answer

    Before: ratio 1, pH = 4.74 (1). After: acid 0.09, acetate 0.11; pH = 4.74 + log(0.11/0.09) = 4.74 + 0.087 = 4.83 (1). In water: [OH⁻] = 0.010 M, pH 12, a change of 5 units from 7, against only 0.09 units for the buffer (1).

11. Q11 · 5 marks · Long answerHOTS / competency

    For the reaction $C(s)+C{O}_2(g)\rightleftharpoons 2CO(g)$, Kp = 3.0 at 1000 K. (a) If the total pressure at equilibrium is 4.0 atm, find the partial pressures of CO and CO₂. (b) How would adding more carbon affect the equilibrium? (c) The reaction is endothermic. Predict the effect of raising the temperature on Kp and on the fraction of CO. (5 marks)

    Hint

    Let p(CO) = x; p(CO₂) = 4 − x.

    Answer

    (a) x²/(4.0 − x) = 3.0, so x² + 3x − 12 = 0; x = [−3 + √(9 + 48)] ÷ 2 = (−3 + 7.55) ÷ 2 = 2.27 atm (2). p(CO) ≈ 2.27 atm; p(CO₂) ≈ 1.73 atm (1). (b) No effect: carbon is a pure solid and does not appear in Kp (1). (c) Endothermic forward: Kp increases with temperature, so the fraction of CO increases (1).

12. Q12 · 1 mark · Assertion-reasonHOTS / competency

    Assertion (A): Diluting a 1.0 × 10⁻⁶ M HCl solution a thousand times does not give a solution of pH 9.
    Reason (R): At very low acid concentrations, the H⁺ from the ionisation of water must also be taken into account.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

    Hint

    An acid cannot have a basic pH.

    Answer

    (a) After dilution the acid gives only 10⁻⁹ M H⁺, far less than the 10⁻⁷ M from water, so the solution stays just below pH 7; diluting an acid can never make it basic.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Why is equilibrium called dynamic?Show answer

   Forward and backward reactions continue at equal rates, so concentrations stay constant though reactions do not stop.

2. 2.Write Kp for $CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)$.Show answer

   Kp = p(CO₂).

3. 3.When is Kp equal to Kc?Show answer

   When Δn_g = 0.

4. 4.What happens to K if the reaction is reversed?Show answer

   It becomes 1/K.

5. 5.Which factor changes the value of K?Show answer

   Temperature only.

6. 6.Define a Lewis acid with an example.Show answer

   An electron-pair acceptor, e.g. BF₃ or AlCl₃.

7. 7.What is the value of Kw at 298 K?Show answer

   1.0 × 10⁻¹⁴.

8. 8.Give an example of an acidic buffer.Show answer

   Acetic acid with sodium acetate.

9. 9.What is the common ion effect?Show answer

   Suppression of a weak electrolyte's ionisation (or a salt's solubility) by adding a strong electrolyte with a common ion.

10. 10.When does a precipitate form?Show answer

    When the ionic product exceeds Ksp.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write Kc and Kp for homogeneous and heterogeneous reactions?
• ☐Can you convert Kc to Kp for a reaction with Δn = −2?
• ☐Can you predict the direction of a reaction from Q and K?
• ☐Can you solve an ICE table for equilibrium concentrations?
• ☐Can you predict the effect of P, T, inert gas and catalyst on an equilibrium?
• ☐Can you find the pH of a weak acid, a weak base and a salt solution?
• ☐Can you calculate the pH of a buffer and explain its action?
• ☐Can you get solubility from Ksp for AB and AB₂ salts, with and without a common ion?

Equilibrium: frequently asked questions