Why the ionic half deserves more of your time
Gaseous equilibrium questions tend to be single-concept: write the expression, apply Le Chatelier, or convert between Kp and Kc. Ionic equilibrium contains pH calculations, weak-acid dissociation, buffers and solubility product - four distinct question types, each with its own procedure and each appearing regularly.
The practical consequence is that a student who has mastered Kc and Kp but treats ionic equilibrium as an afterthought has prepared the smaller half thoroughly. Reversing that emphasis is the single most useful adjustment to make in this chapter.
How JEE actually asks Chemical Equilibrium
NTA publishes no chapter-wise weightage, so any figure online is a coaching estimate from past papers; confirm your session's syllabus and pattern in the official NTA information bulletin.
Observable patterns: Le Chatelier questions are usually conceptual and quick. Kp-to-Kc conversions test whether you computed Δn correctly. On the ionic side, pH of strong and weak acids, buffer calculations via Henderson-Hasselbalch, and solubility product with a common ion are the four recurring types.
Key concepts, compressed
- ✓At equilibrium the forward and reverse rates are equal, so concentrations stop changing - it is dynamic, not static.
- ✓The equilibrium constant is fixed for a given reaction at a given temperature, and nothing but temperature alters it.
- ✓The reaction quotient Q has the same form as K but uses current concentrations; comparing them predicts the direction of change.
- ✓Le Chatelier's principle: a system at equilibrium responds to a disturbance in the direction that partially offsets it.
- ✓A weak acid dissociates only partially, and its degree of dissociation increases on dilution.
- ✓A buffer resists pH change because added acid or base is consumed by its conjugate partner.
Relations you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Equilibrium constant | Kc = [products]^coeff / [reactants]^coeff | solids and liquids omitted |
| Kp and Kc | Kp = Kc(RT)^Δn(g) | Δn from gas moles only |
| Reversed reaction | K′ = 1/K | |
| Doubled equation | K′ = K² | |
| Direction of change | Q < K forward, Q > K reverse | |
| Ostwald dilution law | Ka = Cα²/(1 − α) ≈ Cα² for small α | |
| Degree of dissociation | α ≈ √(Ka/C) | weak electrolyte |
| pH | pH = −log[H⁺] | |
| pH and pOH | pH + pOH = 14 at 25 °C | |
| Henderson-Hasselbalch | pH = pKa + log([salt]/[acid]) | |
| Solubility product, AB | Ksp = s² | |
| Solubility product, A₂B | Ksp = 4s³ | |
| Precipitation | occurs when Q > Ksp |
The five mistakes that cost the most marks
- ✓Including pure solids or liquids in the equilibrium expression. Their concentrations are effectively constant and absorbed into K.
- ✓Believing a catalyst shifts the equilibrium. It changes only how fast equilibrium is reached, never where it lies.
- ✓Applying a pressure change when Δn is zero. With equal gas moles on both sides, pressure has no effect on the position.
- ✓Confusing Q with K. K is the fixed value at equilibrium; Q is the same expression evaluated now, and comparing them gives the direction.
- ✓Assuming dilution changes a buffer's pH toward neutral. The salt-to-acid ratio is unchanged, so the pH barely moves.
Practice set 1: the equilibrium constant
1. Write the Kc expression for N₂ + 3H₂ ⇌ 2NH₃.
Kc = [NH₃]² / ([N₂][H₂]³). Products go on top and reactants below, each raised to its stoichiometric coefficient. The cube on hydrogen is the detail most often dropped, and it changes the numerical answer by orders of magnitude.
2. Why are pure solids and liquids omitted from an equilibrium expression?
Because their concentrations - determined by density and molar mass - do not change as the reaction proceeds. Being constant, they are folded into the value of K itself rather than appearing explicitly. This is why the expression for a heterogeneous equilibrium such as CaCO₃ ⇌ CaO + CO₂ is simply Kp = p(CO₂).
3. For N₂ + 3H₂ ⇌ 2NH₃, what is Δn and how does Kp relate to Kc?
Δn = −2, so Kp = Kc(RT)⁻². Count gas moles: 2 on the product side, 1 + 3 = 4 on the reactant side, giving 2 − 4 = −2. Only gaseous species are counted. A negative Δn means Kp is smaller than Kc at ordinary temperatures.
4. When are Kp and Kc numerically equal?
When Δn = 0, since (RT)⁰ = 1. This happens whenever the number of gas moles is the same on both sides - for instance H₂ + I₂ ⇌ 2HI, where there are two on each side. That same condition also makes the equilibrium insensitive to pressure, which is question 13.
5. If the equilibrium constant of a reaction is K, what is it for the reverse reaction?
1/K. Reversing the reaction swaps products and reactants, which inverts the expression. So a reaction with K = 100 has a reverse constant of 0.01 - strongly favouring the forward direction as written and strongly disfavouring the reverse.
6. If every coefficient in an equation is doubled, what happens to K?
It becomes K². Doubling the coefficients squares every concentration term in the expression. Similarly, halving the coefficients gives √K. This is why an equilibrium constant is meaningless without the equation it refers to - the same chemistry written differently gives a different number.
7. What does a very large value of K tell you?
That the equilibrium position lies far toward the products - at equilibrium the mixture is mostly product. A very small K means the reverse. Note that K says nothing about how fast equilibrium is reached; a reaction can have an enormous K and still be imperceptibly slow without a catalyst.
Practice this now
Practice set 2: reaction quotient and Le Chatelier
8. What happens when Q is less than K?
The reaction proceeds forward, converting reactants into products until Q rises to equal K. Q has the same algebraic form as K but uses the concentrations present at that moment, so comparing them tells you which way the system must move. This is the general method for predicting direction.
9. What happens when Q is greater than K?
The reaction proceeds in reverse, consuming products until Q falls to K. There are too many products relative to equilibrium, so the system corrects by running backwards. Note that the reaction does not stop - both directions continue, but the net change is reverse.
10. What does Q = K mean?
The system is at equilibrium and there is no net change in composition. Forward and reverse reactions continue at equal rates, which is why equilibrium is described as dynamic rather than static. Isotopic labelling experiments confirm that both directions are still running.
11. What happens when more reactant is added to a system at equilibrium?
The equilibrium shifts forward to consume part of the added reactant. Adding reactant makes Q smaller than K, and the system responds by moving forward until Q returns to K. Note that K itself is unchanged - only the position has moved.
12. How does increasing pressure affect a gaseous equilibrium?
It shifts the equilibrium toward the side with fewer gas moles, which reduces the volume and partially offsets the increase. For N₂ + 3H₂ ⇌ 2NH₃ that means shifting toward ammonia, since 2 moles occupy less volume than 4. This is why the Haber process runs at high pressure.
13. When does a pressure change have no effect on the equilibrium position?
When Δn is zero - equal numbers of gas moles on both sides. For H₂ + I₂ ⇌ 2HI, compressing the system raises all partial pressures equally and Q is unchanged, so there is no shift. Options claiming a shift in this case are a standard distractor.
14. Does a catalyst shift the position of equilibrium?
No. A catalyst lowers the activation energy for both the forward and reverse reactions by the same amount, so both rates increase equally. Equilibrium is reached sooner but at exactly the same composition, and K is unchanged. This is one of the most reliably examined points in the chapter.
15. Which single factor changes the value of K?
Temperature, and only temperature. Concentration, pressure and catalysts move the position of equilibrium or the rate of approach but leave K itself alone. For an exothermic reaction, raising the temperature decreases K; for an endothermic one it increases K.
Practice set 3: degree of dissociation
16. State Ostwald's dilution law.
Ka = Cα²/(1 − α), where C is the concentration and α the degree of dissociation. For a weak electrolyte α is small, so (1 − α) is close to 1 and the expression simplifies to Ka ≈ Cα², giving α ≈ √(Ka/C). That approximation is what makes weak-acid calculations quick.
17. A weak acid has Ka = 10⁻⁵ at a concentration of 0.1 M. What is its degree of dissociation?
0.01, or 1%. Using α ≈ √(Ka/C) = √(10⁻⁵/0.1) = √(10⁻⁴) = 10⁻². The approximation is safe here because α is far below 1, so neglecting it in the denominator introduces an error of about one per cent. For α above roughly 0.05 the full quadratic should be used.
18. What is the pH of that solution?
3. The hydrogen ion concentration is Cα = 0.1 × 0.01 = 10⁻³ M, so pH = −log(10⁻³) = 3. Note the two-step structure: find α first, then use it to get [H⁺], then take the logarithm. Jumping straight to pH = −log(Ka) is a common shortcut that is simply wrong.
19. What happens to the degree of dissociation on dilution?
It increases. From α ≈ √(Ka/C), reducing C raises α - diluting a weak acid makes a larger fraction of it dissociate. The hydrogen ion concentration still falls, because Cα decreases overall, so the pH rises toward 7. Both facts are true simultaneously and questions test whether you can hold them together.
20. How does a strong acid differ from a weak one in this treatment?
A strong acid dissociates essentially completely, so α ≈ 1 and no equilibrium calculation is needed - the hydrogen ion concentration equals the acid concentration directly. Weak acids need Ka and the dissociation law. Applying the weak-acid formula to a strong acid is unnecessary work and gives a wrong answer.
Practice this now
Practice set 4: pH and buffers
21. Define pH.
pH = −log[H⁺], the negative logarithm to base ten of the hydrogen ion concentration in mol/L. The logarithmic scale means each unit represents a tenfold change in acidity, so a solution of pH 3 is a hundred times more acidic than one of pH 5.
22. What is the relation between pH and pOH?
pH + pOH = 14 at 25 °C, following from the ionic product of water Kw = 10⁻¹⁴. The value 14 is temperature-dependent, since Kw increases with temperature - so at higher temperatures neutral water has a pH below 7 while remaining neutral.
23. What is the pH of 0.01 M hydrochloric acid?
2. Hydrochloric acid is strong and dissociates completely, so [H⁺] = 0.01 = 10⁻² M and pH = 2. No equilibrium calculation is required, which is the practical difference between strong and weak acids highlighted in question 20.
24. State the Henderson-Hasselbalch equation.
pH = pKa + log([salt]/[acid]) for an acidic buffer. It follows from rearranging the weak-acid equilibrium expression and taking logarithms. The corresponding form for a basic buffer is pOH = pKb + log([salt]/[base]).
25. What is the pH of a buffer containing equal concentrations of a weak acid and its salt?
pH = pKa. When the salt and acid concentrations are equal their ratio is 1, log 1 = 0, and the equation reduces to pH = pKa. This is why a buffer is most effective near the pKa of its acid - that is where it has equal capacity to absorb added acid and added base.
26. What is the solubility product?
The equilibrium constant for the dissolution of a sparingly soluble salt, written as the product of the ion concentrations each raised to its coefficient. For AgCl ⇌ Ag⁺ + Cl⁻ it is Ksp = [Ag⁺][Cl⁻]. The undissolved solid does not appear, for the same reason as in question 2.
Practice set 5: solubility and precipitation
27. The solubility of AgCl is 10⁻⁵ M. What is its Ksp?
10⁻¹⁰. Each formula unit gives one Ag⁺ and one Cl⁻, so both ion concentrations equal the solubility s. Then Ksp = s² = (10⁻⁵)² = 10⁻¹⁰. The one-to-one stoichiometry is what makes this the simplest case; the next question shows what changes otherwise.
28. What is the common ion effect?
The suppression of a salt's solubility by adding an ion it already contains. Adding sodium chloride to a saturated silver chloride solution raises [Cl⁻], so [Ag⁺] must fall to keep the product equal to Ksp, and silver chloride precipitates. It is Le Chatelier's principle applied to a solubility equilibrium.
29. Write Ksp for a salt of the type A₂B in terms of its solubility.
Ksp = 4s³. Dissolving gives 2A⁺ and B²⁻, so [A⁺] = 2s and [B²⁻] = s. Then Ksp = (2s)²(s) = 4s³. The coefficient becomes both a multiplier and an exponent, and forgetting the factor of 4 is the standard error for non-one-to-one salts.
30. When does precipitation occur on mixing two solutions?
When the ionic product Q exceeds Ksp. If Q is less than Ksp the solution is unsaturated and no precipitate forms; if Q equals Ksp it is exactly saturated. This is the same Q-versus-K comparison as question 8, applied to a solubility equilibrium - remember to account for the dilution that mixing causes before computing Q.
How to study this chapter efficiently
- ✓Weight your preparation toward ionic equilibrium. It supplies more question types than the gaseous half.
- ✓Write the equilibrium expression before anything else, and check that no solids or liquids have crept in.
- ✓For Kp-to-Kc conversions, compute Δn explicitly from gas moles only, and write it down.
- ✓For any Le Chatelier question, ask which way the system moves to partially undo the change.
- ✓For weak acids, follow the fixed order: α from Ka and C, then [H⁺] = Cα, then pH.
- ✓Learn the four Ksp forms - AB, AB₂, A₂B and A₂B₃ - since the coefficient appears twice in each.
Turn this into active practice
Le Chatelier questions feel obvious while reading and catch people under time pressure, particularly the two cases where nothing happens - a catalyst, and a pressure change with equal gas moles. Those two account for a disproportionate share of lost marks in this chapter.
The JEE Chemical Equilibrium quiz on QUFF generates fresh questions across equilibrium constants, Le Chatelier, weak acids, buffers and solubility, marks them instantly and explains each answer. Do a mixed set and check specifically whether you correctly identified the cases where the equilibrium does not move at all.