Equilibrium describes reactions that never actually stop — forward and reverse reactions continue at equal rates indefinitely, and the entire chapter is about predicting exactly where that balance point settles and how to shift it deliberately.
The equilibrium constant K captures the ratio of products to reactants at balance, and its value alone tells you how far a reaction proceeds before settling. Le Chatelier's principle then answers the practical question: if you disturb that balance — add a reactant, change pressure, change temperature — which direction does the equilibrium shift to partially counteract the disturbance. The chapter's second half applies this same balance thinking specifically to acids and bases, where pH, dissociation constants and buffer solutions are all just equilibrium calculations wearing a different name.
Core equilibrium and acid-base formulae
| Quantity | Formula |
|---|---|
| Equilibrium constant (concentration) | Kc = [products]/[reactants] |
| Relation between Kp and Kc | Kp = Kc(RT)Δn |
| pH | pH = −log[H⁺] |
| pOH | pOH = −log[OH⁻] |
| pH + pOH relation | pH + pOH = 14 (at 25°C) |
| Ionisation constant of water | Kw = [H⁺][OH⁻] = 10⁻¹⁴ |
| Henderson-Hasselbalch equation | pH = pKa + log([salt]/[acid]) |
| Degree of dissociation (weak acid) | α = √(Ka/C) |
Topics covered
- Dynamic nature of equilibrium — why equilibrium is a balance of ongoing opposing rates, not a static stopping point
- Equilibrium constant — Kc and Kp, and the relationship connecting them for gas-phase reactions
- Factors affecting equilibrium — Le Chatelier's principle applied to concentration, pressure, volume and temperature changes
- Acids and bases — Arrhenius, Bronsted-Lowry and Lewis definitions, and when each is most useful
- Ionisation of weak acids and bases — dissociation constants K<sub>a</sub> and K<sub>b</sub>, and degree of dissociation
- pH scale — and calculating pH for strong acids, strong bases, weak acids and weak bases
- Buffer solutions — how they resist pH change, and the Henderson-Hasselbalch equation
- Common ion effect — and its use in controlling the solubility or dissociation of a weak electrolyte
- Solubility product — and predicting whether a precipitate will form
Why this chapter matters
In NEET and JEE, Le Chatelier's principle conceptual questions and equilibrium constant calculations are consistently among the highest-frequency question types in physical chemistry, and pH calculation for various acid-base combinations, including buffers, appears reliably every year. Common ion effect and solubility product questions are a close secondary favourite.
For Class 11 boards, expect a Le Chatelier's principle application question and a pH or buffer calculation numerical.
Mistakes students make in this chapter
- Forgetting Le Chatelier's principle only describes the direction of shift, not a change in K itself. Concentration, pressure and volume changes shift equilibrium position but leave K unchanged at constant temperature — only a temperature change actually alters the value of K.
- Applying Le Chatelier's principle to predict a temperature effect without knowing whether the reaction is exothermic or endothermic. Increasing temperature favours the endothermic direction of a reaction, which could be forward or reverse depending on the specific reaction — this must be checked, not assumed.
- Confusing pH and pOH formulas. pH = −log[H⁺] and pOH = −log[OH⁻] are easy to swap under exam pressure, especially in strong base calculations where [OH⁻] is the directly known quantity.
- Assuming a buffer solution keeps pH completely constant. A buffer resists significant pH change upon small additions of acid or base, but it does not keep pH perfectly fixed — its buffering capacity is also limited and can be overwhelmed by a large enough addition.
- Forgetting the relation Kp = Kc(RT)^Δn requires Δn to be calculated from gas-phase moles only. Δn is moles of gaseous products minus moles of gaseous reactants — solid or liquid species in the equation are excluded from this calculation.
Worked example: equilibrium constant from a degree of dissociation
- Write the balanced equation and set up an ICE table. Let the amount dissociated be 0.20 mol, since 20% of 1 mol reacts:N₂O₄ ⇌ 2NO₂Initial : 1.00 0
- Fill in the change row using the stoichiometry. Every mole of N₂O₄ consumed produces two moles of NO₂, so the change doubles on the product side:Change : −0.20 +0.40Equilib : 0.80 0.40
- Convert to concentrations. The flask volume is 1 L, so the mole values are already molar concentrations — a convenience the question has built in deliberately:[N₂O₄] = 0.80 M · [NO₂] = 0.40 M
- Write and evaluate the equilibrium expression. The coefficient 2 becomes the power on NO₂, not a multiplier:Kc = [NO₂]² / [N₂O₄] = (0.40)² / 0.80Kc = 0.16 / 0.80 = 0.20
- Relate it to Kp. With Δn = (moles of gaseous product) − (moles of gaseous reactant) = 2 − 1 = 1:Kp = Kc(RT)^Δn = 0.20 × RT
What the notes contain
The PDF works through Le Chatelier's principle applications for every type of disturbance, derives the relationship between Kp and Kc, and covers pH calculations for strong and weak acids/bases together with buffer solution numericals using the Henderson-Hasselbalch equation.
How to use this: build the concept from these notes first, then do one quick NCERT pass before the exam for exact wording and diagram labels, and spend whatever time is left on previous year questions.
Questions
Does Le Chatelier's principle change the value of the equilibrium constant?
No — concentration, pressure and volume changes only shift the position of equilibrium at a fixed temperature; the equilibrium constant K itself remains unchanged. Only a change in temperature actually alters the value of K, and the PDF explains this distinction clearly.
How is pH calculated for a weak acid, differently from a strong acid?
For a strong acid, [H⁺] equals the acid's given concentration directly, since it fully dissociates. For a weak acid, you first need the degree of dissociation using Ka, since only a fraction of the acid actually ionises — the full method with a worked example is in the PDF.
What's the Henderson-Hasselbalch equation used for?
It directly calculates the pH of a buffer solution from the ratio of the conjugate base (salt) to the weak acid present, without needing to solve the full equilibrium expression each time — the derivation and several worked buffer numericals are included.
Does this cover the common ion effect and solubility product together?
Yes, with worked examples showing how adding a common ion suppresses the dissociation of a weak electrolyte, and how the solubility product is used to predict whether a precipitate forms in a given solution.
All other chapters
Every chapter of Class 11 and Class 12 Physics, Chemistry and Biology — each with its own notes and PDF.