Equilibrium for NEET UG
Many reactions never go to completion; they settle at a balance. Equilibrium constants, Le Chatelier's principle, pH and buffers describe that balance and how to shift it.
ChemistryClass 1173 MCQs3 lessons13 flashcards4 memory tricks
What you'll learn
- Equilibrium constant and Le Chatelier's principle
Write K_c and K_p, relate them, and predict how changes shift equilibrium. - Acids, bases and pH
Calculate pH for strong and weak acids and bases. - Buffers, hydrolysis and solubility product
Use the Henderson equation, and K_sp to find solubility.
Key formulas
| Concentration constant | K_c = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ |
|---|---|
| Pressure constant | K_p = K_c(RT)^Δn |
| Reaction quotient | Q < K: goes forward; Q > K: goes backward; Q = K: at equilibrium |
| pH | pH = −log[H⁺] |
| Water at 25 °C | K_w = [H⁺][OH⁻] = 10⁻¹⁴, so pH + pOH = 14 |
| Weak acid | [H⁺] = √(K_a × C), degree of ionisation α = √(K_a/C) |
| Conjugate pair | K_a × K_b = K_w |
| Henderson–Hasselbalch (acidic buffer) | pH = pK_a + log([salt]/[acid]) |
Memory tricks
Add reactant: shifts forward. Raise pressure: shifts towards fewer moles of gas. Heat an exothermic reaction: shifts backward.
A catalyst speeds up the forward and backward reactions equally. It does not change K or the equilibrium composition.
pH = −log[H⁺]. One pH unit is a tenfold change in [H⁺]. At 25 °C, pH + pOH = 14.
Acidic buffer: CH₃COOH + CH₃COONa. Basic buffer: NH₄OH + NH₄Cl. A strong acid with its salt (HCl + NaCl) is not a buffer.
Try 3 NEET-style questions
Q1. For A ⇌ B, the equilibrium constant is 4 at a certain temperature. At the same temperature, the equilibrium constant for 2B ⇌ 2A is
- 0.0625
- 0.25
- 0.5
- 16
Show answer
Answer: (1) 0.0625
Reversing a reaction inverts K (1/4). Doubling all coefficients squares it: (1/4)² = 1/16 = 0.0625.
Q2. A weak monobasic acid with Ka = 1.0 × 10⁻⁵ is at a concentration of 0.1 M. Its approximate degree of dissociation is
- 0.01
- 0.1
- 0.001
- 1 × 10⁻⁵
Show answer
Answer: (1) 0.01
Ostwald's dilution law: α ≈ √(Ka/C) = √(1.0 × 10⁻⁵/0.1) = √(10⁻⁴) = 0.01 (1%).
Q3. Match List I with List II. List I A. Kp for CaCO₃(s) ⇌ CaO(s) + CO₂(g) B. ΔG at equilibrium C. Ka × Kb for a conjugate pair D. Ksp of an AB₂ salt with solubility s List II I. 4s³ II. Kw III. Zero IV. p(CO₂) Choose the correct answer from the options given below.
- A-III, B-IV, C-II, D-I
- A-IV, B-III, C-II, D-I
- A-IV, B-II, C-III, D-I
- A-IV, B-III, C-I, D-II
Show answer
Answer: (2) A-IV, B-III, C-II, D-I
Kp for CaCO₃ decomposition is p(CO₂) (A-IV). ΔG = 0 at equilibrium (B-III). Ka × Kb = Kw (C-II). Ksp of AB₂ = 4s³ (D-I).
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