Chemical and ionic equilibrium for NEET

Equilibrium is really two chapters in one. The first half, chemical equilibrium, is about how far a reaction goes: equilibrium constants and what happens when you disturb the system. The second half, ionic equilibrium, applies the same ideas to acids, bases and salts in water, and it is where most of the calculations live.

NEET questions from this chapter are usually short: relate Kp and Kc, predict the direction of a shift, find the pH of a weak acid or a buffer, or get Ksp from solubility. Each needs a formula and one clean line of logarithms, so learn the formulas with their conditions and practise the log values of 2, 3 and 5.

Dynamic equilibrium and physical processes

At equilibrium, the forward and backward processes still happen, but at equal rates, so the concentrations stop changing. That is why it is called dynamic equilibrium. It can be reached from either side, and only in a closed system.

The NMC syllabus starts with physical equilibria:

• Solid ⇌ liquid: ice and water at 273 K and 1 atm. The temperature stays constant as long as both phases are present.
• Liquid ⇌ vapour: water in a closed vessel. The vapour pressure becomes constant at a given temperature.
• Solid ⇌ vapour: sublimation of iodine or camphor in a closed vessel.
• Gas in a liquid (Henry's law): the solubility of a gas in a liquid is proportional to the partial pressure of the gas above it, p = K_H × x. This is why a soda bottle fizzes when opened: the CO₂ pressure drops, so the dissolved CO₂ comes out.

The equilibrium constant, Kc and Kp

For aA + bB ⇌ cC + dD, the law of chemical equilibrium gives:

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Pure solids and pure liquids are left out, because their "concentration" is constant. So for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp is simply the pressure of CO₂.

For gases, Kp uses partial pressures, and the two are linked by:

Kp = Kc(RT)ˣ, where x = Δn_g = moles of gaseous products − moles of gaseous reactants

How K changes when you rewrite the equation:

• Reverse the reaction: K becomes 1/K.
• Multiply the equation by n: K becomes Kⁿ. Halve it: K becomes √K.
• Add two equations: multiply their K values.

Reaction quotient Q has the same form as K but uses the concentrations at any moment. If Q < K, the reaction moves forward; if Q > K, it moves backward; if Q = K, it is at equilibrium.

Link to thermodynamics: ΔG° = −RT ln K, and ΔG = ΔG° + RT ln Q. A large K means a negative ΔG°. See chemical thermodynamics.

Le Chatelier's principle

If a system at equilibrium is disturbed, it shifts in the direction that partly undoes the disturbance. For the exothermic Haber reaction, N₂ + 3H₂ ⇌ 2NH₃:

Acids, bases and the pH scale

The syllabus names three definitions:

A strong acid has a weak conjugate base, and the reverse is also true. For a conjugate pair in water, Ka × Kb = Kw.

Ionic product of water: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K. Kw rises with temperature, so pure water at 50 °C has a pH below 7 but is still neutral, because [H⁺] = [OH⁻].

pH = −log[H⁺] and pH + pOH = pKw = 14 at 298 K.

Weak acids (Ostwald's dilution law): for a weak acid HA of concentration C with small degree of ionisation α:

• α = √(Ka/C)
• [H⁺] = Cα = √(Ka × C)
• pH = ½(pKa − log C)

Multistage ionisation: polyprotic acids such as H₂S and H₃PO₄ ionise in steps, and Ka1 > Ka2 > Ka3. It is harder to pull a proton off a species that is already negative.

Salt hydrolysis, buffers and the common ion effect

A salt's pH depends on which parent is weak:

A buffer resists changes in pH when a little acid or base is added. Use the Henderson–Hasselbalch equation:

• Acidic buffer (CH₃COOH + CH₃COONa): pH = pKa + log([salt]/[acid])
• Basic buffer (NH₄OH + NH₄Cl): pOH = pKb + log([salt]/[base])

When [salt] = [acid], pH = pKa. Buffering is best within about one pH unit of pKa.

The common ion effect is the suppression of a weak electrolyte's ionisation, or a salt's solubility, by adding an ion it already contains. Adding CH₃COONa to acetic acid lowers α, which is exactly why the buffer works.

Solubility product (Ksp)

For a sparingly soluble salt with molar solubility s:

A precipitate forms when the ionic product exceeds Ksp.

Worked NEET-style questions

Example 1: rewriting K

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium constant is K. Find K for NH₃(g) ⇌ ½N₂(g) + 3/2 H₂(g).

• The new equation is the original reversed and halved.
• Reversing gives 1/K; halving takes the square root.
• K' = 1/√K.

Example 2: pH of a weak acid

Find the pH of 0.1 M CH₃COOH. Ka = 1.8 × 10⁻⁵.

• [H⁺] = √(Ka × C) = √(1.8 × 10⁻⁶) ≈ 1.34 × 10⁻³ M.
• pH = 3 − log 1.34 ≈ 3 − 0.13 = 2.87.
• α = 1.34 × 10⁻³/0.1 ≈ 1.3%, small enough for the approximation.

Example 3: buffer pH

A solution is 0.2 M in CH₃COONa and 0.02 M in CH₃COOH. pKa = 4.74. Find the pH.

• pH = pKa + log(0.2/0.02) = 4.74 + log 10 = 5.74.

Example 4: Ksp and the common ion effect

(a) CaF₂ has a solubility of 2 × 10⁻⁴ mol L⁻¹. Find Ksp. (b) AgCl has Ksp = 1.8 × 10⁻¹⁰. Find its solubility in 0.1 M NaCl.

• (a) Ksp = 4s³ = 4 × (2 × 10⁻⁴)³ = 4 × 8 × 10⁻¹² = 3.2 × 10⁻¹¹.
• (b) [Cl⁻] ≈ 0.1 M from NaCl, so s = Ksp/[Cl⁻] = 1.8 × 10⁻¹⁰/0.1 = 1.8 × 10⁻⁹ M. That is about ten thousand times less than in pure water (≈ 1.34 × 10⁻⁵ M).

Practice MCQs

1. For which reaction is Kp = Kc? (a) H₂ + I₂ ⇌ 2HI (b) PCl₅ ⇌ PCl₃ + Cl₂ (c) N₂ + 3H₂ ⇌ 2NH₃ (d) 2SO₂ + O₂ ⇌ 2SO₃
2. The pH of 0.005 M H₂SO₄ (fully ionised) is: (a) 2.3 (b) 2 (c) 3 (d) 1.7
3. For A ⇌ B, K = 4. For 2B ⇌ 2A, K is: (a) 16 (b) 1/4 (c) 8 (d) 1/16
4. In terms of molar solubility s, Ksp of Ag₂CrO₄ is: (a) s² (b) 2s³ (c) 4s³ (d) 27s⁴
5. Argon is added at constant volume to N₂ + 3H₂ ⇌ 2NH₃ at equilibrium. The equilibrium: (a) does not shift (b) shifts backward (c) shifts forward (d) K increases
6. Which is a Lewis acid but not a Brønsted acid? (a) HCl (b) BF₃ (c) NH₄⁺ (d) H₂O
7. An aqueous solution of CH₃COONa is: (a) acidic (b) neutral (c) basic (d) neutral only when hot
8. A solution is 0.1 M in NH₄OH and 0.1 M in NH₄Cl. pKb = 4.74. Its pH is: (a) 4.74 (b) 7 (c) 9.26 (d) 10.26

Answers

1. (a) Δn_g = 2 − 2 = 0.
2. (b) [H⁺] = 2 × 0.005 = 0.01 M, so pH = 2.
3. (d) Reversing gives 1/4; doubling squares it to 1/16.
4. (c) [Ag⁺] = 2s and [CrO₄²⁻] = s, so Ksp = (2s)² × s = 4s³.
5. (a) Partial pressures of the reacting gases do not change at constant volume.
6. (b) BF₃ accepts an electron pair but has no proton to donate.
7. (c) Acetate ions hydrolyse to give OH⁻.
8. (c) pOH = pKb + log 1 = 4.74, so pH = 14 − 4.74 = 9.26.

What to do next

• Write the Kp–Kc relation and the three "rewrite the equation" rules on one card, and solve five questions from it.
• Learn log 2 = 0.30, log 3 = 0.48 and log 5 = 0.70, then do ten pH questions without a calculator.
• Fill in the salt hydrolysis table from memory, including the pH formulas.
• Carry Ka, Kw and ΔG° = −RT ln K into electrochemistry, and revise concentration terms if molarity still slows you down.