For the reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g); ΔH < 0, the yield of NH3 is increased by:

Adding inert gas at constant volume

Equilibrium, Buffers and Solubility (AAMC 5E)

Chemical equilibrium describes the dynamic balance reached in reversible reactions when forward and reverse rates become equal. The AAMC MCAT content outline expects you to compute K_c/K_p, apply Le Chatelier's principle, calculate buffer pH, recognize the common-ion effect, and explain industrial processes such as Haber synthesis.

AAMC content category 5E. Six subtopics: Chemical Equilibrium, Le Chatelier's Principle, Buffer Solution, Common Ion Effect, Solubility Products, Haber's Process.

Chemical Equilibrium

A reaction is at chemical equilibrium when the forward and reverse reactions occur at the same rate, so that macroscopic concentrations of reactants and products remain constant. Equilibrium is dynamic, molecules continue to interconvert, but no net change is observed.

Equilibrium constants

For the general reaction aA + bB ⇌ cC + dD:

K_c (concentration)

K_c = [C]^c[D]^d / ([A]^a[B]^b). Concentrations in mol/dm³. Units depend on Δn; dimensionless when Δn = 0.

K_p (partial pressure)

K_p = P_C^cP_D^d / (P_A^aP_B^b). Units atm^Δn. Relation: K_p = K_c(RT)^Δn where Δn = (c+d) − (a+b).

K_w (water)

Self-ionization: H₂O ⇌ H⁺ + OH⁻. K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C. Hence pH + pOH = 14.

Reading the value of K

K >> 1 → equilibrium lies to the right (products favored).
K << 1 → equilibrium lies to the left (reactants favored).
K depends only on temperature: not on initial concentrations or pressures.

Le Chatelier's Principle

If a system at equilibrium is subjected to a stress (change in concentration, pressure or temperature), the equilibrium shifts in the direction that opposes the stress.

Effect of changes

Le Chatelier's Principle, how each stress shifts equilibrium
Stress applied	Direction of shift	Effect on K	Example: N₂ + 3H₂ ⇌ 2NH₃ (ΔH < 0)
Add reactant	Toward products (forward)	Unchanged	Add N₂ → more NH₃
Add product	Toward reactants (backward)	Unchanged	Add NH₃ → backward
Remove product	Toward products (forward)	Unchanged	Liquefy NH₃ → pulls forward
Increase pressure	Toward side with fewer moles of gas	Unchanged	Forward (4 mol gas → 2 mol)
Decrease pressure	Toward side with more moles of gas	Unchanged	Backward
Pressure if Δn_gas = 0	No shift	Unchanged	(N/A)
Increase T, exothermic fwd	Backward	K decreases	Less NH₃ at higher T
Increase T, endothermic fwd	Forward	K increases	(opposite scenario)
Add catalyst	No shift	Unchanged	Same final composition, reached faster
Add inert gas, constant V	No shift	Unchanged	Partial pressures of reactive gases unchanged
Add inert gas, constant P	Toward side with more moles of gas	Unchanged	Backward shift (volume must rise)

Big rules: only T changes K; catalyst never shifts equilibrium; inert gas at constant V never shifts equilibrium.

Common trap. A catalyst never shifts the position of an equilibrium. Examiners exploit this almost every year, if a catalyst is mentioned, the position of equilibrium is unchanged. Only K (= temperature-dependent) and Le Chatelier stresses shift the equilibrium.

Haber's Process

Industrial synthesis of ammonia from atmospheric N₂ and hydrogen produced by steam reforming. The reaction is exothermic and goes with a decrease in moles of gas, so favorable thermodynamically at high pressure and low temperature, but very slow at low T.

Haber synthesis

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g); ΔH = −92 kJ/mol.
· Pressure: 200 atm (high P shifts equilibrium toward 2 mol of NH₃).
· Temperature: 400–450°C (a compromise, lower T favors equilibrium but is too slow).
· Catalyst: finely divided iron (Fe) with Mo or Al₂O₃/K₂O promoters.
· The unreacted N₂/H₂ is recycled; ammonia is condensed out.

Other important industrial equilibria

Contact process: 2 SO₂ + O₂ ⇌ 2 SO₃; V₂O₅ catalyst, ~ 450°C, 1–2 atm. (For sulfuric acid manufacture.)
Ostwald process: 4 NH₃ + 5 O₂ ⇌ 4 NO + 6 H₂O; Pt-Rh catalyst. (For HNO₃.)

Buffer Solution

A buffer is a solution that resists change in pH on the addition of small amounts of acid or base. Buffers contain a weak acid + its conjugate base (acidic buffer) or a weak base + its conjugate acid (basic buffer).

Henderson–Hasselbalch equation

For an acidic buffer (HA / A⁻):
pH = pK_a + log([salt]/[acid]).
For a basic buffer (B / BH⁺):
pOH = pK_b + log([salt]/[base]); pH = 14 − pOH.
The buffer works best when [salt] ≈ [acid] (or [base]), giving pH ≈ pK_a.

How a buffer resists pH change

If you add H⁺: the conjugate base A⁻ mops it up: A⁻ + H⁺ → HA. If you add OH⁻: the weak acid HA neutralizes it: HA + OH⁻ → A⁻ + H₂O. Either way the [salt]/[acid] ratio changes only slightly, so pH stays nearly constant.

Examples

Acidic buffer: CH₃COOH + CH₃COONa, pH ≈ 4.74.
Basic buffer: NH₄OH + NH₄Cl, pH ≈ 9.25.
Biological buffer: H₂CO₃/HCO₃⁻ in blood, pH 7.4.
Phosphate buffer: H₂PO₄⁻/HPO₄²⁻, pH 7.2 (intracellular).

Common Ion Effect

The common-ion effect is the suppression of the ionization of a weak electrolyte (or the dissolution of a sparingly soluble salt) when a strong electrolyte providing an ion in common with it is added. It is a direct application of Le Chatelier's principle.

Example 1: weak acid

CH₃COOH ⇌ CH₃COO⁻ + H⁺. Adding CH₃COONa increases [CH₃COO⁻], pushing the equilibrium back to the left. The acid ionizes less, [H⁺] falls, pH rises slightly. This is exactly how an acidic buffer works.

Example 2: sparingly soluble salt

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq); K_sp = [Ag⁺][Cl⁻]. Adding NaCl increases [Cl⁻]; the equilibrium shifts left and AgCl precipitates, its solubility decreases. Used in qualitative analysis to ensure complete precipitation.

Solubility Products

For the dissolution of a sparingly soluble salt M_xA_y at saturation, the solubility-product constant K_sp is defined by:

K_sp definition

M_xA_y(s) ⇌ x M^y+(aq) + y A^x−(aq); K_sp = [M^y+]^x[A^x−]^y. The activity of the solid is unity, so it doesn't appear.

Examples

AgCl: K_sp = [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰ → molar solubility s = √K_sp = 1.34 × 10⁻⁵ M.
BaSO₄: K_sp = 1.1 × 10⁻¹⁰; s ≈ 1.05 × 10⁻⁵ M.
PbI₂: K_sp = [Pb²⁺][I⁻]² = 4s³; s = (K_sp/4)^1/3.
Ag₂CrO₄: K_sp = [Ag⁺]²[CrO₄²⁻] = 4s³.

Predicting precipitation (ion product Q vs K_sp)

If Q < K_sp → unsaturated, no precipitate.
If Q = K_sp → saturated.
If Q > K_sp → supersaturated, precipitate forms until Q = K_sp.

Mnemonic for Le Chatelier shifts. "Push back, fight the change." Add reactant → reaction makes more product to use it up. Heat an exo reaction → reaction reverses to absorb the heat. Increase pressure → reaction moves to the side with fewer gas moles to relieve pressure.

Worked MCQs

Five MCQs that capture the high-yield testing patterns for this chapter. Read the explanation even when you get the answer right, that's where the deeper concept lives.

Q1. For the reaction N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g); ΔH < 0, the yield of NH₃ is increased by:

Increasing temperature
Decreasing pressure
Increasing pressure
Adding inert gas at constant volume

Forward direction has fewer gas moles (4 → 2), so high pressure shifts equilibrium right by Le Chatelier. The reaction is exothermic, so high T would shift it left; inert gas at constant V doesn't affect partial pressures, so no shift.

Q2. The pH of a buffer prepared by mixing 0.1 M acetic acid with 0.1 M sodium acetate (pK_a = 4.74) is:

2.74
3.74
4.74
9.26

pH = pK_a + log([salt]/[acid]) = 4.74 + log(1) = 4.74. When the salt and acid concentrations are equal, the pH equals the pK_a: the buffer is at its point of maximum capacity.

Q3. Which of the following does NOT shift the position of an equilibrium?

Change in temperature
Change in pressure (with Δn ≠ 0)
Change in concentration
Addition of a catalyst

A catalyst lowers the activation energy of forward and reverse reactions equally, so it speeds both up by the same factor and the system reaches the same equilibrium faster. K is unchanged, position is unchanged.

Q4. If K_sp of AgCl at 25°C is 1.6 × 10⁻¹⁰, its molar solubility (mol/L) in pure water is approximately:

1.6 × 10⁻¹⁰
1.6 × 10⁻⁵
1.26 × 10⁻⁵
2.5 × 10⁻⁵

For a 1:1 salt, K_sp = s², so s = √K_sp = √(1.6 × 10⁻¹⁰) = 1.26 × 10⁻⁵ mol/L.

Q5. Adding NaCl to a saturated solution of AgCl will:

Increase the solubility of AgCl
Decrease the solubility of AgCl
Have no effect
Change K_sp of AgCl

NaCl supplies Cl⁻, the common ion. By Le Chatelier, the AgCl(s) ⇌ Ag⁺ + Cl⁻ equilibrium shifts to the left, precipitating more AgCl and reducing its solubility. K_sp itself depends only on temperature, so it is unchanged.

Quick Recap

K_c uses concentrations; K_p uses partial pressures; K_p = K_c(RT)^Δn; K_w = 10⁻¹⁴.
Le Chatelier: equilibrium shifts to oppose change. Catalyst, never shifts.
Haber: N₂ + 3 H₂ ⇌ 2 NH₃; 200 atm, 450°C, Fe catalyst, exothermic.
Buffer pH (acidic): pH = pK_a + log([salt]/[acid]); blood buffer = H₂CO₃/HCO₃⁻.
Common-ion effect = suppression of ionization/solubility by a shared ion (Le Chatelier).
K_sp = product of ion concentrations at saturation; K_sp depends only on T. Q > K_sp → precipitate.

Test yourself. Take a timed practice test or browse topic-wise MCQs to lock these concepts in.

Equilibrium, Buffers and Solubility (AAMC 5E)

Chemical Equilibrium

Equilibrium constants

Reading the value of K

Le Chatelier's Principle

Effect of changes

Haber's Process

Other important industrial equilibria

Buffer Solution

How a buffer resists pH change

Examples

Common Ion Effect

Example 1: weak acid

Example 2: sparingly soluble salt

Solubility Products

Examples

Predicting precipitation (ion product Q vs Ksp)

Worked MCQs

Quick Recap

Predicting precipitation (ion product Q vs K_sp)