Equilibrium — Practice Quiz

HPO₄²⁻

Step 1: According to Bronsted-Lowry theory, an acid donates a proton (H⁺) to form its conjugate base. Step 2: H₂PO₄⁻ acting as an acid: H₂PO₄⁻ → H⁺ + HPO₄²⁻. Step 3: So HPO₄²⁻ is the conjugate base of H₂PO₄⁻ (it has one fewer proton). Step 4: H₃PO₄ is the conjugate acid of H₂PO₄⁻ (has one more proton). PO₄³⁻ would be formed by removing two protons — that is not its direct conjugate base. Option D has a wrong formula (incorrect charge).

3

Step 1: HCl is a strong acid and completely dissociates: HCl → H⁺ + Cl⁻. Step 2: Since [HCl] = 0.001 M = 1.0 × 10⁻³ M, [H⁺] = 1.0 × 10⁻³ mol L⁻¹. Step 3: pH = −log₁₀[H⁺] = −log₁₀(1.0 × 10⁻³) = −(−3) = 3. Step 4: Option B (pH = 11) would be the pOH, a common mistake of confusing pH and pOH. Option C (pH = 4) is wrong — that would correspond to [H⁺] = 10⁻⁴ M. Option D (pH = 7) is for neutral water.

Low temperature and high pressure

Step 1: The reaction is exothermic (releases heat). By Le-Chatelier's principle, lowering temperature shifts equilibrium to the right (forward direction) to release more heat, favouring NH₃ formation. Step 2: The reaction involves a decrease in moles of gas: 4 moles → 2 moles. Increasing pressure shifts equilibrium towards the side with fewer moles (right), again favouring NH₃. Step 3: Therefore, low temperature and high pressure give maximum yield. Step 4: In practice, very low temperatures reduce reaction rate, so a moderate temperature (~450°C) with a catalyst (iron + molybdenum) is used as

Kc = [CO₂]

Step 1: This is a heterogeneous equilibrium involving both solid and gaseous phases. Step 2: By convention, the active mass (concentration) of a pure solid is taken as unity (constant) and is NOT included in the Kc expression. Step 3: Both CaCO₃(s) and CaO(s) are pure solids, so [CaCO₃] = 1 and [CaO] = 1. Step 4: Therefore, Kc = [CaO][CO₂]/[CaCO₃] simplifies to Kc = [CO₂]. This is a very important result — the decomposition of CaCO₃ depends only on the CO₂ pressure, not on the amounts of solid present.