JEE Chemistry

In presence of 0.1 M Cl⁻: Ksp = [Ag⁺][Cl⁻]; 1.8 × 10⁻¹⁰ = s × 0.1; s = 1.8 × 10⁻⁹ M

Using Henderson-Hasselbalch: pH = pKa + log([salt]/[acid]) = 4.74 + log(0.15/0.1) = 4.74 + 0.48 = 5.22

After 3 half-lives (30 min): (1/2)³ = 1/8 = 12.5% remains

2A ⇌ B; Initial: 2, 0; Change: -2x, +x; At equilibrium: 2-2x, x. Kp = x/(2-2x)² = 0.5. Solving: x = 1 atm

Using ΔG° = -RT ln K; -46500 = -8.314 × 298 × ln K; ln K ≈ 6.9; K ≈ 1000

CN⁻ is a neutral ligand (carries -1 charge). Let oxidation state of Fe = x. x + 6(-1) = -3, so x = +3. Fe is in +3 oxidation state.

For CaCO₃ ⇌ Ca²⁺ + CO₃²⁻, Ksp = [Ca²⁺][CO₃²⁻] = s². Therefore, s = √Ksp = √(3.3 × 10⁻⁹) ≈ 5.7 × 10⁻⁵ M.

Using Henderson-Hasselbalch equation: pH = pKa + log([salt]/[acid]) = 4.74 + log(0.2/0.2) = 4.74 + 0 = 4.74. Since equal concentrations, pH = pKa.

Using Ka = [H⁺]²/(0.1 - [H⁺]) ≈ [H⁺]²/0.1. [H⁺] = √(Ka × C) = √(1.8 × 10⁻⁵ × 0.1) = 1.34 × 10⁻³. pH = -log(1.34 × 10⁻³) ≈ 2.87 ≈ 2.74.

For Al³⁺ + 3e⁻ → Al, 3 moles of electrons are needed per mole of Al. Charge = n × F = 3 × 96500 = 289500 C (Faraday constant F = 96500 C/mol).