HA₍ₐq.₎ ⇌ H⁺₍ₐq.₎ + A⁻₍ₐq.₎ The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is Given: Kf(H₂O) = 1.8 K-kg mol⁻¹, molality = molarity
❄️ Freezing Point Depression & Dissociation Constant of a Weak Acid Concept Focus : HA₍ₐq.₎ ⇌ H⁺₍ₐq.₎ + A⁻₍ₐq.₎ We’re diving into a classic problem of colligative properties and equilibrium . Here, we’re given data about the freezing point depression of a weak monobasic acid solution to calculate its dissociation constant (Ka) . 🧪 Problem Statement A 0.1 m aqueous solution of a monobasic weak acid HA shows a freezing point depression of 0.20°C . Calculate the dissociation constant (Ka) for the acid. Given: K f ( water ) = 1.86 K . kg/mol K_f(\text{water}) = 1.86 \, \text{K·kg/mol} Δ T f = 0.20 ∘ C \Delta T_f = 0.20^\circ C Molality (m) = 0.1 mol/kg (assume molarity ≈ molality) Acid: HA ⇌ H⁺ + A⁻ 🧠 Step-by-Step Solution Step 1: Use the Freezing Point Depression Formula Δ T f = i ⋅ K f ⋅ m \Delta T_f = i \cdot K_f \cdot m Where: Δ T f \Delta T_f = depression in freezing point i i = van’t Hoff factor K f K_f = cryoscopic constant (1.86 K·...