0.01 moles of a weak acid HA (K = 2.0 × 10 ) is dissolved in 1.0 L of 0.1 M HCl solution. The degree — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the given information** - Moles of HA = 0.01 mol - Volume = 1.0 L, so [HA]₀ = 0.01 M - [HCl] = 0.1 M (strong acid) - Ka = 2.0 × 10⁻⁵ - Degree of dissociation α << 1 **Step 2: Account for the common ion effect** HCl completely dissociates, providing [H⁺] = 0.1 M initially. This suppresses the dissociation of the weak acid HA. **Step 3: Set up the Ka expression** For HA ⇌ H⁺ + A⁻: Ka = [H⁺][A⁻]/[HA] Since α << 1, the H⁺ from HCl dominates: - [H⁺] ≈ 0.1 M (from HCl) - [A⁻] = α × 0.01 (from HA dissociation) - [HA] ≈ 0.01 M (negligible dissociation) **Step 4: Substitute into Ka expression** 2.0 × 10⁻⁵ = (0.1)(α × 0.01)/(0.01) 2.0 × 10⁻⁵ = (0.1)(α) **Step 5: Solve for α** α = (2.0 × 10⁻⁵)/0.1 α = 2.0 × 10⁻⁴ **Step 6: Express in the required form** α = 2.00 × 10⁻⁴ Therefore, the answer is 2.00 × 10⁻⁴ (or **2.00** when expressed as a × 10⁻⁴).