The conductivity of a weak acid HA of concentration 0.00 l mol L is 2.0 × 10 S cm . If (HA) = 190 S — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Find the degree of ionization (α)** Use the conductivity formula: $$\kappa = c \times \Lambda_m \times \alpha$$ Where: - κ = 2.0 × 10⁻⁵ S cm⁻¹ - c = 0.01 mol L⁻¹ = 10⁻² mol L⁻¹ = 10⁻⁵ mol cm⁻³ - Λ_m = 190 S cm² mol⁻¹ Rearranging: $$\alpha = \frac{\kappa}{c \times \Lambda_m} = \frac{2.0 \times 10^{-5}}{10^{-5} \times 190} = \frac{2.0}{190} = 0.0105$$ **Step 2: Calculate ionization constant using Ka expression** For weak acid HA: $$K_a = \frac{c \times \alpha^2}{1-\alpha}$$ Since α << 1, simplify to: $$K_a \approx c \times \alpha^2$$ $$K_a = 0.01 \times (0.0105)^2 = 10^{-2} \times 1.1 \times 10^{-4}$$ $$K_a = 1.1 \times 10^{-6}$$ **Step 3: Express in required form** $$K_a = 1.1 \times 10^{-6} \approx 12 \times 10^{-7}$$ Or equivalently: **K_a = 1.2 × 10⁻⁶** Rounded to the nearest integer: **12 × 10⁻⁷** (or 1.2 × 10⁻⁶) Therefore, the answer is **12.00**.