K for butyric acid (C H COOH) is 2 × 10 . The pH of 0.2 M solution of butyric acid is ___ × 10 . (Ne — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Write the ionization equation and Ka expression** For butyric acid (CH₃CH₂CH₂COOH): $$K_a = \frac{[H^+][A^-]}{[HA]} = 2 \times 10^{-5}$$ **Step 2: Set up the ICE table** For 0.2 M butyric acid: - Initial: [HA] = 0.2 M, [H⁺] = 0, [A⁻] = 0 - Change: -x, +x, +x - Equilibrium: 0.2-x, x, x **Step 3: Apply Ka expression** $$K_a = \frac{x^2}{0.2-x} = 2 \times 10^{-5}$$ Since Ka is small, assume x << 0.2: $$\frac{x^2}{0.2} = 2 \times 10^{-5}$$ **Step 4: Solve for [H⁺]** $$x^2 = 0.2 \times 2 \times 10^{-5} = 4 \times 10^{-6}$$ $$x = 2 \times 10^{-3} \text{ M} = [H^+]$$ **Step 5: Calculate pH** $$pH = -\log[H^+] = -\log(2 \times 10^{-3})$$ $$pH = -(\log 2 + \log 10^{-3})$$ $$pH = -(0.30 - 3) = -(-2.70) = 2.70$$ **Step 6: Express in required form** $$pH = 2.70 = 27 \times 10^{-1}$$ Therefore, the answer is 27.