Sulphurous acid (H SO ) has Ka =1.7 × 10 and Ka =6.4 × 10 . The pH of 0.588 M H SO is ___. (Round of — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

# Solution: pH of H₂SO₃ Solution **Step 1: Identify the relevant Ka value** Since H₂SO₃ is a diprotic weak acid, use Ka₁ = 1.7 × 10⁻² for the first dissociation (most significant contributor to [H⁺]). **Step 2: Set up the Ka₁ expression** For H₂SO₃ ⇌ H⁺ + HSO₃⁻ $$K_{a1} = \frac{[H^+][HSO_3^-]}{[H_2SO_3]} = 1.7 × 10^{-2}$$ **Step 3: Apply the ICE table** - Initial: [H₂SO₃] = 0.588 M - Change: -x, +x, +x - Equilibrium: (0.588 - x), x, x $$1.7 × 10^{-2} = \frac{x^2}{0.588 - x}$$ **Step 4: Check if approximation is valid** Since Ka₁ is relatively large (1.7 × 10⁻²), solve without approximation: $$1.7 × 10^{-2}(0.588 - x) = x^2$$ $$x^2 + 0.017x - 0.00998 = 0$$ Using the quadratic formula: x ≈ 0.0958 M ≈ 0.10 M **Step 5: Calculate pH** $$[H^+] = 0.10 \text{ M}$$ $$pH = -\log(0.10) = 1.00$$ Therefore, the answer is 1.00.