Question 9

**Setting up the problem** The acid is extremely weak (Ka = 4.00 × 10⁻¹¹), so the H⁺ contributed by the acid (~√(KaC) ≈ 2 × 10⁻⁷ M) is comparable to the H⁺ from water autoionization (10⁻⁷ M). Hence the self-ionization of water must be included. Let [H⁺] = h. - Charge balance: h = [A⁻] + [OH⁻] - Acid equilibrium: [A⁻] = Ka·C/(Ka + h) ≈ Ka·C/h (since h ≫ Ka, and negligible dissociation so [HA] ≈ C) - Water: [OH⁻] = Kw/h Therefore: h = Ka·C/h + Kw/h h² = Ka·C + Kw = (4.00 × 10⁻¹¹)(1.00 × 10⁻³) + 1.00 × 10⁻¹⁴ h² = 4.00 × 10⁻¹⁴ + 1.00 × 10⁻¹⁴ = 5.00 × 10⁻¹⁴ h = √5 × 10⁻⁷ M ≈ 2.236 × 10⁻⁷ M Check: h ≈ 2.2 × 10⁻⁷ ≫ Ka = 4 × 10⁻¹¹, so the approximation [A⁻] ≈ KaC/h is valid. **X ≈ 2.24**

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