For a weak monoprotic acid , under what specific conditions is the simplified formula considered an — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The simplified formula relies on assuming $(1 - \alpha) \approx 1$. This is generally deemed acceptable if $\alpha \le 0.05$ ($5\%$ ionization). Substituting $\alpha = \sqrt{K_a/C} \le 0.05$ gives $K_a/C \le (0.05)^2 = 0.0025 = 2.5 \times 10^{-3}$. If the acid is extremely dilute, $\alpha$ approaches 1, invalidating the approximation, requiring the full quadratic expression and potentially water's contribution.