Question 10:
At 25 °C, the degree of ionization of pure water at equilibrium is:
Water has an extremely low degree of ionization at 25°C, with exactly 2 H⁺ (or OH⁻) ions per 10⁹ water molecules, corresponding to α ≈ 1.8–2 × 10⁻⁹ based on Kw = 1.0 × 10⁻¹⁴.
Correct Answer: (C) 2 per 10⁹ molecules
In pure water, [H⁺] = [OH⁻] = 10⁻⁷ M from Kw = 10⁻¹⁴ at 25°C. Water concentration is ~55.5 M (1000/18 g/L), so α = 10⁻⁷ / 55.5 ≈ 1.8 × 10⁻⁹, or ~2 ions dissociated per 10⁹ molecules.
Options Breakdown
| Option | Value | Explanation |
|---|---|---|
| (A) | 20 per 10⁹ | Too high (~10x actual); would imply [H⁺] ≈ 1.1 × 10⁻⁶ M, Kw ≈ 10⁻¹² (wrong temp or impure water). |
| (B) | 10 per 10⁹ | Still high (~5x); α ≈ 10⁻⁸, inconsistent with standard Kw=10⁻¹⁴. |
| (C) | 2 per 10⁹ | Correct; rounds 1.8 × 10⁻⁹ precisely for MCQs, matching [H⁺]=10⁻⁷ M. |
| (D) | 12 per 10⁹ | Overestimate (~6x); no standard condition fits this value. |
Calculation Insight
α = √(Kw / C_H2O) ≈ √(10⁻¹⁴ / 55.5) ≈ 1.8 × 10⁻⁹; multiply by 10⁹ molecules gives ~2 ions. This underscores water’s neutrality (pH=7) despite autoionization.
