Final answer:
The correct statement for pure water at 5.0°C, based on the given values of Kw and pKw, is that [H₃O⁺] is equal to the square root of Kw, or √(1.9×10⁻¹⁵). Options B, C, and D are incorrect interpretations of the concepts of pH, pOH, and the relationship between hydronium and hydroxide ion concentrations at equilibrium.
Step-by-step explanation:
The autoionization of water can be represented by the equation H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq). At 5.0°C, the equilibrium constant for this reaction, Kw, is 1.9×10⁻¹⁵, and the negative logarithm of Kw, or pKw, is 14.73. The ion product, Kw, is the product of the molar concentrations of hydronium ions [H₃O⁺] and hydroxide ions [OH⁻]. In pure water at equilibrium, [H₃O⁺] and [OH⁻] are equal because water is neutral, thus:
[H₃O⁺] = [OH⁻] = √(1.9×10⁻¹⁵) = 1.38×10⁻⁸ M
Therefore, option A is correct: [H₃O⁺] = √(1.9×10⁻¹⁵). Option B would give the pKw, not the pH, since pH is defined as –log[H₃O⁺]. For option C, the correct relation is pKw = –log([H₃O⁺]eq[OH⁻]eq), not the concentration values themselves. Lastly, option D is incorrect because the sum of pH and pOH equals pKw (which is 14.73 at 5.0°C), not that pOH equals pH plus 14.73.