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MCQ

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The ionic product of water at 298 K is 1.0 × 10⁻¹⁴ . What is the pH of pure water at this temperature?

Choose one · Correct answer highlighted

6 7 Correct 8 14

Kw = [H+][OH-] = 1.0 × 10⁻¹⁴ . In pure water, [H+] = [OH-] = sqrt1.0 × 10⁻¹⁴ = 1.0 × 10⁻⁷ , so pH = -log(1.0 × 10⁻⁷) = 7 .

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