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