Practice question
Question
The Kw of water at 373 K is 1.0 × 10⁻¹² . What is the pH of pure water at this temperature?
Explanation
Kw = [H+][OH-] = 1.0 × 10⁻¹² , [H+] = sqrt1.0 × 10⁻¹² = 1.0 × 10⁻⁶ , pH = -log(1.0 × 10⁻⁶) = 6 .
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