Practice question
Question
The Kw of water at 310 K is 2.9 × 10⁻¹⁴ . What is the pH of pure water at this temperature?
Explanation
Kw = [H+][OH-] = 2.9 × 10⁻¹⁴ . In pure water, [H+] = sqrt2.9 × 10⁻¹⁴ ≈ 1.7 × 10⁻⁷ , pH = -log(1.7 × 10⁻⁷) ≈ 6.77 .
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