What is the ph of pure water at 40.0°c if the kw at this temperature is 2.92 * 10-14?

Answer: pH = 6.77

Explanation:

1) Chemical equilibrium

2 H₂O (l) ⇄ H₃O⁺ (aq) + OH⁻ (aq)

2) Equilibrium constant, Kw

Kw = [H₃O⁺] * [OH⁻] By stoichiometry [H₃O⁺] = [OH⁻]. Call it x Kw = x² x² = 2.92 * 10⁻¹⁴ M² x = √ (2.92 * 10⁻¹⁴) = 1.709 * 10⁻⁷ M = [H₃O⁺]

3) pH

pH = - log [H₃O⁺] = - log (1.709 * 10⁻⁷) = 6.77