Question

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

Answer 1

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