Question

At 20°C, the ion-product constant of water, Kw, is 6.88 x 10-15 . What is the pH of pure water at 20°C?

A) 7.000
B) 6.501
C) 7.181
D) 7.081
E) none of these

Answer 1

The correct answer is D) 7.081

Step-by-step explanation:

The water equilibrium is the following:

H₂O ⇄ H⁺ + OH⁻ Kw

Where Kw= [H⁺] x [OH⁻]

In pure water: [H⁺] = [OH⁻]= C

If we introduce C in the expression for Kw:

Kw= [H⁺] x [OH⁻]= C x C= C²

⇒ C=
`√(Kw)`=
`\sqrt{6.88 x 10^(-15) }`= 8.29 x 10⁻⁸

pH= -log [H⁺] = -log C = -log (8.29 x 10⁻⁸) = 7.081