Question

In class, it was mentioned that at 25 °C, the pH of pure water is 7.00 becuase Kw = 1.0 × 10−14. At other temperatures, the pH value of pure water will be different than 7.00 becuase the value of Kw changes. Suppose that at a temperature different than 25 °C, the pH of pure water was measured and found to be 7.27. What must the value of Kw be equal to at this second temperature?

Answer 1

Kw = 2.88 × 10⁻¹⁵

Step-by-step explanation:

Let's consider the dissociation of water.

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

The equilibrium constant Kw is:

Kw = [H⁺].[OH⁻]

If pH = 7.27, we can find [H⁺]:

pH = -log [H⁺]

H⁺ = anti log (-pH) = anti log (-7.27) = 5.37 × 10⁻⁸ M

According to the balanced equation, 1 mole of H⁺ is produced per mole of OH⁻. So, [H⁺] = [OH⁻] = 5.37 × 10⁻⁸ M

Then,

Kw = [H⁺].[OH⁻]= (5.37 × 10⁻⁸)² = 2.88 × 10⁻¹⁵