At 60 °C the value of Kw is 9.5 x 10-14.

Considering this, what is the pH of a 5.00 x 10-2 M Ba (OH) 2 (aq) solution at this temperature?

12

Explanation:

Ba (OH) ₂ is a strong base that dissociates according to the following equation.

Ba (OH) ₂ (aq) → Ba²⁺ (aq) + 2 OH⁻ (aq)

The molar ratio of Ba (OH) ₂ to OH⁻ is 1:2. Then, the concentration of OH⁻ is 2 * 5.00 * 10⁻² M = 0.100 M

The ionic product of water (Kw) is:

Kw = 9.5 * 10⁻¹⁴ = [H⁺].[OH⁻] = [H⁺]. (0.100)

[H⁺] = 9.5 * 10⁻¹³ M

The pH is

pH = - log [H⁺] = - log 9.5 * 10⁻¹³ = 12