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

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