A chemist dissolves of pure barium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (The temperature of the solution is.) Round your answer to significant decimal places.

12.45

Explanation:

There is some info missing. I think this is the original question.

A chemist dissolves 169 mg of pure barium hydroxide in enough water to make up 70 mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Round your answer to 2 significant decimal places.

First, we will calculate the molarity of barium hydroxide.

M = mass / molar mass * liters of solution

M = 0.169 g / 171.34 g/mol * 0.070 L

M = 0.014 M

Barium hydroxide is a strong base that dissociates according to the following equation.

Ba (OH) ₂ → Ba²⁺ + 2 OH⁻

The molar ratio of Ba (OH) ₂ to OH⁻ is 1:2. The concentration of OH⁻ is 2 * 0.014 M = 0.028 M

The pOH is:

pOH = - log [OH⁻] = - log 0.028 = 1.55

The pH is:

pH = 14 - pOH = 14 - 1.55 = 12.45