Answer to Question #114961 in Chemistry for tristan

Question

Suppose 0.0100 mole of Ba(OH)2 is dissolved in enough water to give 469.5 mL of solution.
What is the OH− concentration?
What is the pOH?
What is the pH?

Answer 1

Solution:

Molarity of Ba(OH)₂ = Moles of Ba(OH)₂ / Volume of solution

Molarity of Ba(OH)₂ = (0.0100 mol) / (0.4695 L) = 0.0213 M.

Molarity of Ba(OH)₂ = C_M(Ba(OH)₂) = 0.0213 M

Barium hydroxide is a strong base for both stages of dissociation:

Ba(OH)₂(s) → Ba²⁺ + 2OH⁻

According to the equation: [OH⁻] = 2 × C_M(Ba(OH)₂)

[OH⁻] = 2 × (0.0213 M) = 0.0426 M

[OH⁻] = 0.0426 M

The pOH is calculated using the expression:

pOH = − log[OH⁻]

pOH = − log(0.0426) = 1.37

pOH = 1.37

The pH and pOH of a water solution at 25^oC are related by the following equation:

pH + pOH = 14.

Then,

pH = 14 - pOH

pH = 14 - 1.37 = 12.63

pH = 12.63

Answer:

[OH⁻] = 0.0426 M;

pOH = 1.37;

pH = 12.63.