Question

Calculate the pH of 1.5 × 10⁻³ M solution of Ba(OH)₂.

Hint:

Always remember the stoichiometry! For group 2 hydroxides like Ba(OH)₂ or Ca(OH)₂, the concentration of OH^- is double the concentration of the base.

Answer 1

Concept: Barium hydroxide, Ba(OH)₂, is a strong base that dissociates completely in aqueous solution. To find the pH, we first need to determine the concentration of hydroxide ions (OH^-), then calculate the pOH, and finally use the relationship between pH and pOH.
Each molecule of Ba(OH)₂ releases two OH^- ions upon dissociation: Ba(OH)₂(aq) Ba²⁺(aq) + 2OH^-(aq) Given concentration of Ba(OH)₂ = 1.5 × 10⁻³ M. [OH^-] = 2 × [Ba(OH)₂] = 2 × 1.5 × 10⁻³ M = 3.0 × 10⁻³ M
The pOH is defined as the negative logarithm of the hydroxide ion concentration: pOH = -[OH^-] pOH = -(3.0 × 10⁻³) pOH = -( 3.0 + 10⁻³) = -(0.4771 - 3) = -(-2.5229) pOH ≈ 2.52
At 298 K (25C), the sum of pH and pOH is always 14: pH + pOH = 14 pH = 14 - pOH = 14 - 2.52 pH = 11.48