Answer to Question #167848 in Chemistry for Gail

Question

Calculate the pH and the pOH of the solution at 25 c for 0.0031 M Ca(OH)2

Answer 1

Solution:

When Ca(OH)₂ molecules dissolve they dissociate into Ca²⁺ ions and OH^- ions.

Ca(OH)₂ → Ca²⁺ + 2OH^-

For every Ca(OH)₂ you produce 2OH^-.

Therefore, the [OH^-] equals 0.0062 M:

[OH^-] = 2 × C_M(Ca(OH)₂) = 2 × 0.0031 M = 0.0062 M

We can convert between [OH^-] and pOH using the following equations:

pOH = - log[OH^-]

pOH = -log(0.0062) = 2.21

pOH = 2.21

For any aqueous solution at 25^∘C:

pH + pOH = 14

pH = 14 - pOH = 14 - 2.21 = 11.79

pH = 11.79

Answer: pOH = 2.21; pH = 11.79