calculate the pH in a 0.025 M Ca(OH)2 solution

Answer 1

Ca(OH)₂ ==> Ca²⁺ + 2OH^-

Since 1 mol Ca(OH)₂ produces 2 mols OH^-, the [OH^-] in this example will be 2 x 0.025 M = 0.050 M

pOH = -log [OH-] = -log 0.050 = 1.30

And recall the pH + pOH = 14,... therefore....

pH = 14 - pOH = 14 - 1.3

pH = 12.7