Acid base and pH

Question

What is the [OH^−] of a 0.00033 M solution of Ca(OH)2?

What is the [H3O^ +] of the solution?

Answer 1

Ca(OH)₂ ==> Ca²⁺ + 2OH^- (dissociation of Ca(OH)₂ in water)

0.00033 M Ca(OH)₂ produces 2 x 0.00033 M OH^- = 0.00066 M OH^- = 6.6x10^-4 M = [OH^-]

[H₃O⁺][OH^-] = Kw = 1x10^-14

[H₃O⁺] = 1x10^-14 / 6.6x10^-4

[H₃O⁺] = 1.5x10^-11 M