Finding pOH of a solution

Question

The pOH of 0.0280 M solution of Sr(OH)₂ is

I tried doing -log(Molarity) and that did not work.

I also tried taking the number I found from above and subtracting it from 14. I am thoroughly confused about the steps in finding pOH. Any help would be appreciated.

Answer 1

The pOH is -log molarity of the OH^- concentration. Note that when you have Sr(OH)₂, the molarity of the OH^- is TWICE the molarity of the Sr(OH)₂ because it dissociates to Sr²⁺ + 2OH^-

[OH^-] = 2 x 0.0280 M = 0.0560 M

pOH = -log 0.0560

pOH = 1.252