Calculate the pOH of each of the following aqueous solutions

Question

(1) 1.75 M NaOH
(2) 0.045 M Ba (OH)2

Answer 1

pOH = -log(OH^-)
For 1 the (OH^-) = (NaOH)
for 2 the (OH^-) = 2*[Ba(OH)2]