J.R. S. answered 07/26/23
Ph.D. University Professor with 10+ years Tutoring Experience
0.02 M NaOH => 0.02 M OH-
pOH = -log 0.02 = 1.70
pH = 14 - 1.70 = 12.30
0.5 M NaOH => 0.5 M OH-
pOH = -log 0.5 = 0.30
pH = 14 - 0.30 = 13.7
0.04 M Mg(OH)2 => 2x0.04 M OH- = 0.08 M OH-
pOH = -log 0.08 = 1.10
pH = 14 - 1.1 = 12.9
0.5 M HClO4
This is a strong acid and produce a pH significantly less than 7, so rule this out.
0.5 M NaOH WILL PRODUCE A SOLUTION WITH A pH OF 13.7