Verified answer

[OH-][H+]=10^-14

=> 10^-9 x [H+] = 10^-14

=> [H+] = 10^-5 M

pH = -log[H+] = 5

pOH = -log[OH-] = 9

[OH-] x [H+] = Kw <=> [H+] = Kw/[OH-] = 10^-14/10^-9 = 10^-5 M

pH = -log10[H+] = -log10(10^-5) = 5

pKw = pH + pOH <=> pOH = pKw - pH = 14 - 5 = 9

Hope I helped! If so, thumbs up! :)