How to caculate hydroxide ion concentrations

Question

So i have some problems that ask me to solve the hydronium and hydroxide ion concentration of 4 substances:

rainwater: pH=5.3

household ammonia: pH= 11.9

vinegar: pH=2.7

seawater: pH= 7.6

Ok, so i know that you just take 10^-pH to get hydronium ions, but i'm not exactly sure how to get hydroxide ions. Can someone help please? Thank you

Answer 1

pH + pOH = 14.0

[OH-] = 10^-pOH

[H+]=10^-pH

{H+] x [OH-] = 1x10^-14

Answer 2

pOH = 14 - 5.3 = 8.7

[OH-] = 10^- 8.7 =2 x 10^-9 M

There is another way :

[H+] = 10^-5.3 = 5.0 x 10^-6 M

Since [H+][OH-] = 1.0 x 10^-14

[OH-] = 1.0 x 10^-14 / 5.0 x 10^-6 = 2 x 10^-9

in the same way

pOH = 14 - 11.9 = 2.1

[OH-] = 10^-2.1 =0.0079 M

pOH = 14 - 2.7 =11.3

[OH-] = 10^-11.3 =5.0 x 10^-12 M

pOH = 14 - 7.6 =6.4

[OH-] = 10^-6.4 =4.0 x 10^-7 M