Question

help please, how can I get the [OH-],[H3O+],pH, pOH. Given 0.18 M NaOH

Answer 1

[OH\(^-\)]= \(\sf \color{}{10^{-pOH}}\)
if you want to solve for pH, you can log both sides to get: \(\sf \color{red}{-log[OH^-]=pOH}\)

Answer 2

Also note, \(\sf \color{}{k_w = [OH^-][H_3O^+]=1 \times 10^{-14}}\) at 25ºC

Answer 3

pOH=-log[OH-]
=-log[0.18]
now,
pOH+pH=14
find the value of pOH first and put it in dis formula u wil het pH