Question

What is the solution to the question; The hydrogen ion concentration of a sample of orange juice is `2.0×10^-11mol` `dm^-3`. What is its `"pOH"`?

Answer 1

`"pOH"=3.30`

Explanation:

`"pOH"=-log([OH^-])=-log(K_w/([H^+]))`

`K_w=1*10^-14`
`[H^+]=2.0*10^-11`

`"pOH"=-log((1*10^-14)/(2.0*10^-11))=3.30`