If ... [OH -] = 9.1E-3 M ... then ... pOH = ?

Question

... [OH -] = 10 - pOH

Answer 1

take -log(9.1E-3 M) = 2.04

There are as many numbers after the decimal place in the pOH as there are significant figures in your molarity.

No, btw hydroxide concentration does not relate to 10 and for that matter minus the pOH.

pOH = 14 - pH

Answer 2

9.1E-3 M  ---> log ---> change the sign

9.1E-3 M  ---> log ---> -2.04  --> +/- --> +2.04

your answer is : POH = 2.04

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& yes you are right , if by "... [OH -] = 10 - pOH" ,...

you are stating that :

[OH-] = 10^(-pOH)

because we do take the pOH , & change the sign to "-pOH",...

& the do a "10^x" with it to get the [OH-]