What is the #"pOH"# of a solution if the #["OH"^-] = 33.5 xx 10^-2# #"M"#?

Question

What is the #"pOH"# of a solution if the #["OH"^-] = 33.5 xx 10^-2# #"M"#?

1 Answer

Impact of this question

1695 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-15T23:52:21

#"pOH"=0.475#

Explanation:

The chemical definition of #"pOH"# or in fact, the #"p"# of any value/constant is given as the cologarithm of that thing. That is the negative (or reciprocal) logarithm of that thing.

#therefore# #"pOH"=-log["OH"^-]=log(1/(["OH"^-]))#

#"pOH"=-log(3.35xx10^-1)=0.475#

Science

Math

Humanities

... and beyond

What is the #"pOH"# of a solution if the #["OH"^-] = 33.5 xx 10^-2# #"M"#?

1 Answer

Explanation:

Related questions

Impact of this question