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

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

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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#

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

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Explanation:

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