What is the $pOH$ $"pOH"$ of a solution if the $[{OH}^{-}] = 33.5 \times 10^{- 2}$ $["OH"^-] = 33.5 xx 10^-2$ $M$ $"M"$ ?

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What is the $pOH$ $"pOH"$ of a solution if the $[{OH}^{-}] = 33.5 \times 10^{- 2}$ $["OH"^-] = 33.5 xx 10^-2$ $M$ $"M"$ ?

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Answer 1 · 2017-04-15T23:52:21

$pOH = 0.475$ $"pOH"=0.475$

Explanation:

The chemical definition of $pOH$ $"pOH"$ or in fact, the $p$ $"p"$ 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 $pOH$ $"pOH"$ of a solution if the $[{OH}^{-}] = 33.5 \times 10^{- 2}$ $["OH"^-] = 33.5 xx 10^-2$ $M$ $"M"$ ?

1 Answer

Explanation:

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

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What is the $pOH$ $"pOH"$ of a solution if the $[{OH}^{-}] = 33.5 \times 10^{- 2}$ $["OH"^-] = 33.5 xx 10^-2$ $M$ $"M"$ ?