What is the pOH of a $8.4 x 10^{- 11} M$ $8.4x10^-11 M$ $O H^{-}$ $OH^-$ solution?

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Answer 1 · 2016-08-28T19:46:27

$p O H = 10.08$ $pOH=10.08$

The $p O H$ $pOH$ of a solution is defined in terms of the concentration of hydroxide ions in solution as follows:

$p O H = - log [O H^{-}]$ $pOH=-log[OH^(-)]$ .

In this question, $[O H^{-}] = 8.4 \times 10^{- 11} M$ $[OH^(-)]=8.4xx10^(-11)M$ , thus the $p O H$ $pOH$ can be calculated by:

$p O H = - log (8.4 \times 10^{- 11}) = 10.0757 ≅ 10.08$ $pOH=-log(8.4xx10^(-11))=10.0757~=10.08$

Note that if the concentration is given with 2 significant figures , the $p O H$ $pOH$ value is written with two decimal places.

What is the pOH of a 8.4x10−11M8.4x10^-11 M OH−OH^- solution?