If $H O^{\equiv} 2.7 \times 10^{- 11} \cdot m o l \cdot L^{- 1}$ $HO^-=2.7xx10^-11molL^-1$ , what is $p O H$ $pOH$ of this solution?

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Answer 1 · 2016-08-25T17:42:07

$p O H$ $pOH$ $=$ $=$ $10.57$ $10.57$

$p O H$ $pOH$ $=$ $=$ $- {log}_{10} [H O^{-}]$ $-log_10[HO^-]$ .

Thus $p O H$ $pOH$ $=$ $=$ $- {log}_{10} (2.7 \times 10^{- 11})$ $-log_10(2.7xx10^-11)$ .

Is this solution acidic or basic? How do you know?

If HO≡2.7×10−11⋅mol⋅L−1HO^-=2.7xx10^-11*mol*L^-1, what is pOHpOH of this solution?