What is the $p O H$ $pOH$ of a $2.1 \cdot 10^{- 6}$ $2.1*10^-6$ $M$ $M$ solution?

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Answer 1 · 2016-07-02T13:04:43

We assume that you mean a $2.1 \times 10^{- 6} \cdot m o l \cdot L^{- 1}$ $2.1xx10^-6*mol*L^-1$ solution of $K O H$ $KOH$ .

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

$=$ $=$ $- {log}_{10} (2.1 \times 10^{- 6})$ $-log_10(2.1xx10^-6)$ $=$ $=$ $- (- 5.68)$ $-(-5.68)$ $=$ $=$ $5.68$ $5.68$ .

What is $p H$ $pH$ of this solution? You should be able to answer this easily.

What is the pOHpOH of a 2.1⋅10−62.1*10^-6 MM solution?