Calculate the concentration of ${O H}^{-}$ ions in solution when $[H^{+}] = 6.2 \times 10^{- 2} m o l L^{- 1}$ .

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Solution

We know, $[H^{+}] \times [O H^{-}] = 1 \times 10^{- 14}$ at $25^{o} C$
(Given: $[H^{+}] = 6.2 \times 10^{- 2} m o l L^{- 1}$ )
Thus,
$[O H^{-}] = \frac{1 \times 10^{- 14}}{6.2 \times 10^{- 2}} = 1.61 \times 10^{- 13} m o l L^{- 1}$
$H_{2} O ⟶ {O H}^{-} + H^{+}$
$[O H^{-}] = \frac{[H_{2} O]}{[H^{+}]}$

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