If $H^{+}$ ion concentration in water is $10^{- 2}$ mol/l then $O H^{-}$ ion concentration

17 \times 10^{- 3}

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17 \times 10^{- 2}

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17 \times 10^{- 1}

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Solution

The correct option is A $17 \times 10^{- 3}$
$Concentration of H^{+} = 10^{- 2}$
$Concentration of H^{+} + Concentration of O H^{-} = 10^{- 14}$

$O H_{(m g / l)}^{-} = 10^{- 12} \times 17 \times 1000 \times 10^{6}$

$= 17 \times 10^{- 3}$

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