The iron product of water $p K_{w}$ is 14. If a rain water sample has a pH of 5.6, the concentration of $O H^{-}$ in the sample (in $10^{- 9}$ mol/litre, round off to one decimal place), is____
3.981

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Solution

The correct option is A 3.981
We know that,

$p k_{w} = p H + p O H$

$\Rightarrow p O H = (14 - 5.6) = 8.4$

$\Rightarrow p O H = - l o g [O H^{-}]$

[ $O H^{-}$ concentration of $O H^{-}$ ion in $m o l / l t]$

$\Rightarrow 8.4 = - l o g [O H^{-}]$

$\Rightarrow [O H^{-}] = 3.981 \times 10^{- 9} m o l / l t$

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Similar questions

Alkalinity of water, in equivalent/litre (eq/litre), is given by {H C O_{3}^{-}} + 2 {C O_{3}^{- 2}} + {O H^{-}} - {H^{+}}

H C O_{3}^{-} = 2 \times 10^{- 3} m o l / l i t r e,

C O_{3}^{2 -} = 3.04 \times 10^{- 4} m o l / l i t r e

O H^{-}

H^{+}

NEGLECTED, the alkalinity of the water sample (in mg/litre as C a C O_{3}

O H^{-}

25^{\circ}

C a C l_{2}

9.5 g M g C l_{2}

C l^{-}

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