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One litre of solution contains $10^{- 5}$ moles of $H^{+}$ ions at 25° C. Percentage ionization of water in solution is :

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$1.8 \times 10^{- 9} %$

$1.8 \times 10^{- 7} %$

$3.6 \times 10^{- 9} %$

$1.8 \times 10^{- 11} %$

answer is B.

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Detailed Solution

$\begin{array}{l} [{OH}^{-}] = 10^{- 9} = \frac{1000}{18} \times α \\ \Rightarrow α = 1.8 \times 10^{- 11} or 1.8 \times 10^{- 9} % \end{array}$

Thus, ionization of water in solution is $1.8 \times 10^{- 9} %$

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