At certain temperature $H^{\oplus}$ ion concentration of pure water is $5 \times 10^{- 7}$ . The ionic product of water at this temperature is ………. $m o l e s^{2} l i t^{- 2}$

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$1 \times 10^{- 14}$

$5 \times 10^{- 14}$

$2.5 \times 10^{- 10}$

$2.5 \times 10^{- 13}$

answer is B.

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Detailed Solution

In pure water $[H^{+}] = [O H^{-}] = 5 \times 10^{- 7}$

$∴ K_{w}$ of this temperature $= [H^{+}] [O H^{-}] = {[5 \times 10^{- 7}]}^{2}$

$= 25 \times 10^{- 14} = 2.5 \times 10^{- 13} m o l e s^{2} l i t^{- 2}$

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