At $90^{\circ} C$ for pure water, $[H_{3} O^{+}] = 10^{- 6.5}$ mol/litre. The value of $K_{w}$ at this temperature is:

10^{- 6}

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10^{- 13}

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10^{- 14}

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10^{- 8}

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Solution

The correct option is A $10^{- 13}$
Pure water is neutral having equal concentrations of hydrogen ion and hydroxide ion.
$K_{w}$ is a self-ionization constant which is obtained from a self-ionization reaction, as given below:
$H_{2} O + H_{2} O ⟺ H_{3} O^{+} + O H^{-}$ where $[H_{3} O^{+}] = [O H^{-}] = 1 \times 10^{- 6.5}$
$K_{w} = [H_{3} O^{+}] [O H^{-}]$
Hence, $K_{w} = 1 \times 10^{- 6.5} \times 1 \times 10^{- 6.5} = 1 \times 10^{- 13}$ .

Option B is correct.

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