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Auto Protolysis of Water

At 363 K, pur...

Question

At 363K, pure water has $[H_{3} O^{+}] = 10^{- 6} M$ . The value of Kw at this temperature will be

$10^{- 6}$

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$10^{- 12}$

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$10^{- 13}$

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$10^{- 14}$

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Solution

The correct option is B
$10^{- 12}$

Since water is pure, $[H_{3} O^{+}] = [O H^{-}] = 10^{- 6} M$

From, $K_{w} = [H_{3} O^{+}] [O H^{-}]$

= $10^{- 6} \times 10^{- 6} = 10^{- 12} M$

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