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pH the Power of H

At 90∘C pure ...

Question

At $90^{\circ}$ C pure water has [ ${H_{3} O}^{+}$ ] = $10^{- 6}$ mol ${l i t r e}^{- 1}$ . The value of $K_{W}$ at $90^{\circ}$ C is

$10^{- 6}$

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$10^{- 12}$

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$10^{- 14}$

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$10^{- 8}$

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Solution

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

Say x = [ ${H_{3} O}^{+}$ ]=[ ${O H}^{- 1}$ ] = $10^{- 6}$ mol ${l i t r e}^{- 1}$

$K_{W}$ = [ ${H_{3} O}^{+}$ ] $\times$ [ ${O H}^{- 1}$ ] $10^{- 12}$ ${{(m o l l i t r e}^{- 1})}^{- 2}$

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