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$K_{w}$ of pure water is $10^{- 12}$ at $60^{0} C$ . The $p^{H}$ of pure water at $60^{0} C$ is

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answer is 6.

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Detailed Solution

$(H^{+}) (O H^{-}) = K_{w} (H^{+}) = 10^{- 6} (H^{+}) = (O H^{-}) i n p u r e w a t e r p H = 6$

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