The ionic product of water at 60degree C is 9.61*10^-14.The pH of water at 60degree C is

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Solution

$H_{2} O \to H^{+} + O H^{-} K_{w} = [H^{+}] [O H^{-}] = 9.61 \times 10^{- 14} [H^{+}] = [O H^{-}] = 3.1 \times 10^{- 7} p H = - \log [H^{+}] = - \log (3.1 \times 10^{- 7}) = 7 - \log 3.1 = 6.509$

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