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The ionic product of water, Kw at 298K = 1 x 10^-14. Given that the Kw of water at 50 degrees C is 5.47 x 10^-14, calculate the pH of water at this temperature. Please help! thanks
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The ionic product of water, Kw at 298K = 1 x 10^-14.

Given that the Kw of water at 50 degrees C is 5.47 x 10^-14, calculate the pH of water at this temperature.

Please help! thanks

User Jonathan Burley
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1 Answer

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[H+] [OH-]= Kw
[H+] [OH-]=5.47x10^-14
Log [H+]+ Log [OH-]= -14 log 5.47
PH+POH= -13.26
we need more info to solve this
User Kyle Robson
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