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.
Chemistry
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Question
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
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
1 Answer

1. User Answers sciencific
[H+] [OH]= Kw
[H+] [OH]=5.47x10^14
Log [H+]+ Log [OH]= 14 logĀ 5.47
PH+POH= 13.26
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