Chemistry

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

1 Answer

  • [H+] [OH-]= Kw
    [H+] [OH-]=5.47x10^-14
    Log [H+]+ Log [OH-]= -14 logĀ 5.47
    PH+POH= -13.26
    we need more info to solve this
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