chemistry

Calculate the [H+] for a solution with [OH−] = −80 M.

1x10*-14/-80=-log(answer)

How would I do this since I can not take the negative log of the previous anwser

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asked by susan
  1. I can tell you how to do it if you can tell me how you get a concn of anything less than zero. I can see (OH^-) = 1 x 10^-80 but that is still more than zero. Not by much I'll grant you, but still more than zero.

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    posted by DrBob222

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