Ph logh3o ph poh 14 h3o oh- 1.0 x 10-14 ph -log0.001, Ask an Expert

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PH and pOH
Here is the pH of each of the following:
pH = log[H3O+] , pH + pOH = 14, [H3O+] * [OH-] = 1.0 X 10^-14
pH = -log(0.001) = 3
then...
pOH = -log[OH-] = -log(0.001) = 3
pH = 14- pOH = 14 - 3 = 1

Could you show me a few of the steps between line one and two in both of these conclusions?

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