7th Edition 6B.9

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Kathryn Wilhem 1I
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7th Edition 6B.9

Postby Kathryn Wilhem 1I » Wed Jan 16, 2019 8:51 pm

For parts (i) and (ii) of the problem, we are given the [H30+] = 1.50M and [OH-] = 1.50M respectively. When I calculated the pH and pOH, I found that the answer was -0.176. When we know that taking the log of the concentration will give us a positive answer, do we just not put a negative sign in front of it?

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Re: 7th Edition 6B.9

Postby Chem_Mod » Thu Jan 17, 2019 1:52 pm

pH =-log[H3O+], if you plug in the 1.5, you should get about 0.176


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