6B.9a

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Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

6B.9a

Postby Alicia Lin 2F » Wed Jan 15, 2020 2:33 pm

For the first line, when I type in -log(1.50) to try to get the pH, I keep getting -0.176 but the answer is supposed to be positive 0.176. Also, when I multiply the OH- and H3O+ concentrations in the answer key, I don't get 1.0 x 10^-14. Is this an error in the book?

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

Re: 6B.9a

Postby Justin Vayakone 1C » Wed Jan 15, 2020 2:58 pm

Yeah it looks like the answer key is wrong. The pH should be -0.176. With this pH, the pOH would now be 14.176 and =


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