HW 7th edition 6B.9

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Kathryn Wilhem 1I
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Joined: Fri Sep 28, 2018 12:27 am

HW 7th edition 6B.9

Postby Kathryn Wilhem 1I » Sun Dec 02, 2018 9:29 pm

For 6B.9 part a row i and ii, we are given a chart giving just the [H3O+] for i and [OH-] for ii and we have to find the pH, pOH, [H3O+], and [OH-]. For i, we are given [H3O+] = 1.50M. If we use the formula pH = -log[H3O+], then we get a negative pH, and the same goes for using pOH = -log[OH-] in ii using the given [OH-] = 1.50M. How do we know when to use the negative sign in the formula or not?

juleschang16
Posts: 55
Joined: Fri Sep 28, 2018 12:19 am

Re: HW 7th edition 6B.9

Postby juleschang16 » Sun Dec 02, 2018 9:32 pm

You always use the negative sign in the pH formula, regardless of the outcome.

Samantha Hoegl Roy 2C
Posts: 81
Joined: Fri Sep 28, 2018 12:15 am

Re: HW 7th edition 6B.9

Postby Samantha Hoegl Roy 2C » Tue Dec 04, 2018 10:10 pm

For part i) why is the OH- concentration 1.5 x 10^-14? I thought the equation for was a concentration of H3o+ multiplied by the concentration of OH- = 10 x 10^-14
thus OH- conc = 10 x 10^-14 / 1.5 but that seems to be wrong

Julia Lindner 1I
Posts: 35
Joined: Fri Sep 28, 2018 12:17 am

Re: HW 7th edition 6B.9

Postby Julia Lindner 1I » Wed Dec 05, 2018 7:52 pm

I had the same question... both part i) and part ii) run into the same problem, where taking the negative log of 1.50 gives -0.176 instead of positive 0.176. Anyone have an idea what's happening?


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