6B.9 (a)

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AngieGarcia_4F
Posts: 120
Joined: Thu Jul 25, 2019 12:17 am

6B.9 (a)

Postby AngieGarcia_4F » Wed Jan 22, 2020 8:24 pm

For this problem I keep taking the -log of [H3O+] to find the pH and getting -0.176 but the answer book says it's +0.176, what am I doing wrong?

AngieGarcia_4F
Posts: 120
Joined: Thu Jul 25, 2019 12:17 am

Re: 6B.9 (a)

Postby AngieGarcia_4F » Wed Jan 22, 2020 8:25 pm

The [H3O+] is 1.5M

Hope Hyland 2D
Posts: 50
Joined: Wed Feb 20, 2019 12:16 am

Re: 6B.9 (a)

Postby Hope Hyland 2D » Thu Jan 23, 2020 12:24 am

I was also really confused on how to get the answer to parts (i) and (ii) of this question. How come you can't use the given value of [H30+]=1.5 mol.L-1 and then substitute into [H30+][OH-]=10^-14 to find [OH-]? I tried doing that, but I kept getting the wrong answer.

Morgan Carrington 2H
Posts: 54
Joined: Wed Nov 14, 2018 12:22 am

Re: 6B.9 (a)

Postby Morgan Carrington 2H » Thu Jan 23, 2020 1:11 am

I think sometimes the concentrations can produce negative numbers when calculating pH or pOH because I don't see any other way to find the pH without getting a negative concentration.


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