6B.1

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Jasmine 2C
Posts: 119
Joined: Wed Sep 18, 2019 12:18 am

6B.1

Postby Jasmine 2C » Sat Dec 07, 2019 10:27 pm

Why is there a [HCL]0 on the denominator? Isn't the new molarity just going to be 0.12 x [HCL]0 since the molar concentration is reduced to 12% of the initial value?

Jiyoon_Hwang_2I
Posts: 50
Joined: Sat Sep 14, 2019 12:17 am

Re: 6B.1

Postby Jiyoon_Hwang_2I » Sun Dec 08, 2019 3:19 am

I think it's because they're using the log rule that when you do log x - log y you can also write it as log (x/y) and in this problem they were finding the change in pH so they were doing log(final) - log(initial). So, the final would be 0.12 x [HCl]o and the initial is just [HCl]o

MaggieHan1L
Posts: 61
Joined: Thu Jul 11, 2019 12:17 am

Re: 6B.1

Postby MaggieHan1L » Sun Dec 08, 2019 11:28 pm

That's the only way to get a number without knowing the concentrations. Since it's -log (0.12H+)-(-log (H+), you can change it to log (H+)-(log 0.12(H+)) and thus you can take log 1/0.12 since the H+ will cancel.


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