6B.11

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Michelle Song 1G
Posts: 30
Joined: Wed Nov 14, 2018 12:23 am

6B.11

Postby Michelle Song 1G » Wed Jan 16, 2019 11:46 pm

For part ii of a, how do you figure out the concentration of OH- from the original solution? The solutions manual says to multiple the concentration of OH- in the diluted solution by 500.0 mL/5.00 mL but I don't understand why.

Camille Marangi 2E
Posts: 60
Joined: Fri Sep 28, 2018 12:26 am

Re: 6B.11

Postby Camille Marangi 2E » Thu Jan 17, 2019 2:24 pm

So starting with the current concentration of OH- in the diluted solution you do pH+pOH=14 so pOH=14-13.25=0.75. Then 10^-0.75=[OH-] in the diluted solution. This comes out to be 0.178 M or 0.18 M with sig figs. From there, we have to work backwards to get to the original solution. So we know the concentration we have now arose from diluting to 500 mL. Therefore we multiply 0.18mol/L by 0.5L to give us 0.09 mol. Following this, we know that our original concentration was diluted from 5 mL to 500 mL and thus to get back to our original 5mL solution (and therefore our original concentration) we divide our moles by .005 (.09/.005) and this gives us the initial concentration of 18M.

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

Re: 6B.11

Postby Matthew Tran 1H » Thu Jan 17, 2019 5:17 pm

The easiest way to do this is to use the dilution formula that we used a lot in Chem 14A: since moles stay the same (n1=n2), M1V1=M2V2, where M is molarity and V is volume. Plugging in the numbers, you would get M1(5.00mL)=(0.18M)(500.0mL) --> M1=18M.


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