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Author Topic: Ka pKa-calculations  (Read 524 times)

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tubstar98

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Re: Ka pKa-calculations
« Reply #15 on: December 06, 2017, 04:12:05 AM »

And there will be no NH3 before the reaction occurs?
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Borek

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Re: Ka pKa-calculations
« Reply #16 on: December 06, 2017, 05:10:00 AM »

There will be the original amount of NH4 minus the amount NH4 which have given away an H atom to OH- to make H2O. the former posted calculations was wrong.

Yes.

Quote
so how much OH- is now in the liquid? have all of this reacted?

You can't ever use all OH- from the water, as because of water autodissociation there will be always some equilibrium concentration present. But for stoichiometry you can assume all of it reacted (and equilibrium concentration can be easily calculated from known pH).

And there will be no NH3 before the reaction occurs?

Small equilibrium amounts, safe to ignore.
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tubstar98

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Re: Ka pKa-calculations
« Reply #17 on: December 06, 2017, 05:40:22 AM »

I have now calculated my pKa value. It is 0,2 % deviant(spelling?) from what the assignment want me to compare it to. I have also calculated the [OH-] using the [H+].

I think i have understood what this assignment is about and how to calculate the problems,  it has a lot to do with the help you have given me, thank you very much for your time and effort in helping me understand. It is really appreciated!
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