Midterm 2014 Question 8

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Alex Nguyen 3I
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Joined: Fri Sep 25, 2015 3:00 am

Midterm 2014 Question 8

Postby Alex Nguyen 3I » Tue Feb 09, 2016 3:34 pm

Calculate the value of Ka for HF.
Can some explain to me the idea behind Ka again as well as why they assumed that potential difference would be negative?
I'm confused as to why it is assumed that the potential difference of the cell is going to be negative. We are trying to find the Ka for HF. Is it because we need H and F to be products and HF to be a reactant that we flip the half reactions accordingly?

Hailey Donaldson 1E
Posts: 44
Joined: Fri Sep 25, 2015 3:00 am

Re: Midterm 2014 Question 8

Postby Hailey Donaldson 1E » Tue Feb 09, 2016 6:41 pm

Potential difference can be negative, given that this reaction doesn't necessarily need to be spontaneous. Ka is the equilibrium constant expression with respect to HF, which would be given by Ka=[H+][F-]/[HF]. In order to get the Ka of HF, H+ and F- need to be in the reactants, so yes, you would flip the second equation. Assuming this reaction is at equilibrium (because you are solving for Ka), you can use the Nernst Equation (using emf=0 and Q=K) to find the expression for Ka. Hope this helped!

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