Textbook Question 6.57

[Ian_Lee_1E]

For this question

1. Is the given half reaction reversed to oxidation even thought it has a greater reduction potential BECAUSE the question is asking for the value of Ka for HF?

2. What is the relationship between Ka and K? because in the answer, in order to find Ka, the calculated K was square rooted.

[Emily Ding 1J]

I had the exact same questions and only just figured it out! We switch the anode/cathode even though the potential is negative just so that HF is our reactant and the second part works out (will explain below), but I literally made the same mistake. For the second part, I just realized that we're supposed to use the Henderson-Hasselbalch equation which is:

[H+]=Ka[HA]/[A-]

If we rearrange this, we get Ka= [H+][A-]/[HA],

which is the same general format we get for K if we arrange our equation so that HF is the reactant and H+ and F- are the products:

2HF --> 2H+ + 2F-
K= [H+]^2[F-]^-2/[HF]^2

Due to the coefficients from the equation, we see that every term has been squared. So, in order to convert from K to Ka, we just have to square root K.

I hope this made sense!! I completely forgot about this equation too but I hope this was helpful.

[Emily Ding 1J]

Sorry, I didn't mean for there to be a negative in front of the 2 for [F-] in K! I hope the rest still makes sense though.