textbook 6.57

[Elena Chen 2E]

Use the data in Appendix 2B and the fact that, for the half-reaction F2(g) + 2 H+(aq) + 2 e- ----> 2 HF(aq), Eº= +3.03 V, to calculate the value of Ka for HF.

Here's what the answer key says:

So I'm ok up to the point where they say to use lnK = nFEº/RT. What I don't understand is how they got the values after that. I got lnK=-12.461878, which drastically changes the K value. Is this just a rounding difference, or is there something more I'm missing?
Also why do they take the square root of the K value at the end?

Thanks!

[Jessie Hsu 1C]

I'm not sure why the textbook's answer is -12 instead of -12.461878 either, but I think in the end they square root the value because the original K calculated was for 2HF --> 2H+ + 2F-. To convert it to the K for only one HF, you square root the original K.