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### 6.57

Posted: **Tue Feb 25, 2020 10:19 pm**

by **Joseph Saba**

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

How would I do this problem?

### Re: 6.57

Posted: **Wed Feb 26, 2020 1:02 am**

by **Benjamin Feng 1B**

You can relate E to Gnot using Gnot = -nFE. From there, relate Gnot to K using Gnot = -RT lnK. In this case, the reaction given makes K equal to Ka

### Re: 6.57

Posted: **Sat Feb 29, 2020 7:13 pm**

by **705121606**

Benjamin Feng 1B wrote:You can relate E to Gnot using Gnot = -nFE. From there, relate Gnot to K using Gnot = -RT lnK. In this case, the reaction given makes K equal to Ka

I solved up until K but didn't know how to get to Ka. In the solutions manual they took the square root of K to solve for Ka. Do you know why this is the case?

### Re: 6.57

Posted: **Sat Feb 29, 2020 10:23 pm**

by **Emma Popescu 1L**

Using the appendix you find that F2+ 2e- -> 2F- gives a value of Enot of +2.87V. Then by using the equation Enot (cathode)- Enot (anode), you can find Enot cell. Then using the equation InK=nFEnot/RT you can find K. To find Ka, just find the square root of K.