Okay I ran into this question in homework.
A 100.0 ml sample of 0.20M HF is titrated with 0.10 M KOH. Determine the pH of the solution after adding 200 ml of KOH. The Ka of HF is 3.5 x 10-4?
Oh yeah its multiple choice
- 3.46
- 10.54
- 8.14
- 9.62
- 7.00
I've found answers to similar questions.Here
But I come into a different problem with this question. The HF and the KOH cancel out each other for they have the same amount of moles
I assume what ever we're looking at on the other side will have 0.02 moles and is an acid
- HF -> 0.1L * 0.2M = 0.02 mol
- KOH -> 0.2L * 0.1M = 0.02 mol
- HF-KOH = 0
I beleive this means I can't use the hasselbach equation so I did this:
(M of ???) = 0.02 mol / 0.3 L = 0.667*10^(-2)
3.5*10^(-4) = x^(2) / (0.667*10^(-2))
sqrt(3.5*10^(-4)*(0.667*10^(-2))) = x
so there is another way of solving it
I have had some feed back on this reaction: "You assume the first reaction goes to completion when the Ka is not that high"
If it doesn't go to completion what do I do?!
I really just need to see how this works out and know why things are happening
