Finding the pH among HF and KOH

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](https://answers.yahoo.com/question/index?qid=20091208210458AAQNKyR)


But I come into a different problem with this question. T**he 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 




From the comments:



> 
> [OP] Okay Thank you all I finally got it. so K is just thrown out the window. and we are left with the equation (F- + H2O <-> FH + OH-) So i take the Ka to make Kb by (Kw/Ka) and my concentration of F- and OH- was (0.02/0.3) which I plugged into sqrt((Kb)\*(M of OH- or F-)) took the -log of that got the pOH and then converted it back to pH (14 - pH) One final thing why did I K out the window does that mean its neutral. Oh and the answers 8.14
> 
> 
> 


Paraphrased with paragraphs:


A strong base ($\ce{KOH}$) reacts with a weak acid ($\ce{HF}$) at stoichiometric ratio:


$$\ce{KOH + HF -> H2O + F- + K+}$$


The major species is fluoride, a weak base with pKb = 14 - pKa, where pKa is that of hydrofluoric acid. The potassium ion is a spectator. To find the pH, use your favorite strategy for a pure weak base. The result is pH = 8.14.


As MaxW pointed out in the comments, this relies on getting the stoichiometric ratio just right. If $\ce{KOH}$ is even in slight excess (let's say one tenth of a millimole per liter too much), the pH will be strongly basic (pH 10). If you err in the other direction, you will have an $\ce{HF, F-}$ buffer with an acidic pH.


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