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

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
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 &lt;-> 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:

$$\ce{KOH + HF -&gt; H2O + F- + K+}$$

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.