Finding the pH among HF and KOH
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.