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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

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  • $\begingroup$ I would recommend you stick to the assumption that the reaction is complete first, before going on to consider the incomplete case. If the reaction is complete, what products are you going to form? $\endgroup$ – orthocresol♦ Nov 8 '15 at 21:20
  • $\begingroup$ I'm sorry but I wasn't sure really either FH2 + KO or H2O + FK not really sure actually $\endgroup$ – Ryan Henry Nov 8 '15 at 21:48
  • $\begingroup$ Which of HF and KOH is the base? Which is the acid? $\endgroup$ – orthocresol♦ Nov 8 '15 at 22:15
  • $\begingroup$ HF is the acid KOH is the base $\endgroup$ – Ryan Henry Nov 8 '15 at 22:17
  • $\begingroup$ Uhhuh. So the acid is a proton donor right? And the base is a proton acceptor right? What species will they form? Remember KOH exists as $\ce{K+}$ and $\ce{OH-}$ in solution. $\endgroup$ – orthocresol♦ Nov 8 '15 at 22:18
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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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