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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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