Posted by John on Wednesday, April 6, 2011 at 3:15am.
I would do this, all with the HH equation.
You didn't give a pKa for HF. I used 3.14 but you need to use whatever value your text has (or notes).
2.75 = 3.14 + log [(base)/(acid)]
Here is what you do for the (acid)
If we use all of the HF, and we want 250 mL total volume, I would calculate the molarity of that solution. That is
M = moles/L = mmoles/mL = (75mL x 0.25M/250mL) = 0.07500M
(base) = moles/L = (x moles/0.250 L).
Plug those into HH equation and solve for x moles NaF. Then moles NaF x molar mass NaF = grams. I get an answer close to 0.4 g but you need to go through it and do it more accurately. Finally, I suggest you start with the number of grams NaF, convert to moles and M, do the same for the HF solution, plug into the HH equation and see if you end up with a pH of 2.75. If yes you can be confident in your answer; if not there is an error somewhere.
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