Calculate amount of solid NaF required to prepare a 250mL acidic buffer with pH=2.75. The initial molarity of HF is .25M but you only have .075L of it.

I have tried many different ways... the most recent and seems to be simplest was using H-H equaion and solving for the CB/CA ratio. Then equivalating this to mol NaF/ (M HF* V HF).... any suggestions?

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  1. 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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