Chemistry
posted by Jake on .
I can't figure out how to solve this problem:
The Ka for hydroflouric acid is 4.5 *10^4 at 298K. Calculate the amount of sold NaF that is required to prepare a 250. mL acidic buffer with pH = 2.75. The initial molarity of hydroflouric acid is .250M but you have only 75mL.
I looked on here, and someone said they got .2 g. I'm not getting anything close to that. Help please?

I think 0.2 g NaF is close if not correct.
M of the HF = (75mL x 0.25M/250 mL) = 0.075
M of the NaF = (xgrams/molar mass/0.25 L) = (x/41.988/0.25) = ?? whatever that is.
So substitute into the HH equation.
2.75 = 3.35 + log[(x/41.988/0.25)]/0.075 and solve for x = grams NaF. My numbers crunch to 0.198 grams.
If you will post your work I will look for any errors.