posted by Anonymous .
You want to make an acetic acid/acetate buffer, pH 3.8, with a volume of 250 mL, and a final concentration of ([AcOH] + [AcO-]) = 0.1 M. You may only use acetic acid and sodium acetate (no strong acid or base). How many mL of glacial acetic acid would be needed (d = 1.05 g/mL)?
You need to look up pKa for acetic acid in your text or notes. I used pKa = 4.74.
And is that 3.8 or 3.80? It will make a difference in determining the number of significant figures.
pH = pKa + log[(base)/(acid)]
3.80 = 4.74 + log B/A
Solve for B/A which is equation 1.
Then A + B = 0.1 is equation 2.
Solve these two equation simultaneously for A and B which will give you the mols A and mols B. I assume you can take it from here.
using this method i calculated 1.03 mL, but this was not one of the options
the options were:
a. ) 1.28 mL
b. ) 1.41 mL
c. ) 0.15 mL
d. ) 2.18 mL
Sorry. I didn't correct for the volume.
You want 0.250 L (not a liter) of solution.
So A + B = 0.025
and B/A is the same as before. By the way, when I solved for B/A, I get 0.1136. That's too many s.f. but I carried it through. You can round at the end. I ended up with answer a.
I got 1.028mL instead of 1.28mL.
After finding the concentration of the acid, I multiplied by the molar mass. then, i multiplied by L/1050g to cancel out the g. Finally, I multiplied by 0.250L. Clearly, I'm making this too complicated.
B/A = 0.1136 or
B = 0.1136*A
A + B = 0.025
A + 0.1136A = 0.025
A = 0.025/1.1136 = 0.02245 moles
0.02245*60g/mol = 1.347 grams acetic acid.
1.347/1.05 = 1.282 which rounds to 1.28 mL.