Describe the preparation of 500 mL of a 0.100 M propanoate buffer that has a pH of
4.90. Assume you have solid sodium propanoate (NaC3H5O2) and 0.250 M propanoic acid (HC3H5O2).
- hcbiochemLv 74 months ago
Using HPr and Pr- because it's late and I don't want to type the whole formulas.
HPr <--> H+ + Pr-
Use the Henderson-Hasselbalch equation to calculate the ratio of propanoate/propanoic acid in the desired buffer:
pH = pKa + log [Pr-]/[HPr]
4.90 = 4.87 + log [Pr-]/[HPr]
[Pr-]/[HPr] = 1.07
Rearranging this gives:
[Pr-] = 1.07 [HPr]
Now, in the final buffer, [HPr] + [Pr-] = 0.100 M.
Substituting into this equation gives:
[HPr] + 1.07[HPr] = 0.100
2.07[HPr] = 0.10
[HPr] = 0.0483 M
Then, [Pr-] = 0.100 - 0.0483 = 0.0517 M
Mass NaPr required = 0.500 L X 0.0517 mol/L X 96.07 g/mol = 2.48 g sodium propanoate
For the propanoic acid,
M1V1 = M2V2
0.250 M (V1) = 0.100 M (500 mL)
V1 = 200 mL
So, you will take 200. mL of the propanoic acid stock solution, add 2.48 g sodium propanoate to that, and bring the total volume up to 500 mL with water.