neera asked in Science & MathematicsChemistry · 4 months ago

# Help!!!!!!!!!!!!!!?

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).

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• 4 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.

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