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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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