Chemistry

Hello, I have a question that reads: Using the average molarity of your initial acetic acid solutions, the initial volumes, and the volume of NaOH added to reach the equivalence point, calculate the [C2H3O2-] concentration at the equivalence point. You should report 4 significant figures, e.g. 0.06753. You Scored 3 points out of 3 Possible Your Answer: 0.04916

Average Molarity of Acetic Acid: 0.07456,
Initial volume of Acetic Acid: 30.0 mL
Vol NaOH needed to reach equivalence: 16.0 mL

I was told that the answer should be [(0.07456)*(0.030)]/(0.030+0.016), but that gives the wrong answer. Apparently it's supposed to be 0.04916. Can someone explain?

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asked by Jens
  1. The formula you have been given is correct for the values of 30.0, 16.0 and 0.07456 M. The answer is not 0.04916, perhaps because the 30.0 and the 16.0 look suspiciously "too even."

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    posted by DrBob222
  2. Hmm...do you think it could have to do with Acetic Acid being a weak acid and NaOH being a strong base? (Incomplete dissociation?)

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    posted by Jens
  3. No.
    (C2H3O2^-) = moles/L
    Moles comes from the acetic acid titrated COMPLETELY to the equivalence point. So
    M x L = moles acetic acid. If all of that is converted to acetate, as it should be, then M x L/final volume = (acetate).
    Two sources of error. The molarity of the acetic acid is not quite right.
    The titration with NaOH not exact.

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    posted by DrBob222
  4. Ah, I figured out the problem. Thanks!

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    posted by Jens

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