Chemistry

A beaker with 100 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.10 mL of a 0.260 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

I got 0.27 as the change, but it says I made a rounding error, and I can't find my rounding error. I also got 0.26 and 0.25, but they didn't work.

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  1. I would assume three significant digits on this.

    work it out to three digits.

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    bobpursley
  2. Sorry, I forgot to say that it told me to round it to two decimal places.

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  3. recheck how you rounded the antilogs.

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    bobpursley
  4. I did everything from the beginning and I didn't round anything at all until my last step where it told me to round to two decimal places. I still can't get the answer, though

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  5. I can't get any of your answers.
    Post your work and let me look for the error. I think the answer is close to 0.7 if I didn't make an error.

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  6. Here's what I did:

    So I set values for:
    [Acetic Acid] = y
    [H3O+] = 10^-5
    [Acetate] = 0.1M - y
    Ka = 10^-4.76

    And from the equation: Ka = [Acetate][H3O+]/[Acetic Acid], I found y, or the molarity of acetic acid, to be 10^-6/(10^-5 + 10^-7.46). And from this, I also got the molarity of acetate: 0.1 - y.

    Then, I found the total molarity. I converted 10^-5 M acetic acid to 10^-6 M/0.1 L. I then took 0.260 M HCl and converted it to 0.001586 M/0.0061 L. I added the top and the bottom and it turned out to be 0.001587 moles/.1061 L. Let this value be A.

    And since the Ka value is low, it goes to completion to the left. So, [Acetic Acid] = y + A. And [Acetate] = 0.1 - y -A. And now for the Henderson/Hasselbach equation:

    pH = 4.76 + log((y + A)/(0.1 - y - A))
    And I plugged in:
    y = 10^-6/(10^-5 + 10^-7.46)
    A = 0.001587/.1061
    Then, I subtracted 5 from this and got my answer.

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  7. I said I would find your error but I've changed my mind after I see how you worked the problem. Have you seen the Henderson-Hasselbalch equation? Have you used it? Why not use it here? It is MUCH simpler and it uses exactly the same chemistry; just a different form of it.
    pH = pKa + log[(base)/(acid)] and all of that comes from Ka = (H^+)(acetate)/(acetic acid) and pH = -log(H^+).
    From pH = 5.00 and pKa = 4.760, find (base)/(acid) ratio. (Two unknowns there). Then from the problem you know base + acid = 0.1 (two of the same unknowns). Solve for (base) and (acid). Then set up an ICE chart, add the 6.10 mL and plug that information into a new HH equation and solve for pH. calculate delta pH.

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  8. Okay, so I found the ion concentrations through the Henderson-Hasselbalch equation, and the concentrations I got:

    [Acetic Acid] = 0.036525666
    [Acetate] = 0.063474334

    I got the same concentrations as when I did my ice chart, and I also got the same delta pH, which was -0.266.

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  9. I still don't get it.
    pH = pKa + log(B/A)
    5.00 = 4.76 + log(B/A)
    B/A = 1.7378
    A + B = 0.100
    Two equation in two unknowns.
    A = your number
    B = your number.
    You started with 100 mL; therefore, you started with 3.653 mmoles HAc (which may be the difference) and 6.347 mmoles Ac^-

    When I did the ice chart I ended up with this.

    ...........Ac^- + H^+ ==> HAc
    ..........6.347....0......3.653
    add..............1.586...........
    change.....-1.586..-1.586..+1.586
    final.....4.761.....0......5.239

    pH = 4.760 + log(4.761/5.239)
    pH = 4.71845
    delta pH =0.28155 to 2 s.f. is 0.28 and you data base MAY want to see -0.28
    I did find an error in my original calculation in which I thought it was closer to 0.7. Interestingly enough, I check every one of my steps but one and I didn't check that one because it was too simple (just before I posted the 0.7 number). Guess where the error was? I subtracted incorrectly. But I think 0.28 is right. I hope this helps.

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  10. Thanks a lot! I realized that I forgot to account for the volume change when I calculated my molarities for the base and the acid, and that was enough to get the pH off by 0.01.

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