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I need to produce a buffer solution that has pH 5.28. I have already 10 mmol of acetic acid. How much solid sodium acetate will i need to add to this solution? pKa of acetic acid is 4.74.

I plugged in all my info to the Hendersen/Hasslebalch equation but I keep getting the wrong answer......!!!! I also calculated the [A-]/[HA]= 3.47....Please help me

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7 answers

  1. If you posted your work I could have already found the error and posted it.
    pH = pKa + log (base/acid)
    5.28 = 4.74 + (acetate/10).
    solve for base which will be in millimols.
    millimols sodium acetate = M x mL.
    You know millimols, you know mL of acetic acid, calculate M. Then, mols = g/molar mass. You may be able to skip one step of this but make sure the dilution of the acetate by the volume of acetic acid doesn't get in your way. Or post your quick numbers and I can work through it. The volume of acetic acid, I think, is needed.

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  2. So far this is the equation I have 0.54= log(x/10) and I cant solve for x...I forgot how to do it correctly

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  3. What is the volume of acetic acid you have for the 10 millimols acetic acid.

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  4. This is all they gave me:

    You need to produce a buffer solution that has pH 5.28. You already have a solution that contains 10 millimoles of acetic acid. How many millimoles of solid sodium acetate will you need to add to this solution? The of acetic acid is 4.74.
    Then they gave me this equation:

    pH= pKa + log (salt/acid)

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  5. 0.54 = log B/10
    punch in 0.54 on your calculator. then hit the 10^x key. You should obtain 3.467 which is the antilog of 0.54 so
    3.467 = B/10 and B = 3.467 x 10 = 34.67 millimols NaC2H3O2.
    Then 3.467 millimols x 82 mg/mol = 284.4 mg NaC2H3O2. Check my work. Plug 284.4 mg back into B and 10 for Acid, and see if the left side and the right side of the HH equation are equal.

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  6. By the way, we haven't been strictly correct the way we did this. Technically, the B and the A must go in in concentration units or molarity. We have plugged in millimols. And we got away with it. WHY? Because the log (base/acid) term is
    log (molarity base)/(molarity acid) =
    log (millimols base/total volume/(millimols acid/total volume) and since the total volume is the same, it really doesn't matter what goes there because it cancels. Therefore, when you use the HH equation, MOST of the time we can use mols (or millimols). Before I retired, however, I always took off a few points if a student used mols and didn't show mols/v/mols/v. That way I knew the student recognized that these were concentration terms. Picky? There are some times when one must know the volume.

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  7. OMG, Thank you so much! I dunno what i was doing! Thank you Thank YOU!!!

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