Describe preparation of 5 liters of a 0.3 M acetate buffer, pH 4.47, starting from a 2 M solution of acetic acid and a 2.5 M solution of KOH.

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  1. I will use millimoles to work this problem although technically the Henderson-Hasselbalch equation uses concentrations in mols/L (M). However, it's easier for me to use moles or millimoles and not mols/L. The answer is the same so here goes.
    You want 5 L of a 0.3 M acetate buffer solution of pH 4.47 made with 2 M HAc and 2.5 M KOH. Using the HH equation acid = a = HAc and base = b = Ac^-. pH = pKa + log (base)/(acid)
    4.47 = 4.77 + log b/a
    log b/a = -0.3 and
    b/a = 0.501 which I will call 0.5. You may want to redo it with 0.501 and you will a lightly different answer but not by much. So b = 0.5a
    You want 5 L of a 0.3 M buffer so you want 1500 millimoles. The next equation you need is
    a + b = 1500. Substituting frm above for b = 0.5a
    a + 0.5a = 1500 and
    a = 1000 mmoles and b = 500 mmoles.How we get that with 2 M HAc and 2.5 M KOH.
    ....................HAc + KOH ==> KAc + H2O
    I...................x mmol................0............0
    Yiy want x-y to be 1000 mmols. You want y to be 500. So x initially must be 1500 mmols acid and you want to add 500 KOH to make 500 mmols acetate. How do you do that.
    mL x M = millimoles.
    mL x 2 = 1000 mmols HAc
    mL HAc = 1000/2 = 500 mL
    mL x M = millimoles KOH
    mL x 2.5 = 500
    mL KOH = 200
    So you place 500 mL of the acid and 200 mL of the KOH into a 5 L volumetric flask (be careful--it will get VERY hot)--let it cool, make to the mark with DI water. stopper, mix thoroughly, label. Done.
    You may want to put those numbers into the HH equation to see if the pH is 4.47. pH = 4.77 + log (500/1000) = ?
    I've done this with such detail you shouldn't have any doubt but post our work if you get stuck somewhere.

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  2. Sir,
    In the HH equation, you assumed that PKA is 4.77 although it is not given, how did you acquire that number because I have tried to calculate it but there is insufficient data.
    Thanks in advance.

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