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    Calculate the pH of a buffer solution produced by adding 4.00g of sodium ethanoate to 1dm^3 of 0.01M ethanoic acid. The Ka of ethanoic acid is 1.84x10^-5 mol/dm^3 at 300K.

    Calculate the pH of this buffer if 5cm^3 of 0.1M HCl is added.


    I tried using this pka=ph-log HA/H+ but couldn't do it!

    +rep
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    (Original post by arvin_infinity)
    Calculate the pH of a buffer solution produced by adding 4.00g of sodium ethanoate to 1dm^3 of 0.01M ethanoic acid. The Ka of ethanoic acid is 1.84x10^-5 mol/dm^3 at 300K.

    Calculate the pH of this buffer if 5cm^3 of 0.1M HCl is added.


    I tried using this pka=ph-log HA/H+ but couldn't do it!

    +rep
    find the concentration of A- by finding the moles of ethanoate and then dividing by volume, then stick this into the Ka formula.

    Spoiler:
    Show

    mr of sodium ethanoate = 82
    n=4/82 = 0.0488mol
    conc of A-= 0.0488 / 1 = 0.0488 moldm-3

    Ka=[H+][A-]/[HA]
    1.84x10^-5 = [H+] x 0.0488/0.01
    so [H+] = 3.77x10^-6
    pH = -log [H+]
    so pH = 5.42



    The volume of acid added in the second part is very small, so by the definition of a buffer, the pH shouldn't change.
  3. Online

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    (Original post by clownfish)
    find the concentration of A- by finding the moles of ethanoate and then dividing by volume, then stick this into the Ka formula.

    Spoiler:
    Show

    mr of sodium ethanoate = 82
    n=4/82 = 0.0488mol
    conc of A-= 0.0488 / 1 = 0.0488 moldm-3

    Ka=[H+][A-]/[HA]
    1.84x10^-5 = [H+] x 0.0488/0.01
    so [H+] = 3.77x10^-6
    pH = -log [H+]
    so pH = 5.42



    The volume of acid added in the second part is very small, so by the definition of a buffer, the pH shouldn't change.
    Buffers do change pH only not by very much. The question is designed to show you that, but you do have to do the calculation.

    moles of acid (H+) added = 0.1 x 0.005 = 0.0005

    This is absorbed by the ethanoate ions increasing the moles of HA by the same amount, hence new moles of HA = 0.01 + 0.0005 = 0.0105

    The ethanoate moles are decreased by the same amount = 0.0488 - 0.0005 = 0.0483 mol

    1.84 x 10-5 = [H+]*0.0483/0.0105

    pH = 5.40
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    (Original post by charco)
    Buffers do change pH only not by very much. The question is designed to show you that, but you do have to do the calculation.

    moles of acid (H+) added = 0.1 x 0.005 = 0.0005

    This is absorbed by the ethanoate ions increasing the moles of HA by the same amount, hence new moles of HA = 0.01 + 0.0005 = 0.0105

    The ethanoate moles are decreased by the same amount = 0.0488 - 0.0005 = 0.0483 mol

    1.84 x 10-5 = [H+]*0.0483/0.0105

    pH = 5.40
    Just did the same method using pka=ph+log acid/salt
    and I got 4.07..I checked it twice ! wondered why that is?!

    EDITbs I got the equation wrong way around!

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