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    Hi guys,

    I was wondering whether someone could help me out with this question. Part c is what I need help on

    2. a) Calculate the pH of 0.12 moldm-3 ethanoic acid (Ka = 1.7 x 10-5 moldm-3).
    Answer is 2.85

    b) Calculate the mass of sodium ethanoate (CH3COONa) which must be added to 500 cm3 this solution to give a buffer solution of pH = 4.60.
    Answer is 3.33g

    c) Calculate the pH of this solution after 0.01 moles of HCl are added.
    I keep getting an answer of 1.76 by pretty sure this is wrong as the answer should be close to the original because this is a buffer solution?

    Thanks in advance!
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    (Original post by Mazza2000)
    Hi guys,

    I was wondering whether someone could help me out with this question. Part c is what I need help on

    2. a) Calculate the pH of 0.12 moldm-3 ethanoic acid (Ka = 1.7 x 10-5 moldm-3).
    Answer is 2.85

    b) Calculate the mass of sodium ethanoate (CH3COONa) which must be added to 500 cm3 this solution to give a buffer solution of pH = 4.60.
    Answer is 3.33g
    ka = [H+][A-]/[HA]

    (1.7 x 10-5 x 0.12)/2.5 x 10-5 = [A-] = 0.0816

    This is in 500 ml so mol ethanoate = 0.0816/2 = 0.0408 mol
    relative mass of sodium ethanoate = 82
    therefore mass required = 0.0408 x 82 = 3.3 g

    Spoiler:
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    we agree



    c) Calculate the pH of this solution after 0.01 moles of HCl are added.
    I keep getting an answer of 1.76 by pretty sure this is wrong as the answer should be close to the original because this is a buffer solution?

    Thanks in advance!
    If you add 0.01 mol HCl this is absorbed by the ethanoate ions in equal quantity (approximation)

    hence mol of ethanoate is reduced by 0.01 mol
    new mol of ethanoate = 0.0406 - 0.01 = 0.0306
    and the mol of HA increases by 0.01 = 0.07

    ka = [H+][A-]/[HA]

    hence [H+] = ka[HA]/[A-]

    [H+] =(1.7 x 10-5 x 0.14)/0.0612 = 3.89 x 10-5

    pH= 4.41
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    (Original post by charco)
    ka = [H+][A-]/[HA]

    (1.7 x 10-5 x 0.12)/2.5 x 10-5 = [A-] = 0.0816

    This is in 500 ml so mol ethanoate = 0.0816/2 = 0.0408 mol
    relative mass of sodium ethanoate = 82
    therefore mass required = 0.0408 x 82 = 3.3 g

    Spoiler:
    Show




    we agree






    If you add 0.01 mol HCl this is absorbed by the ethanoate ions in equal quantity (approximation)

    hence mol of ethanoate is reduced by 0.01 mol
    new mol of ethanoate = 0.0406 - 0.01 = 0.0306
    and the mol of HA increases by 0.01 = 0.07

    ka = [H+][A-]/[HA]

    hence [H+] = ka[HA]/[A-]

    [H+] =(1.7 x 10-5 x 0.14)/0.0612 = 3.89 x 10-5

    pH= 4.41
    Ahh ok Thank you so much for your help
 
 
 
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