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degree of ionisation

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    How do i work this out for a weak acid when knowing values for KA and Initiall concentration.


    could someone please show me with an example.

    any help most apprecitaed.

    Thank You
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    (Original post by JUSTME1)
    How do i work this out for a weak acid when knowing values for KA and Initiall concentration.


    could someone please show me with an example.

    any help most apprecitaed.

    Thank You
    ok, just say you have ethanoic acid (a weak acid) here is what you do:

    you have for instance 0.1M CH3COOH in this equilibria reaction:

    CH3COOH <---> CH3COO- + H+

    ok, this reaction has a Ka value (which is just a Kc value for weak acids) of 1.78x10-5moldm-3

    for equilibria, Ka = [CH3COO-][H+]/[CH3COOH]

    because the products are in a 1:1 ration so in eq, the [CH3COO-]=[H+]

    so Ka = [H+]^2/[CH3COOH]

    now in comes equilibria stuff:

    CH3COOH <---> H+ + CH3COO-
    Start: 0.1M 0M 0M
    Eq: 0.1-x xM xM

    as the value of x is soooo small it can be considered negligable so 0.1-x=0.1

    no substitute int he equation

    1.78x10-5 = [H+]^2/0.1
    1.78x10-6 = [H+]^2
    [H+]=1.33x10^-3moldm-3

    degree of ionisation = 1.33x10^-3/0.1 = 0.0133 = 1.33%

    (so pH = -log(1.33x10^-3) = 2.87)
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    (i just posted it again to notify you by quotation )
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    QUOTE=smetin93]ok, just say you have ethanoic acid (a weak acid) here is what you do:

    you have for instance 0.1M CH3COOH in this equilibria reaction:

    CH3COOH <---> CH3COO- + H+

    ok, this reaction has a Ka value (which is just a Kc value for weak acids) of 1.78x10-5moldm-3

    for equilibria, Ka = [CH3COO-][H+]/[CH3COOH]

    because the products are in a 1:1 ration so in eq, the [CH3COO-]=[H+]

    so Ka = [H+]^2/[CH3COOH]

    now in comes equilibria stuff:

    CH3COOH <---> H+ + CH3COO-
    Start: 0.1M 0M 0M
    Eq: 0.1-x xM xM

    as the value of x is soooo small it can be considered negligable so 0.1-x=0.1

    no substitute int he equation

    1.78x10-5 = [H+]^2/0.1
    1.78x10-6 = [H+]^2
    [H+]=1.33x10^-3moldm-3

    degree of ionisation = 1.33x10^-3/0.1 = 0.0133 = 1.33%

    (so pH = -log(1.33x10^-3) = 2.87)
    Edit/Delete Message

    (i just posted it again to notify you by quotation )[/QUOTE]


    Firstly thanks for your reply

    ok i get it before the equillibrum stuff comes.0.1 - x wat is x

    and for the degree of ionisation why did you divide by 0.1

    Thanks
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    (Original post by JUSTME1)
    QUOTE=smetin93]ok, just say you have ethanoic acid (a weak acid) here is what you do:

    you have for instance 0.1M CH3COOH in this equilibria reaction:

    CH3COOH <---> CH3COO- + H+

    ok, this reaction has a Ka value (which is just a Kc value for weak acids) of 1.78x10-5moldm-3

    for equilibria, Ka = [CH3COO-][H+]/[CH3COOH]

    because the products are in a 1:1 ration so in eq, the [CH3COO-]=[H+]

    so Ka = [H+]^2/[CH3COOH]

    now in comes equilibria stuff:

    CH3COOH <---> H+ + CH3COO-
    Start: 0.1M 0M 0M
    Eq: 0.1-x xM xM

    as the value of x is soooo small it can be considered negligable so 0.1-x=0.1

    no substitute int he equation

    1.78x10-5 = [H+]^2/0.1
    1.78x10-6 = [H+]^2
    [H+]=1.33x10^-3moldm-3

    degree of ionisation = 1.33x10^-3/0.1 = 0.0133 = 1.33%

    (so pH = -log(1.33x10^-3) = 2.87)
    Edit/Delete Message

    (i just posted it again to notify you by quotation )

    Firstly thanks for your reply

    ok i get it before the equillibrum stuff comes.0.1 - x wat is x

    and for the degree of ionisation why did you divide by 0.1

    Thanks[/QUOTE]

    x is the concentration of [H+] at equilibria. so to get the [acid] at eq, you do the original - the amount of [H+], it is just molar ratios. for the other question, you divide[H+]/[acid] x 100, because it is a percentage of [H+] made for [acid], it is a simple percentage calculation. if u wanna find th degree of ionisation you just do not multiply by 100

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