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    My chemistry textbook uses the simplified version for the dissociation of a weak acid:

    HA (aq) <--> H+ (aq) + A- (aq)

    so Ka for this reaction would be Ka=[H+][A-]/[HA]

    Firstly, it says that when the acid is dissociated [H+] [A-] exist equally so you can use [H+]^2.

    Secondly it also says that the bottom of the Ka expression should actually be [HA] - [H+] but as little [H+] dissociates we just use [HA]. This is my first question as surely this is assuming you are given the concentration of undissociated HA and not the equilibrium concentration?

    The second step appears to be applying an approximation to Ka and i dont know whether the first step is also applying an approxiamation to Ka. My book appears to suggest this but it is unclear.

    Thanks for your help
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    (Original post by 111davey1)
    My chemistry textbook uses the simplified version for the dissociation of a weak acid:

    HA (aq) <--> H+ (aq) + A- (aq)

    so Ka for this reaction would be Ka=[H+][A-]/[HA]

    Firstly, it says that when the acid is dissociated [H+] [A-] exist equally so you can use [H+]^2.

    Secondly it also says that the bottom of the Ka expression should actually be [HA] - [H+] but as little [H+] dissociates we just use [HA]. This is my first question as surely this is assuming you are given the concentration of undissociated HA and not the equilibrium concentration?

    The second step appears to be applying an approximation to Ka and i dont know whether the first step is also applying an approxiamation to Ka. My book appears to suggest this but it is unclear.

    Thanks for your help
    1. You would only ever refer to the concentration of the (undissociated) acid.

    Hydrochloric acid is almost 100% dissociated and yet we only ever talk about the concentration of the "acid" not the conjugate base or hydrogen ions.

    Likewise we would only express the concentration of ethanoic acid as 0.1M (or whatever) as it would be determined when the solution is prepared, by mass of ethanoic acid weighed out, or by titration of the prepared solution.

    2. The approximation refers to the use of the ka equation when finding out the pH of a solution OR using the pH to find the ka value.

    However, as Ka values are available from data books this second usage becomes fairly academic unless the acid is unknown.

    The approximation is a good one as it obviates the need for quadratic equations and gives an answer with an approximately 0.1% inaccuracy for most weak acid solutions.
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    HA -----> H+ and A-
    Initial c 0 0
    Change
    over -ca +ca +ca
    time
    (a= acid dissociation constant alpha)
    Final c(1-a) ca ca

    Equilibrium constant = c^2 a^2 / c(1-a) = ca^2 / 1-a

    usually a is very small so is ignored so Ka = ca^2

    Dissociation constant is usually given and with its help equilibrium constant can be calculated
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    (Original post by charco)
    1. You would only ever refer to the concentration of the (undissociated) acid.

    Hydrochloric acid is almost 100% dissociated and yet we only ever talk about the concentration of the "acid" not the conjugate base or hydrogen ions.

    Likewise we would only express the concentration of ethanoic acid as 0.1M (or whatever) as it would be determined when the solution is prepared, by mass of ethanoic acid weighed out, or by titration of the prepared solution.

    2. The approximation refers to the use of the ka equation when finding out the pH of a solution OR using the pH to find the ka value.

    However, as Ka values are available from data books this second usage becomes fairly academic unless the acid is unknown.

    The approximation is a good one as it obviates the need for quadratic equations and gives an answer with an approximately 0.1% inaccuracy for most weak acid solutions.
    Thanks, i just need to clarify this:

    a strong acid HA --> H+ + A-

    Say the concentration of HA is 1 moldm-3 then [H+] and [A-] is both 1 moldm-3 once dissociated. My question here is that surely you end up with a solution which is of double the initial concentration.

    I then found a question which was: Find the values of Ka for the following weak acids:
    0.13moldm-3 solution with a PH of 3.52

    and i didnt know if the solution value was referring to the concentration of undissociated weak acid or undissociated weak acid + the concentraion of [H+] ions or undissociated weak acid + the concentration of [H+] ions + the concentration of [A-] ions.

    i know how to answer it just unsure on basically how the reaction equations work
    Thanks
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    (Original post by 111davey1)
    Thanks, i just need to clarify this:

    a strong acid HA --> H+ + A-

    Say the concentration of HA is 1 moldm-3 then [H+] and [A-] is both 1 moldm-3 once dissociated. My question here is that surely you end up with a solution which is of double the initial concentration.
    Concentrations are only quoted with respect to a certain species.

    Yes, the solution is 1 mol dm-3 in both hydrogen ions AND A-ide ions


    I then found a question which was: Find the values of Ka for the following weak acids:
    0.13moldm-3 solution with a PH of 3.52

    and i didnt know if the solution value was referring to the concentration of undissociated weak acid or undissociated weak acid + the concentraion of [H+] ions or undissociated weak acid + the concentration of [H+] ions + the concentration of [A-] ions.

    i know how to answer it just unsure on basically how the reaction equations work
    Thanks
    This has already been explained. You only ever state the concentration of "acid" as assumed to be undissociated.
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    (Original post by charco)
    Concentrations are only quoted with respect to a certain species.

    Yes, the solution is 1 mol dm-3 in both hydrogen ions AND A-ide ions



    This has already been explained. You only ever state the concentration of "acid" as assumed to be undissociated.
    Oh ok thankyou very much
 
 
 
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