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Calculating the concentration of H+ watch

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    If I want to calculate [H+] in the CH3CHOOH/CH3COONa using the following formula:

    Ka(Hln) = [H+][ln- / [Hln]

    Provided that:

    Hln(aq) <-----> H+ + ln(aq)

    And, provided that I'm adding bromocresol green to the mixture, and given that ka(Hln) for bromocresol green is [I]2 x 10-5 mol dm-3.

    Now, since CH3CHOOH = CH3COO-, Ka = [H+].

    So, in this case, will [H+] = 1.8 x 10-5 (Ka for ethanoic acid), or, will it be [H+] = 2.0 x 10-5 (Ka for bromocresol green)?

    Thanks.
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    (Original post by SWEngineer)
    If I want to calculate [H+] in the CH3CHOOH/CH3COONa using the following formula:

    Ka(Hln) = [H+][ln- / [Hln]

    Provided that:

    Hln(aq) <-----> H+ + ln(aq)

    And, provided that I'm adding bromocresol green to the mixture, and given that ka(Hln) for bromocresol green is [I]2 x 10-5 mol dm-3.

    Now, since CH3CHOOH = CH3COO-, Ka = [H+].

    So, in this case, will [H+] = 1.8 x 10-5 (Ka for ethanoic acid), or, will it be [H+] = 2.0 x 10-5 (Ka for bromocresol green)?

    Thanks.
    I am really not sure what you're asking here. Your notation is rather vague.

    I presume you are wanting to calculate the pH of an ethanoate buffer, yes? In which case you use the Ka for ethanoic acid.

    I have no idea what this is supposed to mean:

    Hln(aq) <-----> H+ + ln(aq)

    but your charges don't balance for a start.
 
 
 
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