Finding the concertration of a acid when given the Ka and Ph

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    I don't understand how u find the concertration using Ka. Also how to calculate buffer solutions.

    Any examples would be helpful. Thank you
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    (Original post by Hazel99)
    I don't understand how u find the concertration using Ka. Also how to calculate buffer solutions.

    Any examples would be helpful. Thank you
    Do you know the definitions of Ka and pH?
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    (Original post by MexicanKeith)
    Do you know the definitions of Ka and pH?
    Acid dissociation constant
    Nurmerical scale of acidity and alkalinity
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    (Original post by Hazel99)
    Acid dissociation constant
    Nurmerical scale of acidity and alkalinity
    That's true, but they both also have specific formula which define them

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

    pH = -log[H+]

    are those familiar?
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    (Original post by MexicanKeith)
    That's true, but they both also have specific formula which define them

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

    pH = -log[H+]

    are those familiar?

    Yeah, I have used them before but sometimes in the question it will give a concertration (e.g. 0.100 mol dm-3). So where does that concertration go in the formula ?. Also when's it saids a time 298k that doesn't need to be in the equation?
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    (Original post by Hazel99)
    Yeah, I have used them before but sometimes in the question it will give a concertration (e.g. 0.100 mol dm-3). So where does that concertration go in the formula ?. Also when's it saids a time 298k that doesn't need to be in the equation?
    Do you know what the different terms in the two equations mean, particularly the parts in square brackets?
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    (Original post by MexicanKeith)
    Do you know what the different terms in the two equations mean, particularly the parts in square brackets?
    Concertration of hydrogen ions?
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    (Original post by Hazel99)
    Concertration of hydrogen ions?
    That's along the right lines, [H+] is the concentration of protons
    [A-] is concentration of conjugate base
    [HA] is the concentration of the undissociated acid

    I'll give you an example

    Ka of ethanoic acid is 1.8x10-5
    pH of a solution of ethanoic acid is 2.523
    find the concentration of the acid solution?

    So the concentration its asking you for is [HA]

    First you can use pH to find [H+]
    pH = -log[H+]
    rearranges to [H+] = 10-pH

    so here [H+] = 10-2.523 = 0.003 mol dm-3

    Than you can use the definition of Ka to find what you want.
    First you have to realise that all the protons (H+) and all the conjugate base molecules (A-) in this question come from dissociation of (HA) molecules so there must always be the same amount of H+ and A- ( ie [H+] = [A-] )

    so
    Ka = [H+][A-] / [HA]
    becomes
    Ka = [H+]2 / [HA]

    putting in the numbers you know for Ka and for [H+]
    1.8x10-5= (0.003)2 / [HA]

    rearranging gives

    [HA] = (0.003)2 / 1.8x10-5 = 0.5 mol dm-3

    So the acid solution is 0.5 mol dm-3

    Hope from that example you can see that if you know two out of three of the following bits of information:
    [HA]
    pH
    Ka

    than you can use the two equations too work out the third bit of information!

    Sorry for the long post, hope it helped!
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    (Original post by MexicanKeith)
    That's along the right lines, [H+] is the concentration of protons
    [A-] is concentration of conjugate base
    [HA] is the concentration of the undissociated acid

    I'll give you an example

    Ka of ethanoic acid is 1.8x10-5
    pH of a solution of ethanoic acid is 2.523
    find the concentration of the acid solution?

    So the concentration its asking you for is [HA]

    First you can use pH to find [H+]
    pH = -log[H+]
    rearranges to [H+] = 10-pH

    so here [H+] = 10-2.523 = 0.003 mol dm-3

    Than you can use the definition of Ka to find what you want.
    First you have to realise that all the protons (H+) and all the conjugate base molecules (A-) in this question come from dissociation of (HA) molecules so there must always be the same amount of H+ and A- ( ie [H+] = [A-] )

    so
    Ka = [H+][A-] / [HA]
    becomes
    Ka = [H+]2 / [HA]

    putting in the numbers you know for Ka and for [H+]
    1.8x10-5= (0.003)2 / [HA]

    rearranging gives

    [HA] = (0.003)2 / 1.8x10-5 = 0.5 mol dm-3

    So the acid solution is 0.5 mol dm-3

    Hope from that example you can see that if you know two out of three of the following bits of information:
    [HA]
    pH
    Ka

    than you can use the two equations too work out the third bit of information!

    Sorry for the long post, hope it helped!
    Thank you it helped a lot
    I just get a bit confused on the steps
 
 
 
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