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    The value of the acid dissociation constant, Ka, for a different weak acid HY is
    2.65 × 10–4 mol dm–3 at 298 K.
    Calculate the pH of the buffer solution formed when a 0.0300 mol sample of the solid
    salt NaY is dissolved in 500 cm3 of a 0.250 mol dm–3 solution of the acid HY.



    Jan 09 Q2b on AQA
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    Use pH = pKa + log ([NaY]/[HY])
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    (Original post by EierVonSatan)
    Use pH = pKa + log ([NaY]/[HY])
    I havnt seen it done like that before. Is that quicker than just using the fraction?
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    (Original post by Tomato_Soup1992)
    I havnt seen it done like that before. Is that quicker than just using the fraction?
    It's just a manipulation of 'the fraction' and I find it a lot quicker :yes:
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    (Original post by EierVonSatan)
    Use pH = pKa + log ([NaY]/[HY])
    I learned that the notation [A] can only be used if A is actually in the solution (e.g. [NaCl] would be wrong). Am I wrong or was it just a little mistake you made in this post?
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    Ka= [H+][Y-]/[HY]
    Then you have the values you need to put in, rearrange for [H+], but remember there is 500cm3 (0.5dm3) and all the [H+] values are moldm-3. So convert to Moldm-3. Then use pH=-log[H+].

    Hope this helps you.
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    (Original post by phen)
    I learned that the notation [A] can only be used if A is actually in the solution (e.g. [NaCl] would be wrong). Am I wrong or was it just a little mistake you made in this post?
    My, you're picky :hmmm:
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    (Original post by EierVonSatan)
    My, you're picky :hmmm:
    :P Any mistakes I make on tests are usually that kind of small mistakes. I've learned to watch out for them in particular.. (managed to not let one slip in this entire year so far.)

    So that's why I'm picky on those things.
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    The value of the acid dissociation constant, Ka, for a different weak acid HY is
    2.65 × 10–4 mol dm–3 at 298 K.
    Calculate the pH of the buffer solution formed when a 0.0300 mol sample of the solid
    salt NaY is dissolved in 500 cm3 of a 0.250 mol dm–3 solution of the acid HY.

    ---
    2.65 x 10^-4 = [0.03/0.5][H+] / [0.250]
    [H+] = 1.104 x 10^-3

    -log{10}[H+] = 2.96 PH
 
 
 
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