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Ph and buffer solutions help

  1. Jun 10, 2006 #1
    Hey I have been having a lot of trouble in solving these problems and I have exams soon and I really dont know where to start, I know u aint meant to give the answers so instead would someone please go through all of the steps to finish this!
    I really need some kinda help, This question was in a quiz and I was so confused as to wat I needa do... I am gonna get help from a lecturer as well but inorder to study it I need some help please.

    A buffer solution of pH = 5.30 can be prepared by dissolving acetic acid and sodium acetate in water. How many moles of sodium acetate must be added to 1.0 L of 0.25 M acetic acid to prepare the buffer?
    The answer is meant to be 0.90mol but I have no Idea how to get to that!

    Also one more We mix 100 mL of 0.20 M HBr and 50.0 mL of 0.40 M NaC1O. What is the pH of the resulting solution? Ka(HC1O) = 3.5 x 10-8

    for this one i Wrote out the equation... and found that the concentration of HClO was 0.1333 or something like that... then I just had no idea where to go from there, I no that pH=pka- log(acid/base) but I just didnt no how to get to that point... I changed Ka(HClO) to pKa by using -log3.5 x 10- 8 but as I sed not sure where to go.

    Could anyone please help
     
  2. jcsd
  3. Jun 10, 2006 #2

    Hootenanny

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    Do you know how to calculate the acidity consant for buffer solutions?
     
  4. Jun 10, 2006 #3
    do you mean Kw and Kb coz I think so
     
  5. Jun 10, 2006 #4

    Hootenanny

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    I mean Ka perhaps you use different notantion? Anyway, the relationship between Ka and [H+ is given by;

    [tex]K_{a} = [H^{+}] \times \frac{[\text{salt}]}{[\text{acid}]}[/tex]

    Where Ka is the acidity constant of the acid.
     
    Last edited: Jun 10, 2006
  6. Jun 10, 2006 #5
    I will give it a go... but i wat has me a little stumped is what happens to the pH, I obviously have to use that somewhere dont I?
     
  7. Jun 10, 2006 #6
    wait a second I understand now I think, I will work it all out and if I have a problem I will just post it on this agen, thanks a lot.
     
  8. Jun 10, 2006 #7

    Hootenanny

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    You'll need that to work out the hydrogen ion concentration, you'll also need to look up Ka for acetic acid.
     
  9. Jun 10, 2006 #8
    okay sorry just one more question, I am a little confused... just coz wat will be the [H+], I wrote this, 1.8E-5=[H+] x [salt]/[o.25]

    Then after I find the [salt] then I will just use n=c* volume right?
    If you cant give me anymore help thats fine!
     
  10. Jun 10, 2006 #9

    Hootenanny

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    That looks good to me. You need to use you pH value to calculate the [H+], remember that;

    [tex]pH = -log[H^{+}] \Leftrightarrow [H^{+}] = 10^{-pH}[/tex]

    Yes, you would usually do that, but since the solution is dissolve in one litre, the calculation becomes n = c x 1 and therefore in this case; n = c

    Do you follow?
     
    Last edited: Jun 10, 2006
  11. Jun 10, 2006 #10

    Borek

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  12. Jun 10, 2006 #11
    thanks very much that was heaps helpful and now I understand wat to do, dunno y I didnt think of that at the time.
     
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