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anonymous

  • one year ago

Can anyone please help me with my lab report data? Desperate college freshman student looking for help with the topic of buffers.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @acxbox22 @andrewjordan833 @alurahenderson @badmood @chycora @cupcakerain @desiringeden @desiraem @ElonaSushchik @freckles @GeniousCreation @haleyelizabeth2017 @HackberryAbby18 @ikram002p @JoKeR0331 @khalilforthewin @Lady.Liv1776 @mathmate @Nnesha

  3. haleyelizabeth2017
    • one year ago
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    I don't believe I know how to do this. What science is this?

  4. anonymous
    • one year ago
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    Chemistry, if you know anyone who can. Please let me know!

  5. anonymous
    • one year ago
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    @alurahenderson @background43 @jadaaaa

  6. anonymous
    • one year ago
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    @abbi5169 @Brandonsnider222

  7. aaronq
    • one year ago
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    I'll take a look at it if you post it in a pdf format (rather than word)

  8. anonymous
    • one year ago
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    okay

  9. anonymous
    • one year ago
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    @aaronq - Here you go. If I am not on. Email is jdalloo97@gmail.com.

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  10. aaronq
    • one year ago
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    you should stay on if you wanna work through it. I'm only going to tell you how to do it, but i'm not gonna do it for you.

  11. anonymous
    • one year ago
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    Okay, sure. Whatever's best.

  12. anonymous
    • one year ago
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    I need to know theoretical pH, also but I don't know how to do that. Finding the missing columns are the most important ones.

  13. aaronq
    • one year ago
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    do you have other data regarding the composition of the buffers, or anything like that? there isn't enough information on these sheets to complete this

  14. anonymous
    • one year ago
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    Sodium Acetate and Acetic Acid are the substances that I used.

  15. aaronq
    • one year ago
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    Okay cool. For the theoretical pH, we can use the Henderson-Hasselbalch equation: \(\sf \large pH=pKa+log\dfrac{[A^-]}{[HA]}\) so we need to know the initial concentrations of the components (sodium acetate and acetic acid) and how much acid or base you added.

  16. anonymous
    • one year ago
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    1.0 M for each solution, which was labeled on the Erlenmeyer flask for the lab.

  17. anonymous
    • one year ago
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    Regarding the amount added, it should be on the pdf file.

  18. aaronq
    • one year ago
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    Is the data one second for the first table? it doesnt seem like it. For the first table, beaker 1. you added equal amounts of Na acetate and acetic acid? how much acid did you add, at what concentration?

  19. anonymous
    • one year ago
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    What do you mean?

  20. anonymous
    • one year ago
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    I could send you a pic of my lab scanned....is that fine?

  21. aaronq
    • one year ago
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    On the first table, where it says "beaker, solution, measured pH..", what are the amounts of everything that you used?

  22. aaronq
    • one year ago
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    Lets just work with the data for the first line, "Beaker 1" where you used buffer A + HCl

  23. anonymous
    • one year ago
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    Okay, so first I added 25 mL of Buffer Solution "A" into Beakers 1 +3. I added 25 mL of distilled water to beakers 2 +4. Moreover, I transfered 1.0 mL of 1 M HCl to beakers 1+2, and 1 mL NaOH to beakers 3+4. I measured the new pH after that was done.

  24. aaronq
    • one year ago
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    Hm nothing on the concentrations of the buffer? or the volumes? i was working through it and can't go forth without those

  25. anonymous
    • one year ago
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    No sir. That's why i am stuck.

  26. anonymous
    • one year ago
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    the volumes of the buffers + additional total to 26 mL.

  27. aaronq
    • one year ago
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    We could assume that they're equal (both at 0.5 M to add up to 1 M buffer), since the pH (4.41) is close to the pKa (4.75) and it could be off solely because of experimental error.

  28. anonymous
    • one year ago
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    Okay sure. Would that work, or cause any error in the calculations?

  29. aaronq
    • one year ago
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    Well it would be wrong if the concentrations of the components of the buffers are anything but 0.5 M each.

  30. anonymous
    • one year ago
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    True.

  31. aaronq
    • one year ago
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    I mean i can show you how to do it with 0.5 M and if you come across the actual concentrations you can just rework through it later?

  32. anonymous
    • one year ago
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    That would be great. Anything that's close is good for me. I had a headache trying to figure this out all day.

  33. aaronq
    • one year ago
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    okay cool. we need to make an ICE table to keep things organized. It stands for Initial, Change and Equilibrium (concentrations). The reaction is: \(NaCH_3COO^-+HCl\rightarrow NaCl+ CH_3COOH \) we're only interested in the concentrations of sodium acetate and acetic acid. \([NaCH_3COO^-]\) \([CH_3COOH]\) I C E Initially, we had 25 mL of \([NaCH_3COO^-]\)=0.5 M and \([CH_3COOH]\)=0.5 M we have: \(moles=Molarity*volume=0.5~M*0.025~L=0.0125~moles\) for each \([NaCH_3COO^-]\) \([CH_3COOH]\) I 0.0125 moles 0.0125 moles C E The "change" was adding acid, \(moles=1~M*0.001~L=0.001~moles \) \([NaCH_3COO^-]\) \([CH_3COOH]\) I 0.0125 moles 0.0125 moles C -0.001 +0.001 E this makes sense because you neutralized some base (sodium acetate and made the same amount of acetic acid) \([NaCH_3COO^-]\) \([CH_3COOH]\) I 0.0125 moles 0.0125 moles C -0.001 +0.001 E 0.0125 -0.001 0.0125+0.001 The volume doesnt matter because of the math \(\sf \large pH=4.75+log\dfrac{(0.0125 -0.001)}{(0.0125+0.001)}\) this is the theoretical pH.

  34. anonymous
    • one year ago
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    I don't be mean to sound super rude or anything, but I have to go in a little while. I apologize for the inconvenience.

  35. aaronq
    • one year ago
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    I posted each step individually of the table so you could see how i filled it in

  36. anonymous
    • one year ago
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    Thank you :)

  37. aaronq
    • one year ago
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    This is for the buffers. for the water just use: \(\huge \sf pH=-log[H^+]\)

  38. aaronq
    • one year ago
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    where \([H^+]\) is the concentration of HCl (in the new volume)

  39. aaronq
    • one year ago
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    no problem! let me know if you have questions

  40. aaronq
    • one year ago
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    Also, for the \(\Delta pH\) column, (which means change in pH. you just find the difference (subtract) between the two measured pH's, before and after you added acid or base.

  41. anonymous
    • one year ago
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    Sorry, to be a bother but I had a question about the concentrations of the acids and bases for the first table. Is that for the first step above.

  42. aaronq
    • one year ago
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    hmm i'm not sure i understand which step you're referring to?

  43. anonymous
    • one year ago
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    Where you displayed the ICE table, is that just for theoretical pH?

  44. aaronq
    • one year ago
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    yeah, that's for the theoretical pH. for \(\Delta pH\) just find the differences between the measured pH values

  45. anonymous
    • one year ago
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    and for the [acids] and [bases],, how would i follow from previously mentioned steps.

  46. aaronq
    • one year ago
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    So were talking about the second table right?

  47. anonymous
    • one year ago
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    Yes.

  48. aaronq
    • one year ago
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    Okay, so for that we use the value you mentioned before 1.0 M for each of the components. So for A, we had 25 mL of each base and acid, that is sodium acetate and acetic acid. We need to find the moles of each and the new concentration (molarity). \(\sf moles_{Base}=1~M*0.025~L=0.025~moles\) \(\sf Molarity=\dfrac{0.025~moles}{0.025~L+0.025~L}=0.5~M\)

  49. anonymous
    • one year ago
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    Thanks!

  50. aaronq
    • one year ago
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    no problem!

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