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NotTim

  • one year ago

ADSORPTION OF ACETIC ACID BY ACTIVATED CHARCOAL LAB-"Calculate the exact concentration of your base. To do this, use the volume of base added to 10 mL of your standardized acid solution. Calculate the base concentration for each titration performed then determine the mean concentration."

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  1. NotTim
    • one year ago
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    I'm thinking c1v1=c2v2

  2. NotTim
    • one year ago
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    I have titration volumes of 11.72, 9.49, 10.27, 10.47, 10.27 mL.

  3. NotTim
    • one year ago
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    MY Potassium Hydrogen Phthalate concentration is 0.1024 M

  4. NotTim
    • one year ago
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    1. Freundlich Isotherm 2. Langmuir Isotherm are mentioned in the lab intro.

  5. NotTim
    • one year ago
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    ITs like this lab: http://www.cpp.edu/~sjanz/physical_chemistry/chm_353_labs/adsorption_of_acetic_acid_by_a_solid.pdf

  6. dan815
    • one year ago
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    @hwyl

  7. NotTim
    • one year ago
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    http://www.isca.in/rjcs/Archives/vol2/i9/9.ISCA-RJCS-2012-145.pdf The proper research article, surprisngly not helpful though.

  8. anonymous
    • one year ago
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    is the acid ur titrant or ur base?

  9. NotTim
    • one year ago
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    i think its the base

  10. anonymous
    • one year ago
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    so for ur first one, 11.72 ml of ur base was used to neutralise the acid?

  11. NotTim
    • one year ago
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    Wait base is the thing that is the beaker already right?

  12. anonymous
    • one year ago
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    ive forgotten too haahah, its been a yr since i did titration in uni

  13. anonymous
    • one year ago
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    i should know

  14. anonymous
    • one year ago
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    but i think you can have either acid or base as ur titrant, it just depends what ur experiment tells u, i think the norm is to have the base in the beaker thought (from my vauge memory)

  15. NotTim
    • one year ago
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    Alright, any ideas on figuring out concentrations?

  16. NotTim
    • one year ago
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    test 1

  17. NotTim
    • one year ago
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    test 2

  18. anonymous
    • one year ago
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    best bet is dilution factor c1v1=c2v2

  19. anonymous
    • one year ago
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    give me a sec, i'm working on a maths problem as we speak, i posted the question and need some help haha

  20. NotTim
    • one year ago
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    i dunno what to use for v1 and v2 though i just know that i titrated 11.72 mL into 10mL KHP.

  21. anonymous
    • one year ago
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    KHP is ur acid right?

  22. NotTim
    • one year ago
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    yes

  23. anonymous
    • one year ago
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    so you titrated 11.72ml of base into 10ml KHP?

  24. NotTim
    • one year ago
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    fr my first sample, yes.

  25. anonymous
    • one year ago
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    wait, dilution factor is unessecary, do you know the reaction equation?

  26. NotTim
    • one year ago
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    no, i don't remember what that means

  27. anonymous
    • one year ago
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    A+B>C+D etc

  28. NotTim
    • one year ago
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    oh, no

  29. NotTim
    • one year ago
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    I do not

  30. anonymous
    • one year ago
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    i reckon that the key

  31. anonymous
    • one year ago
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    as in do you know the stoichiometry is it 1:1?

  32. NotTim
    • one year ago
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    I'm looking thru the lab manual, and it says nithing.

  33. anonymous
    • one year ago
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    haha well....

  34. anonymous
    • one year ago
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    http://www.ausetute.com.au/titrcalc.html

  35. anonymous
    • one year ago
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    cause thats sorta what u wanna do, find moles etc then use stoichiometry to find ur base conc

  36. NotTim
    • one year ago
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    you think i was suppose to deriive that from the equation?

  37. anonymous
    • one year ago
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    what is ur base ur using?

  38. NotTim
    • one year ago
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    sorry, was at bathrooom. its naoh

  39. anonymous
    • one year ago
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    well that makes life so much easier ahah

  40. anonymous
    • one year ago
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    CH3COOH + NaOH ----> CH3COONa + H2O

  41. anonymous
    • one year ago
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    so stoichiometry is 1:1

  42. anonymous
    • one year ago
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    C(Acetic)= 0.1024 M V(acetic)=10ml V(naoh)=11.72 C(naoh)=??

  43. NotTim
    • one year ago
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    wait the acid is khp

  44. NotTim
    • one year ago
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    acetic is the part after this section of the lab

  45. anonymous
    • one year ago
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    oh

  46. anonymous
    • one year ago
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    \[KHC _{8}H _{4}O _{4}+NaOH \rightarrow H _{2}O+NaKC _{8}H _{4}O _{4}\]

  47. anonymous
    • one year ago
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    same idea C(KHP)= 0.1024 M VKHP)=10ml V(naoh)=11.72 C(naoh)=??

  48. anonymous
    • one year ago
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    C(KHP)= 0.1024 M VKHP)=0.01L V(naoh)=0.01172L C(naoh)=??M

  49. anonymous
    • one year ago
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    now, n=CV

  50. NotTim
    • one year ago
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    lol im stupid i was trying to figure out how you got 0.01172 L...

  51. anonymous
    • one year ago
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    hahaha

  52. anonymous
    • one year ago
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    >.<

  53. anonymous
    • one year ago
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    n(kHp)=0.1024*0.01=0.001024 moles

  54. anonymous
    • one year ago
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    we know mole ratio is 1:1 therefore number of acid moles=number of base moles

  55. anonymous
    • one year ago
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    n(naoh)=0.001024 moles

  56. anonymous
    • one year ago
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    n=CV C=n/V C(NaOH)=0.001024/0.01172 C(NaOH)=0.087M

  57. anonymous
    • one year ago
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    thats ur first test

  58. NotTim
    • one year ago
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    i don't know where n=cv is coming from; as in, why its being applied to this scenario

  59. anonymous
    • one year ago
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    because u want to find the base concentration...

  60. anonymous
    • one year ago
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    and you can only do that using stoichiometry of what u you (acid)

  61. NotTim
    • one year ago
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    alrihgt.

  62. anonymous
    • one year ago
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    so 11.72, 9.49, 10.27, 10.47, 10.27 mL. Test 1: C(naoh)=0.087M Test 2: C(naoh)= 0.001024/(9.49/1000)=0.108M Test 3: C(naoh)= 0.001024/(10.27/1000)=0.0997M Test 4: C(naoh)= 0.001024/(10.47/1000)=0.0978M test 5: C(naoh)= 0.001024/(10.27/1000)=0.0997M Therefore mean concentration is the average right. C(naoh)= test 1+test2+test3+test4+test5/5runs C(naoh)=0.098M (to be honest, no lab supervisor would make up batches of 0.098M, we can safely say that the sodium hydroxide concentration is 0.1M

  63. NotTim
    • one year ago
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    You just had to explain up to n=CV. But thanks!

  64. anonymous
    • one year ago
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    its because moles of a reaction always is constant, its just concentration changes. so we use stoichiometry (which is in effect molar ratio) so we use a comparison of moles to determine unknown concentrations

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