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Help in deducing a equation for Neutral red

  1. Oct 15, 2009 #1
    Hi to all, hope you can help me with a problem that took me almost all the week
    1. The problem statement, all variables and given/known data
    To find The Pka of Neutral Red, i had to use this expression [tex]\frac{A-A_{HNR}}{A_{NR}-A}[/tex]
    So Now I Have to show that [tex]\frac{[NR]}{[HNR^{+}]}=\frac{A-A_{HNR}}{A_{NR}-A}[/tex]
    i tried backwards and i did it, but starting from the concnetrations is far more dificult

    2. Relevant equations
    i Have to prove that equallity using only this equations
    [tex]C_{HNR^{+}} + C_{NR}=C_{Total}[/tex](1)

    [tex]A_{HNR^{+}}=\epsilon_{HNR^{+}}.b.C_{total}[/tex](2)

    [tex]A_{NR}=\epsilon_{NR}.b.C_{total}[/tex](3)

    [tex]A=\epsilon_{HNR^{+}}.b.C_{HNR^{+}} + \epsilon_{NR}.b.C_{NR}[/tex] (4)

    3. The attempt at a solution

    I tried to solve 4 in order to [tex]C_{HNR^{+}}[/tex] first and then in order to [tex]C_{NR}[/tex].
    I arrive to
    [tex]C_{HNR^{+}}=\frac{A-\epsilon_{NR}.b.C_{NR}}{\epsilon_{HNR^{+}}.b}[/tex]
    [tex]C_{NR}=\frac{A-\epsilon_{HNR^{+}}.b.C_{HNR^{+}}}{\epsilon_{NR}.b}[/tex]

    then i did the following, added and subtracted the same value in the fraction numerator, like this:
    [tex]C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-\epsilon_{HNR^{+}}.b.C_{HNR^{+}}-\epsilon_{HNR^{+}}.b.C_{NR}}{\epsilon_{NR}.b}[/tex]

    so i could do:
    [tex]C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-\epsilon_{HNR^{+}}.b.(C_{HNR^{+}}+C_{NR}}{\epsilon_{NR}.b}[/tex]
    and using equation 1 it came
    [tex]C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-\epsilon_{HNR^{+}}.b.(C_{t})}{\epsilon_{NR}.b}[/tex]
    then using equation 2:
    [tex]C_{NR}=\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-A_{HNR^{+}}}{\epsilon_{NR}.b}[/tex]

    doing the same thing to HNR it came:
    [tex]C_{HNR}=\frac{A+\epsilon_{NR}.b.C_{HNR}-A_{NR}}{\epsilon_{HNR^{+}}.b}[/tex]

    So doing the reason
    [tex]\frac{[NR]}{[HNR^{+}]}[/tex]=[tex]\frac{\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-A_{HNR^{+}}}{\epsilon_{NR}.b}}{\frac{A+\epsilon_{NR}.b.C_{HNR}-A_{NR}}{\epsilon_{HNR^{+}}.b}}[/tex]
    and it's here where i can't see how can i arrive to the final equation
    am i going for the worst way? i thought about this all week and can't find a way to prove what i should :cry:


    hope someone can help me, it's kind of urgent
     
  2. jcsd
  3. Oct 16, 2009 #2
    The backwards trying was easy... but if i try to do the reverse... the multiplying term is not to clear how it showed off from the equations i give cause backwards the final term is:
    [tex]\frac{C_{NR}(\epsilon_{NR}-\epsilon_{HNR})}{C_{HNR^{+}}(\epsilon_{NR}-\epsilon_{HNR})}}[/tex]
    so the thing i don't understand is that term:
    [tex](\epsilon_{NR}-\epsilon_{HNR})[/tex] where can it come from from the equation i have?
     
  4. Oct 16, 2009 #3

    Ygggdrasil

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    Are you sure these are correct? Shouldn't it be:

    [tex]A_{HNR^{+}}=\epsilon_{HNR^{+}}.b.C_{HNR^+}[/tex]
    [tex]A_{NR}=\epsilon_{NR}.b.C_{NR}[/tex]
     
  5. Oct 16, 2009 #4
    well i really had a doubt about that... cause i think In the two case, the absorvance is given in function of total conectration... these equations are giving in a protocol of the journal of chemistry, but they simnply say... we used this 4 equations and arrive to this :S
    I couldn't seem to find that relation even trying very hard this week
     
    Last edited: Oct 16, 2009
  6. Oct 16, 2009 #5

    Ygggdrasil

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    Science Advisor

    Do you have the reference to the journal article? It's hard to figure out what the equations mean without knowledge of the details of the experiment and what is being measured.
     
  7. Oct 16, 2009 #6
    http://jchemed.chem.wisc.edu/Journal/Issues/2001/Mar/PlusSub/JCESupp/JCE2001p0349W.pdf [Broken]
    that's the link to the article... i put the pages needed in the attchments if you can't acess the journal
     

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