Help in deducing a equation for Neutral red

  • Thread starter Thread starter demander
  • Start date Start date
  • Tags Tags
    Neutral
Click For Summary

Discussion Overview

The discussion revolves around deriving an equation related to the concentrations of Neutral Red and its protonated form, HNR+. Participants are attempting to prove a relationship between these concentrations using provided equations and absorbance data. The context is homework-related, focusing on chemical equilibria and absorbance in a laboratory setting.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents an expression to relate the concentrations of Neutral Red and its protonated form, attempting to derive it from given absorbance equations.
  • Another participant questions the correctness of the absorbance equations provided, suggesting they may need to be reformulated based on the concentrations of the individual species rather than total concentration.
  • A different participant expresses confusion about a specific term that appears in the reverse derivation, seeking clarification on its origin.
  • One participant requests a reference to the journal article that contains the equations, indicating that understanding the experimental context is crucial for interpreting the equations correctly.
  • A participant shares a link to the journal article, providing additional context for the discussion.

Areas of Agreement / Disagreement

Participants express uncertainty about the correctness of the absorbance equations and the derivation process. There is no consensus on the proper formulation of the equations or the interpretation of the terms involved.

Contextual Notes

Participants note potential limitations in understanding due to the lack of clarity in the original equations and their dependence on specific definitions from the referenced article. The discussion reflects ongoing attempts to reconcile different approaches to the problem.

demander
Messages
21
Reaction score
0
Hi to all, hope you can help me with a problem that took me almost all the week

Homework Statement


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

Homework 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)

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
 
Physics news on Phys.org
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?
 
demander said:
[tex]A_{HNR^{+}}=\epsilon_{HNR^{+}}.b.C_{total}[/tex](2)
[tex]A_{NR}=\epsilon_{NR}.b.C_{total}[/tex](3)

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]
 
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:
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.
 
http://jchemed.chem.wisc.edu/Journal/Issues/2001/Mar/PlusSub/JCESupp/JCE2001p0349W.pdf
that's the link to the article... i put the pages needed in the attchments if you can't acess the journal
 

Attachments

  • reference1.JPG
    reference1.JPG
    43.5 KB · Views: 448
  • reference2.JPG
    reference2.JPG
    47.2 KB · Views: 426
Last edited by a moderator:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K