Help in deducing a equation for Neutral red

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In summary, the problem presented is to prove the equality \frac{[NR]}{[HNR^{+}]}=\frac{A-A_{HNR}}{A_{NR}-A} using the given equations and information. The attempt at a solution involved manipulating equations and using substitution to solve for C_{HNR^{+}} and C_{NR}. However, the final equation \frac{[NR]}{[HNR^{+}]}=\frac{\frac{A+\epsilon_{HNR^{+}}.b.C_{NR}-A_{HNR^{+}}}{\epsilon_{NR}.b}}{\frac{A+\epsilon_{NR}.b.C_{HNR}-A_{NR}}
  • #1
demander
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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
 
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  • #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?
 
  • #3
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]
 
  • #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:
  • #5
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.
 
  • #6
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
 

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1. What is Neutral red and what is its purpose?

Neutral red is a pH indicator dye that is commonly used in biological and chemical experiments. It changes color based on the pH level of a solution, making it useful in identifying acidic or basic substances.

2. How can Neutral red be used to deduce an equation?

By measuring the color change of Neutral red in a solution with a known pH, we can create a standard curve. This curve can then be used to determine the pH of a solution with an unknown pH based on the color change of Neutral red.

3. What factors can affect the accuracy of deducing an equation using Neutral red?

The accuracy of deducing an equation using Neutral red can be affected by factors such as the concentration of Neutral red used, the temperature and pH of the solution, and the presence of other substances that may interfere with the color change of Neutral red.

4. How do you perform a Neutral red experiment to deduce an equation?

To perform a Neutral red experiment, you will need to prepare a series of solutions with known pH levels and add a standard amount of Neutral red to each solution. Then, measure the color change of each solution and create a standard curve. Finally, use the standard curve to determine the pH of a solution with an unknown pH.

5. What are some alternative methods for deducing an equation for Neutral red?

Apart from using a standard curve, other methods for deducing an equation for Neutral red include using a spectrophotometer to measure the absorbance of the dye at different wavelengths, or using a pH meter to directly measure the pH of a solution. However, these methods may require more advanced equipment and may also be affected by external factors.

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