# Lmax or 1max ?

1. Oct 10, 2008

### ~christina~

lmax or 1max ??

1. The problem statement, all variables and given/known data
I have a graph of concentration vs Absorbance like below, for PNP.
http://img512.imageshack.us/img512/2614/pnpgraphexfa7.jpg [Broken]

problem is that I have to find the unknown concentration, so I use the graph to determine that. The equation to find it is

y= mx
what I don't understand is when they say that the y used is:
y= 1max
Is it the maximum point on the graph?

m= slope
x= concentration of PNP

Can someone tell me what I put for y= 1max so I can find the concentration?
I have the slope, but that's the only thing that is confusing me.

Thank you!

Last edited by a moderator: May 3, 2017
2. Oct 10, 2008

### EnumaElish

Re: lmax or 1max ??

My guess is you are looking to determine the m coefficient from the graph, which does not require focusing on (or solving for) values of individual y points or individual x points. In addition I see that you have defined:

m= slope
x= concentration of PNP

but do not have a similar definition of y. Therefore "y = 1max" is a definition of what y stands for in the equation.

3. Oct 10, 2008

### GCT

Re: lmax or 1max ??

The standard curve equation is able to be used to determine the concentration of an analyte sample with an unknown concentration , you need to state the exact question as your current paraphrasing does not make sense.

Last edited by a moderator: May 3, 2017
4. Oct 10, 2008

### chemisttree

Re: lmax or 1max ??

What is Imax or 1max? Is it the wavelength of light for which the strongest absorption of a given concentration of PNP is measured? Possibly it is 400 nm?

Try solving for x in terms of y and m in your equation. Substitute the value of m you have calculated from your graph. Use the absorbance of your unknown concentration as y and calculate x.

You're gonna kick yourself when you see how close you are to an answer...

5. Oct 10, 2008

### ~christina~

Re: lmax or 1max ??

No I forgot to include:
y= absorbance at lmax

That graph was the example graph given.
My graph of the experiment looked like this below:
http://img183.imageshack.us/img183/9061/actualpnpassayea5.jpg [Broken]

I figured out that it is l for wavlength. I measured it at that wavelength though.
I calculated m and it was 0.003653.
Then I plugged it in but what I got was a huge number.

y= 0.003653x
unknown y= 0.806
0.806/0.003653= x
x= 220 $\mu M$

I think that is high right?
I'm used to unknowns being within the standard range but..maybe this is not so?

Another issue is this:
When analyzing the standards of 10, 20, 30, 50, 75 $\mu M$ solutions that I had to dilute down from a 0.1 $\mu M$ paranitrophenol solution to create, I had to further dilute each when analyzing. I had to use 1ml of standard and 3ml of basic buffer (p.H 9.96) to accomplish this.

When I analyzed the unknown sample, I did the same dilution of 1ml sample and 3ml of basic buffer.

The problem comes in here:
They ask me to calculate BOTH the diluted unknown concentration AND the concentration of the UNDILUTED sample of 4-nitrophenol. (PNP) by using the standard curve that was made.

I have no idea how to calculate the undiluted unknown stock solution that I used to dilute 1:3 with buffer.

Thank you

Last edited by a moderator: May 3, 2017
6. Oct 11, 2008

### Staff: Mentor

Re: lmax or 1max ??

That's the way it should be done, you should not use your calibration curve outside of the range of concentrations used when calibrating. So something is definitely wrong.

I am not sure if I am not missing something, as the answer seems to be too obvious. Why not C1V1=C2V2?

7. Oct 11, 2008

### ~christina~

Re: lmax or 1max ??

That is what I am thinking too. There is a reason we have a calibration curve.
The example graph's slope [1st post] (Molar absorptivity coefficient) when I estimated the values, ended up as being 0.015 and when I used this value to find the concentration, it ended up as a value on the chart.
Could it be the dilution that is causing this? The slope of that graph(one in first post) was much more steep than the one I got.

someone else told me that the concentration should be solved for like this:
c=A/ab 0.806/400= micro M
y=38.95 (0.002015) + .1293 = 0.207

I think that it's wrong though, because a is not supposed to be 400nm but rather the absorbance at 400nm.

But I just wanted to check that, before I dismissed it.

Ah, so when I find the concentration of the unknown dilute, I can solve for the undiluted stock concentration. I get it.

Thanks Borek

8. Oct 11, 2008

### Staff: Mentor

Re: lmax or 1max ??

No idea how you have calculated concentrations and what was the procedure, so it is harrd to tell.

Remember that the slope will be device and cuvette dependent - so unless both calibration curves were made using the same hardware, they can differ. You are not measuring absolute values (like specific coefficient) but relative ones.