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Topic: Standard Addition Confusion!  (Read 8697 times)

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Offline racheltaylor23

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Standard Addition Confusion!
« on: November 04, 2012, 08:24:09 PM »
Hello! It's late on Sunday / early on a Monday and I chose now to attempt a lab report.

I'm just trying to calculate the concentration of quinine sulphate in a Urine sample, after having run 6 standard addition solutions.

I've got a graph of Added Quinine Sulphate vs. fluorescence, I know that the dilution factor of the analyte is 400 (5 into 100 twice)

I haven't really done Standard Addition before so I'm not too sure how to work out the concentration I need.

By extrapolating to where fluorescence is 0, x (added quinine sulphate concentration) is -0.26533

Would I need any more information to calculate the unknown concentration?



My results table is as follows:
Conc (µgcm-3)  /   Fluorescence Reading
0   / 3.112
0.05   / 3.819
0.1   / 4.316
0.15   / 4.91
0.2   / 5.533
0.25   / 6.121

Thanks for your consideration!

Offline Hunter2

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Re: Standard Addition - Working out an unknown concentration
« Reply #1 on: November 05, 2012, 02:44:02 AM »
You shift the graph to 0 fluorescence. What can you see on the x axis?

Offline racheltaylor23

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Re: Standard Addition Confusion!
« Reply #2 on: November 05, 2012, 09:09:08 AM »
When fluorescence is 0, x (added quinine sulphate conc.) = -0.265330526

Offline Hunter2

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Re: Standard Addition Confusion!
« Reply #3 on: November 05, 2012, 09:43:38 AM »
If you now shift the x-axis also to zero, what happens to your value?

Offline racheltaylor23

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Re: Standard Addition Confusion!
« Reply #4 on: November 05, 2012, 10:04:46 AM »
Do you mean what fluorescence reading does x = 0 show? That's 3.1508.

Offline Hunter2

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Re: Standard Addition Confusion!
« Reply #5 on: November 05, 2012, 03:32:55 PM »
No, first you shift the graph to  0 fluorescence.  You get the minus value.
Think its like you built a tower. Basement to ground floor is your sample, then you buildthe first floor,  the second floor, third floor etc.  If you tske a elivator enter from ground floour you have to go down to basement and up to the other floors. Thinky move the basement to ground floor, so there is no basement any more. So you shift the x-axis to 0 of your minus value. So what happend to your value now.

Offline racheltaylor23

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Re: Standard Addition Confusion!
« Reply #6 on: November 06, 2012, 12:30:01 PM »
Sorry I don't really understand what you mean :/

That's my graph!

Offline sjb

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Re: Standard Addition Confusion!
« Reply #7 on: November 06, 2012, 12:35:33 PM »
What would be the fluorescence of a completely blank sample?

Offline Hunter2

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Re: Standard Addition Confusion!
« Reply #8 on: November 06, 2012, 12:52:03 PM »
Sorry I don't really understand what you mean :/

That's my graph!

Great. So shift your scale of the x-axis to the intersection of your graph. Intersection is 0. Similar what SJB says. Blank sample.

You have your Sample + 1. add + 2. add etc.

Offline racheltaylor23

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Re: Standard Addition Confusion!
« Reply #9 on: November 06, 2012, 01:15:44 PM »
So, like this?


Offline Hunter2

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Re: Standard Addition Confusion!
« Reply #10 on: November 06, 2012, 01:21:27 PM »
No. I dont know how to draw here.  Take the picture of your first graph, draw a second x -axis below, but start your scale for x- axis with the intersection. (-0.26..) You got it?

Offline sjb

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Re: Standard Addition Confusion!
« Reply #11 on: November 07, 2012, 03:20:30 AM »
Hello! It's late on Sunday / early on a Monday and I chose now to attempt a lab report.

I'm just trying to calculate the concentration of quinine sulphate in a Urine sample, after having run 6 standard addition solutions.

I've got a graph of Added Quinine Sulphate vs. fluorescence, I know that the dilution factor of the analyte is 400 (5 into 100 twice)

I haven't really done Standard Addition before so I'm not too sure how to work out the concentration I need.

By extrapolating to where fluorescence is 0, x (added quinine sulphate concentration) is -0.26533

Would I need any more information to calculate the unknown concentration?

Sorry, I jumped in too fast earlier. I think you're almost there. Imagine that for instance your initial concentration (before dilution) was 1 g per litre. What would it be after the dilutions? If your initial was x? Can you back calculate given that you know the concentration after dilution is 0.26533 μg / ml

Offline Hunter2

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Re: Standard Addition Confusion!
« Reply #12 on: November 07, 2012, 03:49:28 AM »
Now you solved it for the student. She was not able to figure out what is on the negative x-axis. Of course a positive value after shifting 0 to the intersection.

Offline racheltaylor23

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Re: Standard Addition Confusion!
« Reply #13 on: November 07, 2012, 04:23:12 AM »
Thanks SJB, I had 106.132 as my answer already but wasn't sure if that was right as I was initially using a negative value (-0.26533) as my diluted concentration!
Thanks for your help :)
And Hunter2, thanks for trying!

Offline Hunter2

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Re: Standard Addition Confusion!
« Reply #14 on: November 07, 2012, 04:39:18 AM »
What does the negative value mean? If you change the coordinate system of your graph it turns to positive. Remember the negative value is your result.

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