calibration curves

[year12]

How is everyone going with their exam prep? (Less than 2 weeks to go now...)

I had a question to ask about drawing calibration curves.
Let's say once the data is plotted and it appears to be non-linear,
- do I draw a straight line (a line of best fit) or do I draw a curve, trying to pass through all the given points?
- & when trying to draw a line of best fit, what happens if it doesn't go through the origin? Or should I try to pass it through the origin?
Because it really doesn't make sense especially if they give us an absorbace of 0 when the conentration is 0 ... and my graph does not show this.

Your advice will be helpful!

Thanks.

[stonecold]

I honestly doubt they would get us to draw a calibration curve, and if they did, it would be linear, and would probably be easy to draw in that all the values would fall on the grid lines, and the line would fall on the points.

I couldn't imagine how they would mark us on drawing a non linear calibration curve.

Tbh, I just wish they gave us a calibration equation.  Easier, and wayy more accurate.

[superflya]

u draw a linear line of best fit.. try to get it to cover most of the points.

[Martoman]

This is a good question. If it is non-linear then it has to be DILUTED. The question would almost certainly have to ask you: "Explain why this curve is non linear" or something like that, then ask you to dilute it by some factor and THEN with these transformed data do you graph a calibration curve.

mmmm i smell duuuurty VCAA examiners getting sweaty palms just discussing such a nasty question. They would have a tasty tasty galvanic skin response, im sure.  :idiot2:

[crayolé]

This is a good question. If it is non-linear then it has to be DILUTED. The question would almost certainly have to ask you: "Explain why this curve is non linear" or something like that, then ask you to dilute it by some factor and THEN with these transformed data do you graph a calibration curve.

mmmm i smell duuuurty VCAA examiners getting sweaty palms just discussing such a nasty question. They would have a tasty tasty galvanic skin response, im sure.  :idiot2:

Could you please explain this in more detail? ;]

[Martoman]

Yeah.

Say you had a very concentrated sample. At low concentrations this would still appear linear on a calibration curve. But at higher concentrations... it will start to go all over the shop (ie: most likely exponential). This is why in these types of questions we are usually told that a 20ml sample was diluted to 100ml or 200 ml or whatever so that a linear plot will form.

I think there was a great example of this in the NEAP 2010 paper asking about by what factor do you dilute a sample by to get it in the appropriate region for analysis. hit it up. 

[stonecold]

Isn't that a fairly standard question?

Why does the sample need to be diluted?
SO the absorbance reading for the sample falls within the calibration curve, as it is inaccurate to extrapolate above the curve.

[Martoman]

Its fairly standard to those who know what they are doing

Given the context of this thread and the fact that no one mentioned it I thought it apt to interject.

[olly_s15]

Yeah.

Say you had a very concentrated sample. At low concentrations this would still appear linear on a calibration curve. But at higher concentrations... it will start to go all over the shop (ie: most likely exponential). This is why in these types of questions we are usually told that a 20ml sample was diluted to 100ml or 200 ml or whatever so that a linear plot will form.

I think there was a great example of this in the NEAP 2010 paper asking about by what factor do you dilute a sample by to get it in the appropriate region for analysis. hit it up. 

to me this seems wrong - how can you go from an exponential graph and then multiply it by a dilution factor (which is a constant) and come out with a linear graph? just doesn't make sense; please explain.

[Martoman]

Eh it will turn out to be exponential at high concentrations honey. At LOW concentrations you will see a nice linear fit. Think about a graphics calculator. You have a exponential graphed. Something like . Zoom in between x values (0,1) and y values somewhere like 1 to 2. Notice how it looks as if you could put a straight line through it. This is essentially what you are doing. 

[olly_s15]

but if we dilute a sample to ensure we can obtain information within the calibration graph and then proceed to find out a quantity in the undiluted sample, we just multiply by a dilution factor

what you are saying contradicts this... you're saying that the sample of higher concentration would deviate from a linear shape which essentially means you cannot multiply by a constant

[Martoman]

Of course it would deviate from a linear shape! We are doing fickle VCE stuff here, i wouldn't hold onto it too badly! I'm sure Mao or some other genius can explain from a university standpoint what I mean here 

[stonecold]

Martoman is right.  It is called Beers law I think.  At high concentration, the graph is actually exponential, but because we do a somewhat watered down version of chem, we only get concentrations in the lower part of the graph where it appears linear.  That is why you should NEVER extrapolate above the curve.


And also why it is called a curve

[Martoman]

... because we do a somewhat watered down version of chem....................

 :2funny: :2funny: :2funny: :2funny: :2funny: :2funny: :2funny: :2funny: :2funny: :2funny:

[olly_s15]

Martoman is right.  It is called Beers law I think.  At high concentration, the graph is actually exponential, but because we do a somewhat watered down version of chem, we only get concentrations in the lower part of the graph where it appears linear.  That is why you should NEVER extrapolate above the curve.

"beer's" law, lol!

do you understand about what type of questions i'm talking about though? i'm not saying you guys are wrong, it just contradicts some of the questions i've done - although yes, this is VCE level, watered down etc.