ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: DNAngel on February 27, 2011, 08:49:43 pm
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I was just wondering, how might you determine the concentration of an unknown solution that had a higher absorbance reading than the highest standard?
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Ummmm, get a higher standard :P
Well in the book, they usually follow a linear curve/straight line curve. Therefore, you could determine the gradient from (Absorbance1 - Absorbance2)/(Concentration1 - Concentration2) and then determine your concentration?
EDIT: If you knew it was going to be higher, you can also dilute the sample...
(Credits to cherylim :P)
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Ummmm, get a higher standard :P
Well in the book, they usually follow a linear curve/straight line curve. Therefore, you could determine the gradient from (Absorbance1 - Absorbance2)/(Concentration1 - Concentration2) and then determine your concentration?
EDIT: If you knew it was going to be higher, you can also dilute the sample...
(Credits to cherylim :P)
It kind of depends on HOW high the absorbance reading is higher than the standard. Because the graph becomes exponential as it gets higher. It's usually a VERY bad idea to extrapolate at a HIGHER absorbance.
As luken said, It's better off to dilute the unknown solution and MAKE it inthe range of standards.
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Ummmm, get a higher standard :P
Well in the book, they usually follow a linear curve/straight line curve. Therefore, you could determine the gradient from (Absorbance1 - Absorbance2)/(Concentration1 - Concentration2) and then determine your concentration?
EDIT: If you knew it was going to be higher, you can also dilute the sample...
(Credits to cherylim :P)
It kind of depends on HOW high the absorbance reading is higher than the standard. Because the graph becomes exponential as it gets higher. It's usually a VERY bad idea to extrapolate at a HIGHER absorbance.
As luken said, It's better off to dilute the unknown solution and MAKE it inthe range of standards.
So it doesn't stay linear as shown in the books, or is the book simply making things easy by making it linear?
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Because the graph becomes exponential as it gets higher.
This is not true.
The graph becomes more or less 'logarithmic' as concentration increase, that is adding more molecules does not necessarily lead to more absorption. This is due to matrix effects, such as emissive scattering within the solution. An introductory read can be found here: http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/beers1.htm
As already said by several others, the straight line we see is only valid at low concentrations, thus we always dilute the sample's concentration down to the linear section. One subtle point is that this linear section is a 'property' of the analyte, thus we're diluting to match this range (found experimentally by plotting absorbance vs concentration), not making standards to match the sample. We only measure the standards so we can find the equation of the line.
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Because the graph becomes exponential as it gets higher.
This is not true.
The graph becomes more or less 'logarithmic' as concentration increase, that is adding more molecules does not necessarily lead to more absorption. This is due to matrix effects, such as emissive scattering within the solution. An introductory read can be found here: http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/beers1.htm
As already said by several others, the straight line we see is only valid at low concentrations, thus we always dilute the sample's concentration down to the linear section. One subtle point is that this linear section is a 'property' of the analyte, thus we're diluting to match this range (found experimentally by plotting absorbance vs concentration), not making standards to match the sample. We only measure the standards so we can find the equation of the line.
lol crap my memories are failing me. I was only sure that the graph was curved from high concentration. Thanks for the remainder