400 425 450 475 500 525 550 575 600 625 650 675 700 725 Wavelength nm

400 425 450 475 500 525 550 575 600 625 650 675 700

This preview shows page 5 - 7 out of 15 pages.

400 425 450 475 500 525 550 575 600 625 650 675 700 725 Wavelength (nm) Absorbance (unitless) Abs (1 pkg/L) Abs (dil 1:4) (Figure 3) Visible spectra of strawberry Kool-Aid® (at 2 different concentrations above) show that the high- wavelength (red) end of the spectrum is least absorbed, so the transmitted light is tly g and 635 predominan red. As seen in the lower curve, moderate amounts of short-wavelength (violet-indigo) light are not absorbed, accounting for the tinting to red-violet. Light with wavelengths corresponding to blue (475 nm), through green (510 nm) towards yellow (570 nm) are most heavily absorbed. Each solution has 1 wavelength ( ) at which it absorbs the most light -- its lambda max ( max ). At max the instrument is most sensitive to differences in concentration, since the solute absorbs the most light at this wavelength; small changes in the amount of solute will yield large chan es in the absorbance. For copper sulfate solution, max is 740 nm – see why it could be blue? In designing an experiment you want to use the spectrophotometer in a way that maximizes its sensitivity. The Vernier colorimeter offers four choices of wavelength: 430, 470, 565, nm. Choosing a wavelength beyond 600 nm in the case above would lead to zero absorbance in the concentration range of interest – a poor choice. Of the remaining choices which appears to have the most absorbance? B EER S L AW C ONCENTRATION ( c ) E FFECTS If a substance absorbs light, the more substance present, the more light is absorbed. If you’re teeped for 3 minutes in a cup, tea prepared for the same length of accustomed to one teabag s time using an equivalent teabag, but in a 6-cup teapot full of water will taste weaker and look paler. It has less dissolved solute. This direct relationship between absorbance and amount of dissolved solute will help us determine the concentration of samples of unknown concentration. Take two solutions that differ only in solute concentration. If we measure the intensity of both incident light ( I in ) and transmitted light ( I out ), we can max max I in I out I in I out [2] [1] determine how much light was absorbed. In the figure to the right, the incident light ( I in ) for the samples is the same, but the transmitted light ( I out ) is less for the second sample ( I out [2] ), which is more concentrated and darker than the first ( I out [1]). Using the ratio of the intensities of incident and transmitted light yields percent transmittance: 100% I I %T in out ( 5 )
Image of page 5
This expression tells us the perce 100% means that no light was useful as absorbance, which is derived from it as follows: nt of light that passes though the sample. A transmittance of absorbed. The %T equation is intuitive, but is not as directly out in I I log A ( 6 ) Beer’s Law is unusual in that it hold nonlinear as concentrations solution in the upper curve of Figure 3 is four times that of the lower curve, the absorbance of the s for many substances in dilute solution, but tends to become exceed a threshold. For example, though the concentration of the upper curve is less than that multiple.
Image of page 6
Image of page 7

You've reached the end of your free preview.

Want to read all 15 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes