A student calibrated a UV-V is spectrophotometer at 434 nm using known standards

Question

A student calibrated a UV-V is spectrophotometer at 434 nm using known standards chlorophyll-a and obtained the following data: Prepare a calibration curve showing plusminus 3 sigma error bars with trendline, equation, and R^2 value. Prepare a calibration curve forced through zero with error bars of plusminus 1 sigma, with trendline, eqation, and R^2 value. The concentration of chlorophyll-a extracted from the algae contained in 1.00 L of water was determined based on absorbance. The algae were first filtered from the water determined based on absorbance. The algae were first filtered from the water sample and the chlorophyll-a extracted in acetone. The extract was prepared for analysis by diluting one milliliter of the extract in 25.0 mL of acetone. The sample was analyzed in triplicate the following absorbance values obtained: 0.366, 0.369, and 0.362. Determine the concentration of chlorophyll-a in the extract. What is the concentration of chlorophyll-a due to algae in the entire 1.00 L water sample?

Explanation / Answer

For the calibration curve, all you have to do is plot this data into a excel sheet to get calibration curve. In the same sheet you can put the bars of the error.

How you will do it? just take an average of the trials to do the calibration curve.

That's how you'll do part a and b. However to do part c, you will need some data of the calibration curve, so, I'll put the equation of the calibration curve, basing on your data so I can do part c.

The equation, I just took the average of all trials, and knowing that A = EbC

plotting A vs C, the slope will give you Eb, and assuming b as the length of the cubic as 1 cm, the slope, which is the molar absortivity, is the value of the slope, which we will use to do part c. The equation of this calibration curve is as follow:

y = -0.0004 + 1.1212x r2 = 0.999827567

the value of E = 1.1212 M-1 cm-1

With this value, we can do part c now:

taking an average of the absorbance we have: 0.366+0.369+0.362 / 3 = 0.36567

If A = EbC

C = A / Eb

C = 0.366 / 1.1212*1

C = 0.3264 M

But this concentration is in the extract, let's remember that this was dissolved in 25 mL of acetone so:

C = 0.3264 * (25 mL / 1 mL) = 8.16 M

Hope this helps

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