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A student was given a stock aluminum(III) solution with a concentration of 5.000

ID: 474973 • Letter: A

Question

A student was given a stock aluminum(III) solution with a concentration of 5.000 parts per million (ppm). (ppm are defined as mg/L for dilute aqueous solutions.) The student prepared five 100-mL standard solutions and an unknown as described in the procedure section of the experiment. The absorbance of each solution was read at 565 nm, using the blank to set zero absorbance. The results are tabulated below. For each of the five standard solutions calculate the concentration of Al(III) in ppm. Prelab will ask you to enter the answers for solutions 1 and 2. Use GRAPHICAL ANALYSIS or a sheet of graph paper to prepare a plot of absorbance (y) vs. concentration (x). The plot (standard curve) may be straight, in which case you could use linear least squares to fit it to a line. It may be nonlinear, if so, draw the best possible smooth curve through the points. In either case, notice that the curve does NOT go exactly through the origin. Based upon your curve, was the blank too dark or too light? What is your best estimate of the Al(III) concentration of the unknown sample in the covet based upon where its absorbance falls on the standard curve? (Because of the dilution made by adding reagents during sample preparation, this will not be the original concentration of the unknown.) Use your answer to question 3 to calculate the Al(III) concentration in the original unknown solution by correcting for the dilution made when preparing the sample for analysis. See the procedure section of the experiment for details of the sample preparation used.

Explanation / Answer

Concentration of Al(III) in the unknown solution

4. For a 8 ml stock solution

Al(III) present = 5 ppm x 8 ml/100 ml = 0.4 ppm

For a 6 ml stock solution

Al(III) present = 5 ppm x 6 ml/100 ml = 0.3 ppm

Absorbance of unknown = 0.341

concentration of Al(III) in diluted unknown = 0.4 ppm x 0.341/0.489 = 0.28 ppm

concentration of original unknown solution = 0.28 x 100 = 28 ppm