A scan of the excitation wavelength from 335-435 showed the highest absorption at 399 nm, so the excitation monochromator was set for that value. Then the emission scan was performed from 450-550 nm, and the strongest signal was found to be at 520 nm. These are the wavelengths that are used for all of the samples.

Sample	Fluorescence Intensity	Corrected Fluorescence Intensity
Blank	0.008	0.000
Sample	0.128	0.120
Sample + 1 mL	0.167	0.159
Sample + 2 mL	0.220	0.212
Sample + 3 mL	0.260	0.252
Sample + 4 mL	0.290	0.282

A plot of fluorescence (Figure 3) vs. µg of Al³⁺ added (Figure 4) yielded a least-squares line of:

Fluorescence Intensity = 0.0417 x (µg of Al³⁺ added) + 0.1216

Amount of Al³⁺ = -(Y-Int)/Slope = -0.1216/0.0417 = -2.916 µg/mL

Since the amount of unknown added was 25 mL, then the 2.916 µg/mL value needs to be divided by 25.

Unknown Aluminum Concentration = 2.916 µg/mL / 25.0 mL = 0.117 µg/mL = 0.117 ppm
which is quite close to the actual value of 0.110 ppm (6.4% error).

Figure 1.Graphic representation of method of standard addition.

Figure 2.Determining optimum EX and EM Wavelengths.

Figure 3. Fluorescence of the samples.

Figure 4. The standard addition calibration plot.