Question

In the Fluorometric analysis of Quinine in Tonic Water Lab

Pipette 50 μL of the tonic water into the 10 mL flask and dilute to 10.00 mL with 0.05 M H2SO4. This will be your first sample for analysis. Prepare a second sample by diluting the first solution by a factor of 2. You can do this by pipetting 1.00 mL of solution the first solution and 1.00 mL of 0.05 M H2SO4 into a vial.

From the platereader (fluorescence spectrophotometry):

	Tonic water sample	Tonic water sample (diluted ½)
Mean Intensity	375.26	260.73
concentration (ppm)	0.329	0.202

	standard 1	standard 2	standard 3
[quinine]	0.15	0.30	0.45
intensity	259.315	319.432	407.849

From the calibration curve of standards of 0.15, 0.30, 0.45 ppm (quinine standards (10 ppm) ).

R² = 0.98804

y = 495.11x + 180.33

Calculate the concentration of the quinine in the tonic water as determined by the two sample dilutions. Propagate the error in these results. Determine whether there is a significant difference in the results at 95% confidence.

Answer 1

calculation for the concentration of quinine :

         C₁V1 = C₂V₂

For the first sample     0.05 M x 10 ml = C₂ x 50 x 10^-3 ml

                      C₂ = 0.05 M

for the second sample     0.05 M x 1ml = C₂ x 1 ml

                                                  C₂ = 0.05M

Now, find out LOD (limit of detection) = 3* standard deviation

For concentartion, consider As given :

y = 495.11 x + 180.33

so x = ( y - 180.33) / 495.11

where y is the mean value, putting that find x , and then change it in the unit of mg/L . This will be the concentration of quinine in per lite of tonic water.