Equilibrium 1: 0.500M Na3PO4 was added to 500mL of DI water. Equilibrium 2:0.500

ID: 698808 • Letter: E

Question

Equilibrium 1: 0.500M Na3PO4 was added to 500mL of DI water. Equilibrium 2:0.500M CaSOs in aqueous solution. A standard curve was made with 4 solutions at 1.0ppm, 2.0ppm, 3.0ppm, and 5.0ppm with the absorbance of 0.115, 0.225, 0.350, and 0.580 respectively. The sample was prepped 3 times by taking a 50mL aliquot from a 1L solution. The results were 0.425, 0.432, and 0.422 AU. What was the determined concentration for all 3 analysis? What is the mean and Stdev? An LC run was made for the analysis of tryptamine (160.22g/mol) in a tablet. The tablet was weighed (1.12grams) and dissolved in 1L CHCl3. A 20ml Aliquot was diluted into 500mL CHCl3 and run. The results were an integration of 12367, 12375, 12382, and 12525, what is the %Tryptamine in the tablet for each analysis? Is there a reason for suspect in the analysis? M Int 0.25 10000 0.50 20000 0.75 30000 1.00 40000 4

Explanation / Answer

1) Plot the absorbance against the concentrations (in ppm) of the standard solutions. The plot is shown below.

Use the regression equation to find out the concentrations of the unknown in the three trials. Put y = 0.425 AU, 0.432 AU and 0.422 Au successively and output the concentrations. Tabulate the results.

Absorbance

Concentration in ppm as per the regression equation

0.425

0.425 = 0.152x – 0.0625

====> 0.152x = 0.4875

====> x = 3.2072 3.207

0.432

0.432 = 0.152x – 0.0625

====> 0.152x = 0.4945

====> x = 3.2534 3.253

0.422

0.422 = 0.152x – 0.0625

====> 0.152x = 0.4845

====> x = 3.1875 3.187

The mean concentration of the unknown is given as 1/3*(3.207 + 3.253 + 3.187) ppm = 3.21566 ppm 3.216 ppm (ans).

The standard deviation is given as

SD = [(3.207 – 3.216)2 + (3.253 – 3.216)2 + (3.187 – 3.216)2]/(3-1) ppm = 0.0011455 ppm = 0.03384 ppm 0.034 ppm (ans).