Quinine in a 1.664-g antimalarial tablet was dissolved in sufficient 0.10 M HCI
ID: 493504 • Letter: Q
Question
Quinine in a 1.664-g antimalarial tablet was dissolved in sufficient 0.10 M HCI to give 500 mL of solution. A 20.00-mL aliquot was then diluted to 100.0 mL with the acid. The fluorescence intensity for the diluted sample at 347.5 nm provided a reading of 245 on an arbitrary scale. A standard 100-ppm quinine solution registered 125 when measured under conditions identical to those for the diluted sample. Calculate the mass in milligrams of quinine in the tablet. The determination in Problem 15-9 was modified to use the standard-addition method. In this case, a 4.236-g tablet was dissolved in sufficient 0.10 M HCI to give 1.000 L. Dilution of a 20.00-mL aliquot to 100 mL yielded a solution that gave a reading of 448 at 347.5 nm. A second 20.00-mL aliquot was mixed with 10.0 mL of 50-ppm quinine solution before dilution to 100 mL. The fluorescence intensity of this solution was 525. Calculate the percentage of quinine in the tablet.Explanation / Answer
15-9
A 20 ml aliquot sample gave reading of = 245
A 100 ppm standard sample gave reading = 125
So,
concentration of quinine in 100 ml aliquot = 245 x 100/125
= 196 ppm
Concentration of quinine in 500 ml solution = 196 x 500/100
= 980 ppm
mass of quinine in tablet = 0.980 g
15-10
Concentration of standard in solution = 50 ppm x 10 ml/100 ml
= 5 ppm
Let Cx be concentration of quinine in the diluted sample
Cx/(Cx + 5) = 448/525
525Cx = 448Cx + 2240
Cx = 29.1 ppm
Concentration of quinine in tablet = 29.1 x 1000/100 = 291 ppm
mass of quinine in tablet = 29.1 x 1000/20 = 1.455 mg
Percentage of quinine in the tablet = 1.455 x 100/4.236 = 34.35%
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