Question

A 4.112-g tablet, containing quinine was dissolved in 0.10 M HCl to give 500 mL. Dilution of a 10.0-mL aliquot to 100 mL yielded a reading for fluorescence intensity (at 347.5 nm) of 320 on an arbitrary scale. A second 10.0-mL aliquot was mixed with 10.0 mL of 100-ppm quinine solution before dilution to 100 mL. The fluorescence intensity of this solution was 433. Calculate the percentage of quinine in the tablet.

Answer 1

Quinine in a 1.664 g antimalarial tablet was dissolved in sufficient 0.10 M HCl to give 500 mL of solution. A 15 mL aliquot was then diluted to 100 mL with the acid. The fluorescence intensity for the dilute sample at 347.5 nm provided a reading of 288 on an arbitrary scale. A standard 100 ppm quinine solution registered 180 when measured under conditions identical to those for the diluted sample. Calculate the mass in milligrams of quinine in the tablet.

Your initial dissolution: 4.112 g tablet into 500 mL solution = 8.224 g tablet per L.

10 mL (0.010 L) aliquot @ 8.224 g tablet/L = 0.08224 g tablet in aliquot

0.08224 g tablet into 100 mL solution = 0.8224 g tablet per L.

100 ppm = 100 mg quinine/L = 0.100 g quinine/L

0.100 g quinine/L = 433 absorption units

Absorption is proportional to concentration, so ratio of absorptions will equal ratio of concentrations.

433 / 320 = 0.100 / x, where x is concentration of quinine in sample.

x = (0.100 x 320) / 433

= 0.074 g quinine/L solution.

Now you have 0.074 g quinine/L, and that came from a solution of 0.8224 g tablet/L.
So you have 0.074 g quinine / 0.8224 g tablet, or 0.06 g quinine per gram tablet.

Now just apply that proportion to the original mass of the tablet:

4.112 g tablet * 0.06 g quinine/g tab = 0.25 g quinine in the whole tablet.

So, % of quinine in the tablet = (0.25 g /4.112 g) x 100
= 6.08%