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

Answers Quinime is a 1.664 g Antimalarial tablet was. dissolved in suficient o lom Hel to give 500ml of solution A 15ml Aliquot was thin diluted to looml with the Acid - the fluorescence intensity for the dilute sample at 347.5 nom provided a reading a 288 on an arbitrary scale. A standard looppen quinime solution Registered 180 when measured under conditions identical to those for the Diluted Sample calculating the mass in milligrams of quinine in the tablet. Your intial dissolution: 4.1129 tablet into 500 ml solution – 8.224 a tablet perl.

lome (oolol) Aliquot @ 8.224 9 tablet/L = 0.08224 9 tablet in Aliquot. 0.08224 9 tablet into looml solution = 0.82249 tablet port 100 ppm = 100 mg quinime/l = 0.100g quinime/l. 0.1009 quinine/L= 433 Absorption units Absorption is proportional to concentration, so ratio of absorptions will equal ratio of concentrations. 933/32 - o. 100/8, where x is concentration o quinime in Somple: x= (0-100 x320) 1433 = 0.749 quimine /L solution

Now you have o.074 g quinimell, and that Come from a solution of 0.82249 tablet /L. So You have 0.074 9 quimime 10.82249 tablet, oro obg. quinime per grom tablet. Now Just Apply that proportion to the original mass of the tablet G.112 a tablet *0.069 quinine lg tab = 0.259. quinime in the whole tablet So, lo quimime in tablet = (0.259 14.1129)x100 = 6.08%.

Answer 2

Let the quinine present in tablet = m

then m is dissolved in 500 ml and 10 ml of this taken which means ( m/500) x 10 = 0.02 m present in that 10 ml.

Now this gives reading 320.

Now 2nd aliquot of 10 ml ( containing 0.02m quinine) taken and 100 ppm quinine added.

100 ppm means 100 g per 10^ 6 ml solution. Since 10 ml of 100 ppm used, quinine mass =( 10 ml x 100 g / 10^ 6ml) = 0.001 g

now total quinine present when both added = 0.02m + 0.001 g

This gives reading 433

Absorbance = e x C x l where C is concentrtaion

A2/A1 = C2 /C1

433 / 320 = ( 0.02m+0.001) / ( 0.02m)

1.353 ( 0.02m) =0.02m + 0.001

0.027m =0.02m + 0.001

m = 0.142g

% of quinine = ( mass of quinine / mass of tablet) x 100

= ( 0.142g / 4.122g) x 100

= 3.44 %