Question

Quinine in a $1.553-g$antimalarial tablet was dissolved in sufficient $0.10MHCl$to give $250mL$of solution. A $10.00-mL$aliquot was then diluted to $50.00mL$with the acid. The fluorescence intensity for the diluted sample at $347.5nm$provided a reading of $196$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.

Short answer

The mass of quinine is$196mgQ$

Step-by-step solution

Step 1. Given information

Mass of Quinine is $1.553g$

Volume of aliquot is$10mL$

Step 2. Formula used

In spectroscopy, the fluorescence intensity is mostly employed. The sample's flouroscent intensity is proportional to its concentration in the opposite direction. The following is the link between flouroscent intensity and concentration:

$Q_{\mathrm{A}}=Q_{\mathrm{c}}\times \frac{F_{\mathrm{D}}}{F_{\mathrm{s}}}....\left(1\right)$

Here, $Q_c$represents the quinine concentration, $F_s$represents the fluorescent intensity of the sample, $F_s$represents the fluorescent intensity of the sample, and $Q_A$represents the quantity of quinine.

The mass of quinine is the same as the volume of quinine in millilitres. This is how it's written:

$mass\mathrm{Q}=\left(Q_{\mathrm{A}}\right)\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right).....\left(2\right)$

Here, mass $Q$is equal to the mass of quinine, $V_2$is the volume of sample after dilution, $Q_A$is equal to the amount of quinine and$V_2$ is the volume of sample before dilution.

Step 3. Calculating mass

Substitute the values in equation $\left(1\right)$,

$Q_{\mathrm{A}}=\left(100\mathrm{ppm}\right)\left(\frac{196}{125}\right)=156.8ppm$

Now substitute the values in equation $\left(2\right)$,

$$\begin{gathered} \text{mass}Q=(156.8\mathrm{ppm})\left(\frac{50\mathrm{mL}}{10\mathrm{mL}}\right) \\ (1\mathrm{ppm}=1\mathrm{mgQ}/{10}^3\mathrm{mL}) \\ mass\mathrm{Q}=\left(156.8\mathrm{mgQ}/{10}^3\mathrm{mL}\right)\left(\frac{50\mathrm{mL}}{10\mathrm{mL}}\right) \\ =784\mathrm{mgQ}/{10}^3\mathrm{mL} \end{gathered}$$

For $250mL$,

$$\begin{gathered} mass\mathrm{Q}=\left(784\mathrm{mgQ}/{10}^3\mathrm{mL}\right)(250\mathrm{mL}) \\ =196\mathrm{mg}\mathrm{Q} \end{gathered}$$