Question

The determination in Problem 15-9 was modified to use the standard-addition method. In this case, a 3.925-g tablet was dissolved in sufficient 0.10 M HCl to give 1.000 L. Dilution of a 20.00-mL aliquot to 100 mL yielded a solution that gave a reading of 415 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 503. Calculate the percentage of quinine in the tablet.

Short answer

The percentage of quinine in the tablet is 3 % .

Step-by-step solution

Given Information

The percentage of quinine in the tablet should be determined.

Explanation

The expression of absorbance of unknown concentration is:

$A_x=\frac{\epsilon b{C}_x{V}_x}{V_1}$ ........ (I)

Here, the permittivity is $\epsilon$, the absorbance is b , the absorbance of unknown concentration is $A_x$and the volume of the unknown concentration is $V_x$.

The expression of absorbance of known concentration is:

$A_{x+x}=\frac{\epsilon b\left(c_x{V}_x+c_s{V}_s\right)}{V_1}$........ (II)

Here, the concentration of known concentration is $c_s$, the absorbance of known concentration is $A_{x+x}$and the volume of the known concentration is $V_s$.

Divide Equation (I) by Equation (II).

localid="1646723306722" $$\begin{gathered} \frac{A_x}{A_{x+s}}=\frac{\left({\displaystyle \frac{\epsilon b{c}_x{V}_x}{V_1}}\right)}{\left\{{\displaystyle \frac{\epsilon b(c_x{V}_x+c_s{V}_s}{V_1}}\right\}} \\ \frac{A_x}{A_{x+s}}=\frac{c_x{V}_x}{\left(c_x{V}_x+c_s{V}_s\right)} \\ {A}_{x+s}{c}_x{V}_x=A_x{c}_x{V}_x+A_s{c}_s{V}_s \\ {A}_{x+s}{c}_x{V}_x-A_x{c}_x{V}_x=A_s{c}_s{V}_s \\ {c}_x(A_{x+s}{V}_x-A_x{V}_x)=A_s{c}_s{V}_s \\ {c}_x=\frac{A_s{c}_s{V}_s}{V_x(A_{x+s}-A_x)} \end{gathered}$$......................(III)

The expression of percentage of quinine in the sample of 1000 mL is:

$\%Q=c_x\left(\frac{1000mL}{M}\right)\times 100\%$......................(IV)

Here, the amount of the tablet is M.

Substitute 415 for $A_x$, 50 ppm for $c_s$, 10 mL for localid="1646722705075" $V_s$, 20 mL for $V_x$and 503 for $A_{x+x}$in Equation (III).

$$\begin{gathered} {\mathrm{c}}_{\mathrm{x}}=\frac{\left(415\right)(50\mathrm{ppm})(10\mathrm{mL})}{(20\mathrm{mL})(503-415)} \\ =\frac{\left(415\right)(50\mathrm{ppm})(10\mathrm{mL})}{(20\mathrm{mL})\times 88} \\ =(117.9\mathrm{ppm})\left(\frac{1\mathrm{mg}/\mathrm{L}}{1\mathrm{ppm}}\right)\left(\frac{1\mathrm{L}}{1000\mathrm{mL}}\right)\left(\frac{1\mathrm{g}}{1000\mathrm{mg}}\right) \\ =1.179\times {10}^{-4}\mathrm{g}/\mathrm{mL} \end{gathered}$$

Substitute 1.179 x 10 ^-4 g/mL for $c_x$and 3.925 g for M in Equation (IV).

$$\begin{gathered} \%\mathrm{Q}=\left(1.179\times {10}^{-4}\mathrm{g}/\mathrm{mL}\right)\left(\frac{1000\mathrm{mL}}{3.925\mathrm{g}}\right)\times 100\% \\ =0.03\times 100\% \\ =3\% \end{gathered}$$