Standard addition. Selenium from 0.108 g of Brazil nuts was converted into the fluorescent product in Reaction 18-15, and extracted into 10.0 mL of cyclohexane. Then 2.00 mL of the cyclohexane solution were placed in a cuvet for fluorescence measurement. Standard additions of fluorescent product containingSe/mL are given in the table. Construct a standard addition graph to find the concentration of Se in the 2.00-mL unknown solution. Find the wt%of Se in the nuts and its uncertainty and 95% confidence interval

${\mathrm{I}}_{\mathrm{s}}+\mathrm{X}\to$fluorescence intensity from the unknown + standard addition $V\to$total volume of (unknown + standard)${\mathrm{V}}_0\to$ fluorescence intensity from unknown initial concentration of Unknown initial concentration of standard ${\mathrm{V}}_S\to$volume of standard.

${\mathrm{I}}_{\mathrm{s}+\mathrm{x}}\left(\frac{\mathrm{V}}{{\mathrm{V}}_0}\right)={\mathrm{I}}_{\mathrm{x}}+\frac{{\mathrm{I}}_{\mathrm{x}}}{[\mathrm{X}{]}_i}\cdot [\mathrm{S}{]}_i\cdot \left(\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_0}\right)$

The graph${\mathrm{I}}_{\mathrm{S}}+\mathrm{x}\left(\frac{\mathrm{V}}{{\mathrm{V}}_0}\right)$

Versus is,$[\mathrm{S}{]}_i\left(\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_0}\right)$

The x-intercept is the concentration of Se. We calculate x-intercept by dividing b / a. So, the concentration of Se is$0.03842\mathrm{\mu g}/\mathrm{mL}$

$\begin{array}{r}\mathrm{w}\left(\mathrm{Se}\right)=\frac{\mathrm{m}\left(\mathrm{Se}\right)}{\mathrm{m}\left(\text{nuts}\right)}\\ \text{Where Mass of Se is,}\\ \mathrm{m}\left(\mathrm{Se}\right)=\mathrm{V}\cdot \gamma \left(\mathrm{Se}\right)\\ \mathrm{m}\left(\mathrm{Se}\right)=10\mathrm{mL}\cdot 0.03842\mu \mathrm{g}/\mathrm{mL}\\ \mathrm{m}\left(\mathrm{Se}\right)=0.3842\mu \mathrm{g}\\ \mathrm{w}\left(\mathrm{Se}\right)=\frac{0.3842\cdot {10}^{-6}\mathrm{g}}{0.108\mathrm{g}}\\ \mathrm{w}\left(\mathrm{Se}\right)=3.56\cdot {10}^{-4}\mathrm{\%}\end{array}$

The standard deviation of x-intercept is calculated by formula:

$\frac{s_y}{\left|a\right|}\cdot \sqrt{\frac{1}{n}+\frac{{\overline{y}}^2}{a^2\cdot \mathrm{\Sigma }{\left(x_i-\overline{x}\right)}^2}}$

n is a number of data, y is a average of y,$x_i$are individual values of x, and is average of x.

$\sum {\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{X}}\right)}^2$

The relative uncertainty in the intercept is:

$\frac{0.000732}{0.03842}=1.91\mathrm{\%}$

Uncertainty of :

$\begin{array}{r}0.0192\cdot 3.56\cdot {10}^{-4}\mathrm{\%}=6.84\cdot {10}^{-6}\mathrm{\%}\\ 3.56(\pm 0.068)\cdot {10}^{-4}\mathrm{\%}\end{array}$

95% confidence interval is calculated:

$\pm$t. standard deviation

t is a Student's t. We read value from Table 4-4. The degrees of freedom are:

n - 2 = 5 - 2 =3

For 95% confidence level and 3 degrees of freedom, the value is 3.182.

$\begin{array}{r}3.182\cdot 0.068\cdot {10}^{-4\mathrm{\%}}=0.22\cdot {10}^{-4\mathrm{\%}}\\ 3.56(\pm 0.22)\cdot {10}^{-4}\mathrm{\%}\end{array}$