5-28. Ititandard addition. Lead in dry river sediment was extracted with $\mathbf{25}\mathbf{}\mathbf{wt}\mathbf{\%}\mathbf{}{\mathbf{HNO}}_{\mathbf{3}}\mathbf{}\mathbf{at}\mathbf{}\mathbf{35}\mathbf{°}\mathbf{C}\mathbf{}\mathbf{for}\mathbf{}\mathbf{1}\mathbf{h}\mathbf{.}\mathbf{}\mathbf{then}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{mL}$of filtered extract was mixed with other reagents to bring the total volume to ${\mathbf{V}}_{\mathbf{0}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{60}\mathbf{mL}\mathbf{.}\mathbf{}\mathbf{pb}\mathbf{\left(}\mathbf{ll}\mathbf{\right)}$was measured electrochemically with a series of standard additions of $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{pb}\mathbf{\left(}\mathbf{ll}\mathbf{\right)}\mathbf{.}$

(a) Volume is not constant, so follow the procedure of Figures 5-5 and 5-6 to find $\mathbf{ppm}\mathbf{}\mathbf{pb}\mathbf{\left(}\mathbf{ll}\mathbf{\right)}\mathbf{}\mathbf{in}\mathbf{}\mathbf{the}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{mL}$

extract.

(b) Find the standard uncertainty and $\mathbf{95}\mathbf{\%}$confidence interval for the $\mathbf{x}\mathbf{-}$intercept of the graph. Assuming that uncertainty in intercept is larger than other uncertainties, estimate the uncertainty in ppm $\mathbf{pb}\mathbf{\left(}\mathbf{ll}\mathbf{\right)}\mathbf{}\mathbf{in}\mathbf{}\mathbf{the}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{mL}$extract.

(a)

According to the above chart,a standard addition graph of final pb(II) have to be prepared ,contributed by addition of standard pb(II) solution.

The below graph represents their respective observance

So,the trendline equation for the Abs vs cone graph is $\mathrm{y}=49.52\mathrm{x}+1.094$,which is in the form of $\mathrm{y}=\mathrm{mx}+\mathrm{c}$

In the graph, $\mathrm{y}-\mathrm{axis}\mathrm{is}\mathrm{signal},\mathrm{x}-\mathrm{axis}\mathrm{is}\mathrm{pb}\left(\mathrm{ll}\right)\mathrm{in}\mathrm{ppm},\mathrm{y}-\mathrm{intercept}\mathrm{is}\mathrm{c},\mathrm{slope}\mathrm{is}\mathrm{m}.$

Here,the signal is zero at $\mathrm{y}-\mathrm{intercept}=1.094\mathrm{and}\mathrm{m}=49.52$

$\mathrm{pb}\left(\mathrm{ll}\right)\mathrm{is}\frac{\mathrm{intercept}}{\mathrm{slope}}=\frac{1.094}{49.52}=0.0221\mathrm{ppm}$

Hence,$\mathrm{pb}\left(\mathrm{ll}\right)$inunskippedaliquent$=0.0221\mathrm{ppm}$

(b)

The unskipped aliquant is consist of

$0.5\mathrm{mL}$of original solution

$\mathrm{A}$and diluted to a final volume of

$50\mathrm{ml}.$

$$\begin{gathered} {\mathrm{C}}_1{\mathrm{V}}_1\left(\mathrm{unskipped}\mathrm{aliquent}\right)={\mathrm{C}}_2{\mathrm{V}}_2\left(\mathrm{original}\mathrm{solution}\right) \\ {\mathrm{C}}_2=\frac{0.0221\mathrm{ppm}\times 5.0\mathrm{ml}}{0.50\mathrm{ml}} \\ {\mathrm{C}}_2=0.0221\mathrm{ppm} \end{gathered}$$

So.pb(II) is original solution $=0.0221\mathrm{ppm}$

$\mathrm{ppm}=\frac{\mathrm{mg}\mathrm{of}\mathrm{solute}}{\mathrm{lit}\mathrm{of}\mathrm{solution}}$

So,the pb(II) in original solution is

$$\begin{gathered} \mathrm{A}=0.221\mathrm{ppm} \\ \mathrm{pb}\left(\mathrm{ll}\right)=0.221\mathrm{mg}/\mathrm{L} \\ \mathrm{pb}\left(\mathrm{ll}\right)=221\frac{\mathrm{\mu g}}{\mathrm{Lit}} \end{gathered}$$

So,The pb(II) in original solution is$221\frac{\mathrm{\mu g}}{\mathrm{Lit}}$