Spike recovery and detection limit. Species of arsenic found in drinking water include ${\mathbf{AaO}}_{\mathbf{3}}^{\mathbf{3}\mathbf{-}}$(arsenite), ${\mathbf{AaO}}_{\mathbf{3}}^{\mathbf{3}\mathbf{-}}$(arsenate),role="math" localid="1655008942536" ${\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}{\mathbf{Aso}}_{\mathbf{2}}$(dimethylarsinate), and $\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\right)}{\mathbf{AsO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$(methylarsonate). Pure water containing no arsenic was spiked with $\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{\mu g}$arsenate/L. Seven replicate determinations gave 0.39,0.40,0.38, 0.41,0.36,0.35, and $\mathbf{0}\mathbf{.}\mathbf{39}\mathbf{\mu g}\mathbf{/}{\mathbf{L}}^{\mathbf{15}}$Find the mean percent recovery of the spike and the concentration detection limit $\mathbf{(}\mathbf{\mu G}\mathbf{/}\mathbf{L}\mathbf{)}$.

• The detection limit is (informally) the lowest analyte concentration that can be consistently identified, and it reflects the precision of the instrumental response obtained by the method when the analyte concentration is zero.
• The average is the total of a collection of measurements divided by the number of measurements.

Consider the following information:

There were seven replicate determinations. (in $\mathrm{\mu g}/\mathrm{L}):0.39,0.40,0.38,0.41,0.36,0.35,0.39)$

$$\begin{gathered} {\mathrm{C}}_{\mathrm{tmspiked}\mathrm{sample}}=0.0\mathrm{\mu g}/\mathrm{L} \\ {\mathrm{C}}_{\mathrm{added}}=\sout{\mathrm{D}}0.0\mathrm{\mu g}/\mathrm{L} \\ {\mathrm{y}}_{\mathrm{hlan}\mathrm{k}}=0.0 \end{gathered}$$

Solve for (i) percent recovery and (ii) concentration detection limit in order to solve the problem.

(i) First compute the mean concentration for the spiked sample before solving the percent recovery. To accomplish,

role="math" localid="1655011537189" $$\begin{gathered} \overline{\mathrm{x}}=\frac{{\mathrm{i}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{(0.39+0.40+...+0.35++0.39)\mathrm{\mu g}/\mathrm{L}}{7} \\ =\frac{2.68\mathrm{\mu g}/\mathrm{L}}{7} \\ \overline{x}=0.{38}_{\mathrm{s\mu g}/\mathrm{L}} \end{gathered}$$

With the knowledge of $\overline{\mathrm{x}}$, for percent recovery, use the following formula:

$$\begin{gathered} \mathrm{mean}\%\mathrm{recovery}=\frac{0.383-0.0\mathrm{\mu g}/\mathrm{L}}{0.4\mathrm{\mu g}/\mathrm{L}}\times 100 \\ =95.75\%\mathrm{mean}\%\mathrm{recovery} \\ =96\% \\ \mathrm{So},\mathrm{the}\mathrm{mean}\mathrm{percent}\mathrm{recovery}\mathrm{is}96\% \end{gathered}$$

Calculate the number for standard deviation, s, for component (ii).

To do so, recollect the sample mean value obtained earlier.

$\overline{\mathrm{X}}$.

$$\begin{gathered} s=\sqrt{\frac{\mathrm{i}({\mathrm{x}}_1-\overline{\mathrm{x}{)}^2}}{n-1}} \\ =\sqrt{\frac{{(0.39-0.383)}^2+{(0.40-0.383)}^2+...+{(0.35-0.383)}^2+{(0.39-0.383)}^2}{6}}\mathrm{\mu g}/\mathrm{L} \\ =\sqrt{4.581666667\times {10}^{-3}\mathrm{\mu g}/\mathrm{L}} \\ =0.021381456\mathrm{\mu g}/\mathrm{L} \\ \mathrm{s}=0.{0213}_8\mathrm{\mu g}/\mathrm{L} \end{gathered}$$

Now solve for the detection limit for concentration by substituting the values into the equation:

$$\begin{gathered} \mathrm{concentration}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s} \\ =0.0+3(0.02138)\mathrm{\mu g}/\mathrm{L} \\ =0.06414\mathrm{\mu g}/\mathrm{L} \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.064\mathrm{\mu g}/\mathrm{L} \\ \mathrm{Thus}\mathrm{the}\mathrm{concentration}\mathrm{limit}\mathrm{is}0.064\mathrm{\mu g}/\mathrm{L} \end{gathered}$$