Detection limit. A sensitive chromatographic method was developed to measure sub-part-per-billion levels of the disinfectant by-products iodate$\mathbf{\left(}{\mathbf{IO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\right)}$, chlorite$\mathbf{\left(}{\mathbf{CIO}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{\right)}$, and bromate$\mathbf{\left(}{\mathbf{BrO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\right)}$in drinking water. As the oxyhalides emerge from the column, they absorption at$\mathbf{267}\mathbf{}\mathbf{nm}$. For example, each mole of ${\mathbf{BrO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}\mathbf{8}{\mathbf{Br}}^{\mathbf{-}}\mathbf{+}\mathbf{6}{\mathbf{H}}^{\mathbf{+}}\overline{)\mathrm{R}}\mathbf{3}{\mathbf{Br}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$bromate makes by the reaction

Bromate near its detection limit gave the following chromatographic peak heights and standard deviations (s). For each concentration, estimate the detection limit. Find the mean detection limit. The blank is 0 because chromatographic peak height is measured from the baseline adjacent to the peak. Because blank =0, relative standard deviation applies to both peak height and concentration, which are proportional to each other. Detection limit is 3 s for peak height or concentration.

The following is the solution to this problem:

4 amounts of bromate $$\begin{gathered} (\mathrm{in}\mathrm{\mu g}/\mathrm{L}):0.2,0.5,1.0,20 \\ \mathrm{Relative}\mathrm{standard}\mathrm{deviations}(\%):14.4,0.8,3.2,1.9 \\ {\mathrm{y}}_{\mathrm{blank}}=0 \end{gathered}$$

The concentration detection limits for four different concentrations, as well as the mean concentration detection limit, must be determined.

First solve the standard deviations for each concentration since we are given relative standard deviations, which are depending on the amount of concentration.

For the preliminary concentration,$(0.2\mathrm{\mu g}/\mathrm{L}):$

$$\begin{gathered} s-\mathrm{relative}\mathrm{standard}\mathrm{deviation}\times \mathrm{concentration} \\ \mathrm{d}=0.144\times 0.2\mathrm{\mu g}/\mathrm{L} \\ \mathrm{s}=0.028\mathrm{\mu g}/\mathrm{L} \end{gathered}$$

The second concentration will be $(0.5\mathrm{\mu g}/\mathrm{L}):$

$$\begin{gathered} \mathrm{s}=\mathrm{relative}\mathrm{standard}\mathrm{deviation}\times \mathrm{concentration} \\ =0.068\times 0.5\mathrm{\mu g}/\mathrm{L} \\ \mathrm{s}=0.034\mathrm{\mu g}/\mathrm{L} \\ \mathrm{The}\mathrm{third}\mathrm{concentration}\mathrm{consists}\mathrm{of}(1.0\mathrm{\mu g}/\mathrm{L}): \\ \mathrm{s}=\mathrm{relative}\mathrm{standard}\mathrm{deviation}\times \mathrm{concentration} \\ =0.0.32\times 1.0\mathrm{\mu g}/\mathrm{L} \\ \mathrm{s}=0.032\mathrm{\mu g}/\mathrm{L} \\ \mathrm{The}\mathrm{fourth}\mathrm{concentration}\mathrm{consists}\mathrm{of}(2.0\mathrm{\mu g}/\mathrm{L}): \\ \mathrm{s}=\mathrm{relative}\mathrm{standard}\mathrm{deviation}\times \mathrm{concentration} \\ =0.0.19\times 2.0\mathrm{\mu g}/\mathrm{L} \\ \mathrm{s}=0.0038\mathrm{\mu g}/\mathrm{L} \end{gathered}$$

Find the concentration detection limits for each concentration now that we have the s for each concentration:

For the preliminary concentration,$0.2\mathrm{\mu g}/\mathrm{L}$

role="math" localid="1655018837859" $$\begin{gathered} \mathrm{concentration}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s} \\ =-0.+\left(3\right)(0.0288\mathrm{\mu g}/\mathrm{L}) \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.086\mathrm{\mu g}/\mathrm{L} \\ \mathrm{The}\mathrm{second}\mathrm{concentration}\mathrm{will}\mathrm{be}(0.5\mathrm{\mu g}/\mathrm{L}): \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s} \\ =0+\left(3\right)(0.032\mathrm{\mu g}/\mathrm{L})\mathrm{Step}10 \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.142\mathrm{\mu g}/\mathrm{L} \\ \mathrm{The}\mathrm{third}\mathrm{concentration}\mathrm{consists}\mathrm{of}(1.0\mathrm{\mu g}/\mathrm{L}): \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s} \\ =0+\left(3\right)(0.032\mathrm{\mu g}/\mathrm{L}) \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.096\mathrm{\mu g}/\mathrm{L} \\ \mathrm{The}\mathrm{fourth}\mathrm{concentration}\mathrm{consists}\mathrm{of}(2.0\mathrm{\mu g}/\mathrm{L}): \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s} \\ =0+\left(3\right)(0.038\mathrm{\mu g}/\mathrm{L}) \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.114\mathrm{\mu g}/\mathrm{L} \end{gathered}$$

Find the mean of the four values of detection limits that we previously solved to get the mean concentration detection limit:

$$\begin{gathered} \mathrm{mean}\mathrm{concentration}\mathrm{detection}\mathrm{limit}=\frac{\mathrm{i}}{\mathrm{n}} \\ =\frac{\left(0.086+0.102+0.096+0.114\right)\mathrm{\mu g}/\mathrm{L}}{4} \\ =\frac{0.2954\mathrm{\mu g}/\mathrm{L}}{4} \\ =0.07385\mathrm{\mu g}/\mathrm{L} \\ \mathrm{mean}\mathrm{concentration}\mathrm{detection}\mathrm{limit}=0.01\mathrm{\mu g}/\mathrm{L} \end{gathered}$$