Chapter 5: Q19P (page 115)

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

Short Answer

Expert verified

The required concentration detection limit and mean CDL is

$CDL (for 0.2 μg / L) is 0.86 μg / L CDL (for 0.5 μg / L) is 0.102 μg / L CDL (for 1.0 μg / L) is 0.096 μg / L CDL (for 2.0 μg / L) is 0.114 μg / L meanCDL IS 0.1 μg / L$

Step by step solution

Definition of concentration detection limit

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.

Determine the first concentration

The following is the solution to this problem:

4 amounts of bromate $(in μg / L) : 0.2, 0.5, 1.0, 20 Relative standard deviations (%) : 14.4, 0.8, 3.2, 1.9 y_{blank} = 0$

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, $(0.2 μg / L) :$

$s - relative standard deviation \times concentration d = 0.144 \times 0.2 μg / L s = 0.028 μg / L$

Determine the second and third and fourth concentration

The second concentration will be $(0.5 μg / L) :$

$s = relative standard deviation \times concentration = 0.068 \times 0.5 μg / L s = 0.034 μg / L The third concentration consists of (1.0 μg / L) : s = relative standard deviation \times concentration = 0.0.32 \times 1.0 μg / L s = 0.032 μg / L The fourth concentration consists of (2.0 μg / L) : s = relative standard deviation \times concentration = 0.0.19 \times 2.0 μg / L s = 0.0038 μg / L$

Determine the concentration detection limits

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

For the preliminary concentration, $0.2 μg / L$

role="math" localid="1655018837859" $concentration detection limit = y_{blank} + 3 s = - 0 . + (3) (0.0288 μg / L) concentration detection limit = 0.086 μg / L The second concentration will be (0.5 μg / L) : concentration detection limit = y_{blank} + 3 s = 0 + (3) (0.032 μg / L) Step 10 concentration detection limit = 0.142 μg / L The third concentration consists of (1.0 μg / L) : concentration detection limit = y_{blank} + 3 s = 0 + (3) (0.032 μg / L) concentration detection limit = 0.096 μg / L The fourth concentration consists of (2.0 μg / L) : concentration detection limit = y_{blank} + 3 s = 0 + (3) (0.038 μg / L) concentration detection limit = 0.114 μg / L$

Determine the mean concentration detection limits

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

$mean concentration detection limit = \frac{i}{n} = \frac{(0.086 + 0.102 + 0.096 + 0.114) μg / L}{4} = \frac{0.2954 μg / L}{4} = 0.07385 μg / L mean concentration detection limit = 0.01 μg / L$