Chapter 23, Problem 23.22P

(a)

Interpretation Introduction

Interpretation:

For each concentration limit of detection and the limit of quantitation has to be calculated.

Concept Introduction:

The detection limit is the smallest quantity of analyte that is significantly different from the blank.

Limit of detection or minimum detectable limit is calculated as follows:

    $\text{Limit}\text{of}\text{detection}=\left(3\right)\left(\text{Relative}\text{standard}\text{deviation}\right)$

Quantitation limit is the smallest amount that can be measured with reasonable accuracy.

Quantitation limit is calculated as follows:

    $\text{Limit}\text{of}\text{quantitation}=\left(10\right)\left(\text{Relative}\text{standard}\text{deviation}\right)$

Relation between limit of detection and quantitaion limit is as follows:

    $\text{Limit}\text{of}\text{quantitation}=\left(\frac{10}{3}\right)\left(\text{Limit of detection}\right)$

Explanation of Solution

The relative standard deviation of $14.4\%$ of $0.2\text{μg}{\text{L}}^{-1}$ is calculated as follows:

    $\left(0.2\text{μg}{\text{L}}^{-1}\right)\left(\frac{14.4}{100}\right)=0.0288\text{μg}{\text{L}}^{-1}$

Limit of detection of first concentration value of bromated ion is calculated as follows:

    $\begin{array}{c}\left(3\right)\left(\text{Relative}\text{standard}\text{deviation}\right)=\left(3\right)\left(0.0288\text{μg}{\text{L}}^{-1}\right)\\ =0.0864\text{μg}{\text{L}}^{-1}\end{array}$

The relative standard deviation of $6.8\%$ of $0.5\text{μg}{\text{L}}^{-1}$ is calculated as follows:

    $\left(0.5\text{μg}{\text{L}}^{-1}\right)\left(\frac{6.8}{100}\right)=0.034\text{μg}{\text{L}}^{-1}$

Limit of detection of first concentration value of bromated ion is calculated as follows:

    $\begin{array}{c}\left(3\right)\left(\text{Relative}\text{standard}\text{deviation}\right)=\left(3\right)\left(0.034\text{μg}{\text{L}}^{-1}\right)\\ =0.102\text{μg}{\text{L}}^{-1}\end{array}$

The relative standard deviation of $3.2\%$ of $1.0\text{μg}{\text{L}}^{-1}$ is calculated as follows:

    $\left(1.0\text{μg}{\text{L}}^{-1}\right)\left(\frac{3.2}{100}\right)=0.032\text{μg}{\text{L}}^{-1}$

Limit of detection of first concentration value of bromated ion is calculated as follows:

    $\begin{array}{c}\left(3\right)\left(\text{Relative}\text{standard}\text{deviation}\right)=\left(3\right)\left(0.032\text{μg}{\text{L}}^{-1}\right)\\ =0.096\text{μg}{\text{L}}^{-1}\end{array}$

The relative standard deviation of $1.9\%$ of $2.0\text{μg}{\text{L}}^{-1}$ is calculated as follows:

    $\left(2.0\text{μg}{\text{L}}^{-1}\right)\left(\frac{1.9}{100}\right)=0.038\text{μg}{\text{L}}^{-1}$

Limit of detection of first concentration value of bromated ion is calculated as follows:

    $\begin{array}{c}\left(3\right)\left(\text{Relative}\text{standard}\text{deviation}\right)=\left(3\right)\left(0.038\text{μg}{\text{L}}^{-1}\right)\\ =0.114\text{μg}{\text{L}}^{-1}\end{array}$

Hence mean of these four detection limit of concentration is calculated as follows:

    $\frac{0.0864\text{μg}{\text{L}}^{-1}+0.102\text{μg}{\text{L}}^{-1}+0.096\text{μg}{\text{L}}^{-1}+0.114\text{μg}{\text{L}}^{-1}}{4}=0.0996\text{μg}{\text{L}}^{-1}$

Mean quantitation limit of concentration is calculated as follows:

    $\begin{array}{c}\left(\frac{10}{3}\right)\left(\text{Detection}\text{limit}\right)=\left(\frac{10}{3}\right)\left(\text{0}\text{.0996}\text{μg}{\text{L}}^{-1}\right)\\ =0.332\text{μg}{\text{L}}^{-1}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Absorbance of ${\text{Br}}_3^{-}$ in the $\text{6}\text{.00}\text{mm}$ pathlength detection cell of the chromatograph has to be calculared

Concept Introduction:

The detection limit is the smallest quantity of analyte that is significantly different from the blank.

Limit of detection or minimum detectable limit is calculated as follows:

    $\text{Limit}\text{of}\text{detection}=\left(3\right)\left(\text{Relative}\text{standard}\text{deviation}\right)$

The relation between the absorbance and concentration of the analyte is calculated from the Lambert Beer’s law as follows:

    $\text{A}=\left(\text{ε}\right)\left(\text{c}\right)\left(\text{l}\right)$

Here, $\text{A}$ is the absorbance.

$\text{ε}$ is the molar absorption coefficient.

$\text{c}$ is the concentration of analyte.

$\text{l}$ is the optical path length.

Explanation of Solution

The relative standard deviation of $14.4\%$ of $0.2\text{μg}{\text{L}}^{-1}$ is calculated as follows:

    $\left(0.2\text{μg}{\text{L}}^{-1}\right)\left(\frac{14.4}{100}\right)=0.0288\text{μg}{\text{L}}^{-1}$

Limit of detection of first concentration value of bromated ion is calculated as follows:

    $\begin{array}{c}\left(3\right)\left(\text{Relative}\text{standard}\text{deviation}\right)=\left(3\right)\left(0.0288\text{μg}{\text{L}}^{-1}\right)\\ =0.0864\text{μg}{\text{L}}^{-1}\end{array}$

The relative standard deviation of $6.8\%$ of $0.5\text{μg}{\text{L}}^{-1}$ is calculated as follows:

    $\left(0.5\text{μg}{\text{L}}^{-1}\right)\left(\frac{6.8}{100}\right)=0.034\text{μg}{\text{L}}^{-1}$

Limit of detection of first concentration value of bromated ion is calculated as follows:

    $\begin{array}{c}\left(3\right)\left(\text{Relative}\text{standard}\text{deviation}\right)=\left(3\right)\left(0.034\text{μg}{\text{L}}^{-1}\right)\\ =0.102\text{μg}{\text{L}}^{-1}\end{array}$

The relative standard deviation of $3.2\%$ of $1.0\text{μg}{\text{L}}^{-1}$ is calculated as follows:

    $\left(1.0\text{μg}{\text{L}}^{-1}\right)\left(\frac{3.2}{100}\right)=0.032\text{μg}{\text{L}}^{-1}$

Limit of detection of first concentration value of bromated ion is calculated as follows:

    $\begin{array}{c}\left(3\right)\left(\text{Relative}\text{standard}\text{deviation}\right)=\left(3\right)\left(0.032\text{μg}{\text{L}}^{-1}\right)\\ =0.096\text{μg}{\text{L}}^{-1}\end{array}$

The relative standard deviation of $1.9\%$ of $2.0\text{μg}{\text{L}}^{-1}$ is calculated as follows:

    $\left(2.0\text{μg}{\text{L}}^{-1}\right)\left(\frac{1.9}{100}\right)=0.038\text{μg}{\text{L}}^{-1}$

Limit of detection of first concentration value of bromated ion is calculated as follows:

    $\begin{array}{c}\left(3\right)\left(\text{Relative}\text{standard}\text{deviation}\right)=\left(3\right)\left(0.038\text{μg}{\text{L}}^{-1}\right)\\ =0.114\text{μg}{\text{L}}^{-1}\end{array}$

Hence mean of these four detection limit of concentration is calculated as follows:

    $\frac{0.0864\text{μg}{\text{L}}^{-1}+0.102\text{μg}{\text{L}}^{-1}+0.096\text{μg}{\text{L}}^{-1}+0.114\text{μg}{\text{L}}^{-1}}{4}=0.0996\text{μg}{\text{L}}^{-1}$

The reaction between ${\text{BrO}}_3^{-}$ and ${\text{Br}}^{-}$ is as follows:

    ${\text{BrO}}_3^{-}+{\text{8Br}}^{-}+6{\text{H}}^{+}\to 3{\text{Br}}_3^{-}+3{\text{H}}_{\text{2}}\text{O}$

Hence, molar concentration of ${\text{Br}}_3^{-}$ is 3 times greater than the molar concentration of ${\text{BrO}}_3^{-}$.

Hence molar concentration of ${\text{Br}}_3^{-}$ at mean detection limit is calculated as follows:

    $\begin{array}{c}\text{Molar concentration}{\text{of Br}}_3^{-}\text{at}\text{mean}\text{detection}\text{limit}=\frac{\left(3\right)\left(\text{detection limit}\text{of}{\text{BrO}}_3^{-}\text{in}\text{μg}{\text{L}}^{-1}\right)}{\left(\text{Molar}\text{mass}\text{of}{\text{BrO}}_3^{-}\right)}\\ =\frac{\left(3\right)\left(\text{0}\text{.0996}\text{μg}{\text{L}}^{-1}\right)}{\left(\text{127}\text{.90}\text{g}{\text{mol}}^{-1}\right)}\\ =2.3362\times {10}^{-9}\text{M}\end{array}$

Absorbance of ${\text{Br}}_3^{-}$ at mean detection limit is calculated as follows:

    $\begin{array}{c}\text{Absorbace}=\left(\text{Molar}\text{absorptivity}\right)\left(\text{Molar}\text{concentation}\right)\left(\text{Path}\text{length}\text{in}\text{cm}\right)\\ =\left(2.3362\times {10}^{-9}\text{M}\right)\left(\text{40900}{\text{M}}^{-1}{\text{cm}}^{-1}\right)\left(\text{0}\text{.6}\text{cm}\right)\\ =0.00005733\end{array}$