Chapter 5, Problem 5.11P

(a)

Interpretation Introduction

Interpretation:

The absorbance detection limit has to be calculated.

Concept introduction:

The signal detection limit is given as follows,

    $y_{dl}{\text{ = y}}_{\text{blank}}\text{ + 3s}$

Here, s is standard deviation.

Standard deviation:

The standard deviation of a random variable X of size n is defined as,

  $\text{s = }\sqrt{\frac{{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\left({\text{x}}_{\text{i}}\text{-}\overline{\text{x}}\right)}^{\text{2}}}}{\text{n-1}}}$

where ${\text{x}}_{\text{i}}$ is the i^th observation and $\overline{\text{x}}$ is the sample mean.

The standard deviation measures the absolute variability of a distribution.

Answer to Problem 5.11P

The absorbance detection limit is $\text{3}\text{.019}\times {\text{10}}^{\text{-3}}$.

Explanation of Solution

The standard deviation value for the 9 given samples is calculated as follows,

    $\begin{array}{c}\text{Mean, }\overline{\text{x}}\text{ = }\frac{\text{1}}{\text{9}}\left[\begin{array}{l}\text{0}\text{.0047+0}\text{.0054+0}\text{.0062+0}\text{.0060+0}\text{.0046}\\ \text{+0}\text{.0056+0}\text{.0052+0}\text{.0044+0}\text{.0058}\end{array}\right]\\ \\ =\text{ 0}\text{.00532}\\ \\ \text{s = }\sqrt{\frac{\begin{array}{l}{\left(\text{0}\text{.0047-0}\text{.00532}\right)}^{\text{2}}+{\left(\text{0}\text{.0054-0}\text{.00532}\right)}^{\text{2}}+{\left(\text{0}\text{.0062-0}\text{.00532}\right)}^{\text{2}}\\ +{\left(\text{0}\text{.0060-0}\text{.00532}\right)}^{\text{2}}{\left(\text{0}\text{.0046-0}\text{.00532}\right)}^{\text{2}}+{\left(\text{0}\text{.0056-0}\text{.00532}\right)}^{\text{2}}+\\ {\left(\text{0}\text{.0052-0}\text{.00532}\right)}^{\text{2}}+{\left(\text{0}\text{.0044-0}\text{.00532}\right)}^{\text{2}}+{\left(\text{0}\text{.0058-0}\text{.00532}\right)}^{\text{2}}\end{array}}{\text{8}}}\\ \\ =\sqrt{\frac{\begin{array}{l}\text{0}\text{.0000003844}+0.0000000064+0.0000007744+0.0000004624\\ +0.0000005184+0.0000000784+0.0000000144+0.0000008464\\ +0.0000002304\end{array}}{\text{8}}}\\ \\ s=\text{ 0}\text{.00061}\end{array}$

The mean blank value is ${\text{y}}_{\text{blank}}\text{ = 0}\text{.001189}$

The absorbance detection limit is calculated as follows,

    $\begin{array}{c}{y}_{dl}{\text{ = y}}_{\text{blank}}\text{ + 3s}\\ \\ \text{= 0}\text{.001189+3}\times 0.00061\\ \\ y_{dl}=\text{ 3}\text{.019}\times {\text{10}}^{\text{-3}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The concentration detection limit has to be calculated.

Concept Introduction:

The minimum detectable concentration is given as follows,

    $\text{Minimum detectable concentration = 3s/m}$

Here, m is slope and s is standard deviation.

Answer to Problem 5.11P

The concentration detection limit value is $\text{8}\text{.17}\times {\text{10}}^{\text{-8}}\text{ M}$.

Explanation of Solution

The concentration detection limit is calculated as follows,

    $\begin{array}{c}\text{Minimum detectable concentration = 3s/m}\\ \\ \text{= }\frac{3\times \text{ 0}\text{.00061}}{2.24\times {10}^4{\text{ M}}^{\text{-1}}}\\ \\ =\text{ 8}\text{.17}\times {\text{10}}^{\text{-8}}\text{ M}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The lower limit of quantitation has to be calculated.

Concept Introduction:

The lower limit of quantitation is given as follows,

    $lower\lim itofquantitation\text{ =}\frac{10s}{m}$

Here, s is standard deviation and m is slope.

Answer to Problem 5.11P

The lower limit of quantitation is $\text{2}\text{.77}\times {\text{10}}^{\text{-7}}\text{ M}$.

Explanation of Solution

The lower limit of quantitation as follows,

    $\begin{array}{c}lower\lim itofquantitation\text{ =}\frac{10\left(0.00061\right)}{2.20\times {10}^4{\text{ M}}^{\text{-1}}}\\ \\ =\text{ 2}\text{.77}\times {\text{10}}^{\text{-7}}\text{ M}\end{array}$