Chapter 18, Problem 18.BE

(a)

Interpretation Introduction

Interpretation:

For the given compound A, the value of molar absorptivity has to be calculated.

Concept introduction:

Beer-lambert law:

$\begin{array}{l}\text{A}\text{=}\text{-}\text{log}\frac{\text{P}}{{\text{P}}_{\text{o}}}\text{=}\text{εcb}\\ \text{where,}\\ \text{A}\text{-}\text{absorbance}\\ {\text{P}}_{\text{o}}\text{-}\text{incident}\text{irradiance}\text{and}\text{P}\text{-}\text{transmitted}\text{irradiance}\\ \text{ε}\text{-}\text{molar}\text{absorptivity}\\ \text{c-}\text{concentration}\text{and}\text{b}\text{-}\text{pathlength}\text{.}\end{array}$

Answer to Problem 18.BE

The molar absorptivity $\left(\text{ε}\right)$ value of compound A is $\text{1}\text{.50}\text{×}{\text{10}}^{\text{3}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Concentration}\text{is}\text{3}\text{.96}\text{×}{\text{10}}^{\text{-4}}\text{M}\text{.}\\ \text{Absorbance}\left(\text{A}\right)\text{is}\text{0}\text{.624}\text{at}\text{238}\text{nm}\text{in}\text{1}\text{.000}\text{cm}\text{cuvet}\text{.}\\ \text{Absorbance}\text{for}\text{blank}\text{solution}\text{is}\text{0}\text{.029}\text{at}\text{238}\text{nm}\text{.}\end{array}$

Apply the Beer’s law to calculate the value of molar absorptivity $\left(\text{ε}\right)$

$\begin{array}{l}\text{A}\text{=}\text{εcb}\\ \text{ε}\text{=}\frac{\text{A}}{\text{bc}}\text{=}\frac{\text{0}\text{.624}\text{-}\text{0}\text{.029}}{\left(\text{3}\text{.96}\text{×}{\text{10}}^{\text{-4}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\right)\left(\text{1}\text{.000}\text{cm}\right)}\\ \text{=}\text{1}\text{.50}\text{×}{\text{10}}^{\text{3}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\text{.}\end{array}$

The molar absorptivity $\left(\text{ε}\right)$ value of compound A is $\text{1}\text{.50}\text{×}{\text{10}}^{\text{3}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The concentration of compound A in the unknown solution has to be calculated.

Concept introduction:

Beer-lambert law:

$\begin{array}{l}\text{A}\text{=}\text{-}\text{log}\frac{\text{P}}{{\text{P}}_{\text{o}}}\text{=}\text{εcb}\\ \text{where,}\\ \text{A}\text{-}\text{absorbance}\\ {\text{P}}_{\text{o}}\text{-}\text{incident}\text{irradiance}\text{and}\text{P}\text{-}\text{transmitted}\text{irradiance}\\ \text{ε}\text{-}\text{molar}\text{absorptivity}\\ \text{c-}\text{concentration}\text{and}\text{b}\text{-}\text{pathlength}\text{.}\end{array}$

Answer to Problem 18.BE

The concentration of compound A in the unknown solution is $\text{2}\text{.31}{\text{×10}}^{\text{-4}}\text{M}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Absorbance}\left(\text{A}\right)\text{is}\text{0}\text{.375}\text{at}\text{238}\text{nm}\text{in}\text{1}\text{.000}\text{cm}\text{cuvet}\text{.}\\ \text{Absorbance}\text{for}\text{blank}\text{solution}\text{is}\text{0}\text{.029}\text{at}\text{238}\text{nm}\text{.}\\ \text{Molar}\text{absorptivity}\text{is}\text{1}\text{.50}\text{×}{\text{10}}^{\text{3}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\text{.}\end{array}$

Calculate the concentration of compound A in the unknown solution

$\begin{array}{l}\text{c}\text{=}\frac{\text{A}}{\text{εb}}\text{=}\frac{\text{0}\text{.375}\text{-}\text{0}\text{.029}}{\left(\text{1}\text{.50}\text{×}{\text{10}}^{\text{3}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\right)\left(\text{1}\text{.000}\text{cm}\right)}\\ \text{=}\text{2}\text{.31}\text{×}{\text{10}}^{\text{-4}}\text{M}\text{.}\end{array}$

The concentration of compound A in the unknown solution is $\text{2}\text{.31}{\text{×10}}^{\text{-4}}\text{M}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The concentration of compound A in the concentrated solution has to be calculated.

Concept introduction:

Beer-lambert law:

$\begin{array}{l}\text{A}\text{=}\text{-}\text{log}\frac{\text{P}}{{\text{P}}_{\text{o}}}\text{=}\text{εcb}\\ \text{where,}\\ \text{A}\text{-}\text{absorbance}\\ {\text{P}}_{\text{o}}\text{-}\text{incident}\text{irradiance}\text{and}\text{P}\text{-}\text{transmitted}\text{irradiance}\\ \text{ε}\text{-}\text{molar}\text{absorptivity}\\ \text{c-}\text{concentration}\text{and}\text{b}\text{-}\text{pathlength}\text{.}\end{array}$

Answer to Problem 18.BE

The concentration of compound A in the concentrated solution is $\text{5}\text{.87}{\text{×10}}^{\text{-3}}\text{M}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Initial}\text{volume}\text{is}\text{2}\text{.00}\text{mL,}\text{Final}\text{volume}\text{is}\text{25}\text{.00}\text{mL}\text{.}\\ \text{Absorbance}\left(\text{A}\right)\text{is}\text{0}\text{.733}\text{at}\text{238}\text{nm}\text{in}\text{1}\text{.000}\text{cm}\text{cuvet}\text{.}\\ \text{Absorbance}\text{for}\text{blank}\text{solution}\text{is}\text{0}\text{.029}\text{.}\\ \text{Molar}\text{absorptivity}\text{is}\text{1}\text{.50}\text{×}{\text{10}}^{\text{3}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\text{.}\end{array}$

Calculate the concentration of compound A in the concentrated solution

$\begin{array}{l}\text{c}\text{=}\left(\text{dilution}\text{factor}\right)\frac{\text{A}}{\text{εb}}\\ \text{=}\left(\frac{\text{25}\text{.00}\text{mL}}{\text{2}\text{.00}\text{mL}}\right)\frac{\text{0}\text{.733}\text{-}\text{0}\text{.029}}{\left(\text{1}\text{.50}\text{×}{\text{10}}^{\text{3}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\right)\left(\text{1}\text{.000}\text{cm}\right)}\\ \text{=}\text{5}\text{.87}\text{×}{\text{10}}^{\text{-3}}\text{M}\text{.}\end{array}$

The concentration of compound A in the concentrated solution is $\text{5}\text{.87}{\text{×10}}^{\text{-3}}\text{M}\text{.}$