Chapter 18, Problem 18.AE

(a)

Interpretation Introduction

Interpretation:

For the given transmittance value, the value of absorbance 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.AE

The value of absorbance $\left(\text{A}\right)$ is $\text{0}\text{.347}\text{.}$

Explanation of Solution

Given information:

$\text{Transmittance}\text{=}\text{45}\text{.0}\text{\%}\text{.}$

Apply the Beer’s law to calculate the value of absorbance $\left(\text{A}\right)$

$\begin{array}{l}\text{A}\text{=}\text{-}\text{log}\frac{\text{P}}{{\text{P}}_{\text{o}}}\text{=}\text{-}\text{log}\text{T}\\ \text{=}\text{-}\text{log}\left(\text{0}\text{.45}\right)\\ \text{=}\text{0}\text{.347}\end{array}$

The value of absorbance $\left(\text{A}\right)$ is $\text{0}\text{.347}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The value of percent transmittance for $\text{0}\text{.0200}\text{M}$ 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.AE

The value of percent transmittance for $\text{0}\text{.0200}\text{M}$ solution is $\text{20}\text{.2}\text{\%}\text{.}$

Explanation of Solution

Given information:

$\begin{array}{l}\text{0}\text{.0100}\text{M}\text{solution}\text{exihibits}\text{45}\text{.0}\text{\%}\text{T}\text{.}\\ \text{Absorbance}\left(\text{A}\right)\text{=}\text{0}\text{.347}\\ \text{For}\text{0}\text{.0200}\text{M}\text{solution}\text{\%}\text{T}\text{=}\text{?}\end{array}$

Absorbance $\left(\text{A}\right)$ is proportional to concentration $\left(\text{C}\right)$, here concentration is doubled so the value of $\text{A}$ is doubled.

$\begin{array}{l}\text{A=}\text{-}\text{log}\text{T}\\ \text{T}\text{=}{\text{10}}^{\text{-A}}\text{=}{\text{10}}^{\text{-0}\text{.694}}\\ \text{=}\text{0}\text{.202}\\ \text{\%T}\text{=}\text{20}\text{.2}\text{\%}\text{.}\end{array}$

The value of percent transmittance for $\text{0}\text{.0200}\text{M}$ solution is $\text{20}\text{.2}\text{\%}\text{.}$