Chapter 19, Problem 19.11P

Interpretation Introduction

Interpretation:

Values of $\left[{\text{MnO}}_4^{-}\right]$ and $\left[{\text{Cr}}_2{\text{O}}_7^{2-}\right]$ have to be calculated.

Concept Introduction:

Molar absorptivity is a coefficient that appears in Beers’ law. It measures the extent of light absorbed at a certain wavelength. Molar absorptivity is denoted by the symbol $\epsilon$. SI unit for Molar absorptivity is ${\text{M}}^{-1}{\text{cm}}^{-1}$.

Relation between path length and dimension is as follows:

  $D=b\left(\text{ε}{\text{'}}_{\text{X}}\text{ε}{"}_{\text{Y}}-\text{ε}{\text{'}}_{\text{Y}}\text{ε}{"}_{\text{X}}\right)$

Here,

$\epsilon$ denotes molar absorptivity in ${\text{M}}^{-1}{\text{cm}}^{-1}$.

$b$ denotes path length in centimeters.

$D$ denotes dimension.

Expression to compute absorbance from transmittance is as follows:

  $A=-\log T$

Formula to calculate molar absorptivity from absorbance is as follows:

  $\text{ε}=\frac{A}{b\cdot c\cdot l}$

Here,

$\epsilon$ denotes molar absorptivity of species capable to absorb.

$b$ denotes path length in centimeters.

$c$ denotes concentration of species capable to absorb.

$A$ denotes absorbance.

$T$ denotes transmittance.

Explanation of Solution

Relation between path length and dimension is as follows:

  $D=b\left(\text{ε}{\text{'}}_{\text{X}}\text{ε}{"}_{\text{Y}}-\text{ε}{\text{'}}_{\text{Y}}\text{ε}{"}_{\text{X}}\right)$        (1)

Substitute $420{\text{ M}}^{-1}{\text{cm}}^{-1}$ for $\text{ε}{\text{'}}_{\text{X}}$, $1580{\text{ M}}^{-1}{\text{cm}}^{-1}$ for $\text{ε}{"}_{\text{Y}}$, ${\text{1680 M}}^{-1}{\text{cm}}^{-1}$ for ${\text{ε'}}_{\text{Y}}$, ${\text{4010 M}}^{-1}{\text{cm}}^{-1}$ for $\text{ε}{"}_{\text{X}}$, and $\text{1}\text{.00 cm}$ for $b$ in equation (1).

  $\begin{array}{c}D=\left(1\text{ cm}\right)\left[\left(420\right)\left(1580\right)-\left(\text{1680}\right)\left(\text{4100}\right)\right]{\text{ M}}^{-1}{\text{cm}}^{-1}\\ =-6224400{\text{ M}}^{-1}\end{array}$

Expression to compute $\left[{\text{MnO}}_4^{-}\right]$ is as follows:

  $\left[{\text{MnO}}_4^{-}\right]=\frac{A\text{'}\text{ε}{"}_{{\text{MnO}}_4^{-}}-A"{\text{ε'}}_{{\text{MnO}}_4^{-}}}{D}$        (2)

Substitute 0.766 for $A\text{'}$, 0.422 for $A"$, ${\text{1580 M}}^{-1}{\text{cm}}^{-1}$ for $\text{ε}{"}_{{\text{MnO}}_4^{-}}$, ${\text{4100 M}}^{-1}{\text{cm}}^{-1}$ for ${\text{ε'}}_{{\text{MnO}}_4^{-}}$, and $-6224400{\text{ M}}^{-1}$ for $D$ in equation (2).

  $\begin{array}{c}\left[{\text{MnO}}_4^{-}\right]=\frac{\left(0.766\right)\left({\text{1580 M}}^{-1}{\text{cm}}^{-1}\right)-\left(0.422\right)\left({\text{4100 M}}^{-1}{\text{cm}}^{-1}\right)}{\left(-6224400{\text{ M}}^{-1}\right)}\\ =8.352\times {10}^{-5}\text{ M}\end{array}$

Expression to compute $\left[{\text{Cr}}_2{\text{O}}_7^{2-}\right]$ is as follows:

  $\left[{\text{Cr}}_2{\text{O}}_7^{2-}\right]=\frac{A"{\text{ε'}}_{{\text{Cr}}_2{\text{O}}_7^{2-}}-A\text{'}\text{ε}{"}_{{\text{Cr}}_2{\text{O}}_7^{2-}}}{D}$        (3)

Substitute 0.422 for $A\text{'}$, 0.766 for $A"$, ${\text{420 M}}^{-1}{\text{cm}}^{-1}$ for $\text{ε}{\text{'}}_{\text{X}}$, ${\text{1680 M}}^{-1}{\text{cm}}^{-1}$ for $\text{ε}{"}_{\text{X}}$,and $-6224400{\text{ M}}^{-1}$ for $D$ in equation (3).

  $\begin{array}{c}\left[{\text{Cr}}_2{\text{O}}_7^{2-}\right]=\frac{\left(0.442\right)\left({\text{420 M}}^{-1}{\text{cm}}^{-1}\right)-\left(0.766\right)\left({\text{1680 M}}^{-1}{\text{cm}}^{-1}\right)}{-6224400{\text{ M}}^{-1}}\\ =1.78\times {10}^{-4}\text{ M}\end{array}$

Thus, value of $\left[{\text{MnO}}_4^{-}\right]$ and $\left[{\text{Cr}}_2{\text{O}}_7^{2-}\right]$ are $8.352\times {10}^{-5}\text{ M}$ and $1.78\times {10}^{-4}\text{ M}$ respectively.