The Beer-Lambert law relates the absorbance A of a

Question

Problem 9E

The Beer-Lambert law relates the absorbance $A$ of a solution to the concentration $C$ of a species in solution by A = MLC, where $L$ is the path length and $M$ is the molar absorption coefficient. Assume that $C=1.25\pm0.03\mathrm{\ mol}/\mathrm{cm}^3$, $L=1.2\pm0.1\mathrm{\ cm}$, and $A=1.30 \pm 0.05$.

a. Estimate $M$ and find the uncertainty in the estimate.

b. Which would provide a greater reduction in the uncertainty in $M$: reducing the uncertainty in $C$ to $0.01\mathrm{\ mol}/\mathrm{cm}^3$, reducing the uncertainty in $L$ to 0.05 cm, or reducing the uncertainty in $A$ to 0.01?

Equation Transcription:

$A$

$C$

$L$

$M$

$C=1.52\pm{}0.03\ mol/c{m}^{3}$

$L=1.2\pm{}0.1\ cm$

$A=1.30\pm{}0.05$

$0.01\ mol/c{m}^{3}$

Text Transcription:

A

C

L

M

C = 1.52 pm 0.03 mol/cm^3

L = 1.2 pm 0.1 cm

A = 1.30 pm 0.05

0.01 mol/cm^3

Answer 1

Step-by-Step Solution:

Step 1 of 3

Solution 9E

Step1 of 3:

We have Beer-Lambert law in that A be the absorbance of a solution to the concentration C of a species and Let A = MLC.

Where,

L = Path length.

= 1.2 ± 0.1 cm.

M = the molar absorption coefficient.

C = 1.25 ± 0.03 mol/cm3.

A = 1.30 ± 0.05.

Here our goal is:

a).We need to estimate M and find the uncertainty in the estimate.

b).We need to check Which would provide a greater reduction in the uncertainty in M: reducing the uncertainty in C to 0.01 mol/cm3, reducing the uncertainty in L to 0.05 cm, or reducing the uncertainty in A to 0.01?

Step2 of 3:

a).

We have C = 1.25, ${\sigma{}}_{C}=0.03,$ L = 1.2, ${\sigma{}}_{L}=0.1,$ A = 1.30 and ${\sigma{}}_{A}=0.03$

Let

A = MLC

M = $\frac{A}{LC}$

= $\frac{1.30}{(1.2)(1.25)}$

= $\frac{1.30}{1.5}$

= 0.8667

Hence, M = 0.8667.

1).Consider,

M = $\frac{A}{LC}$

Differentiate above equation with respect to “A” then

$\frac{\partial{}M}{\partial{}A}=\frac{\partial{}}{\partial{}A}(\frac{A}{LC})$

= $\frac{1}{LC}\frac{\partial{}}{\partial{}A}(A)$

= $\frac{1}{LC}$ (1)

= $\frac{1}{(1.2)(1.25)}$

= $\frac{1}{1.5}$

= 0.6667

Hence, $\frac{\partial{}M}{\partial{}A}$ = 0.6667.

2).Consider,

M = $\frac{A}{LC}$

Differentiate above equation with respect to “L” then

$\frac{\partial{}M}{\partial{}L}=\frac{\partial{}}{\partial{}L}(\frac{A}{LC})$

= $\frac{A}{C}\frac{\partial{}}{\partial{}A}(\frac{1}{L})$

= $\frac{A}{C}\frac{\partial{}}{\partial{}A}({L}^{-1})$

= $\frac{A}{C}({-L}^{-1-1})$

= $\frac{-A}{C}$ ( ${L}^{-2}$ )

= $\frac{-A{L}^{-2}}{C}$

= $\frac{-(1.30){(1.2)}^{-2}}{(1.25)}$

= $\frac{-0.90277}{1.25}$

= -0.7222

Hence, $\frac{\partial{}M}{\partial{}L}$ = -0.7222.

3).Consider,

M = $\frac{A}{LC}$

Differentiate above equation with respect to “C” then

$\frac{\partial{}M}{\partial{}C}=\frac{\partial{}}{\partial{}C}(\frac{A}{LC})$

= $\frac{A}{L}\frac{\partial{}}{\partial{}A}(\frac{1}{C})$

= $\frac{A}{L}\frac{\partial{}}{\partial{}A}({C}^{-1})$

= $\frac{A}{L}({-C}^{-1-1})$

= $\frac{-A}{L}$ ( ${C}^{-2}$ )

= $\frac{-A{C}^{-2}}{L}$

= $\frac{-(1.30){(1.25)}^{-2}}{(1.2)}$

= $\frac{-0.832}{1.2}$

= -0.6933

Hence, $\frac{\partial{}M}{\partial{}C}$ = -0.6933.

The estimate of uncertainty is given by

${\sigma{}}_{M}=\sqrt{{(\frac{\partial{}M}{\partial{}A})}^{2}{\sigma{}}_{A}^{2}+{(\frac{\partial{}M}{\partial{}L})}^{2}{\sigma{}}_{L}^{2}+{(\frac{\partial{}M}{\partial{}C})}^{2}{\sigma{}}_{C}^{2}}$

${\sigma{}}_{M}=\sqrt{{(0.6667)}^{2}{(0.03)}_{\ }^{2}+{(-0.7222)}^{2}{(0.1)}_{\ }^{2}+{(-0.6933)}^{2}{(0.03)}_{\ }^{2}}$

$=\sqrt{(0.4445)(0.0009)+(0.5215)(0.01)+(0.4806)(0.0009)}$

= $\sqrt{0.0004+0.005215+0.000432}$

= $\sqrt{0.006047}$

= 0.0777

Hence, ${\sigma{}}_{M}$ = 0.0777.

Therefore, The estimate of uncertainty M = 0.8667 $\pm{}$ 0.0777.

Step3 of 3:

b).

From part (a), we have

${\sigma{}}_{M}=\sqrt{{(\frac{\partial{}M}{\partial{}A})}^{2}{\sigma{}}_{A}^{2}+{(\frac{\partial{}M}{\partial{}L})}^{2}{\sigma{}}_{L}^{2}+{(\frac{\partial{}M}{\partial{}C})}^{2}{\sigma{}}_{C}^{2}}$

Where,

1). $\frac{\partial{}M}{\partial{}A}$ = $\frac{1}{LC}$

= 0.6667

2). $\frac{\partial{}M}{\partial{}L}$ = $\frac{-A{L}^{-2}}{C}$

= -0.7222

3). $\frac{\partial{}M}{\partial{}L}$ = $\frac{-A{C}^{-2}}{L}$

= -0.6933

Here,

If ${\sigma{}}_{C}=0.01,\ {\sigma{}}_{L}=0.1\ and\ \ {\sigma{}}_{A}=0.05$ then ${\sigma{}}_{M}=0.080.$

If ${\sigma{}}_{C}=0.03,\ {\sigma{}}_{L}=0.05\ and\ \ {\sigma{}}_{A}=0.05$ then ${\sigma{}}_{M}=0.053.$

If ${\sigma{}}_{C}=0.03,\ {\sigma{}}_{L}=0.1\ and\ \ {\sigma{}}_{A}=0.01$ then ${\sigma{}}_{M}=0.076.$

As reducing the uncertainty in L to 0.05 cm provides the greatest reduction.

Chapter 3.4, Problem 9E is Solved

Step 2 of 3

Step 3 of 3

Solution for problem 9E Chapter 3.4

The Beer-Lambert law relates the absorbance A of a

Chapter 3.4, Problem 9E is Solved