Principles of Instrumental Analysis
Principles of Instrumental Analysis
7th Edition
ISBN: 9781305577213
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cengage Learning
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Chapter 13, Problem 13.12QAP

The equilibrium constant for the reaction 2 CrO 4 2 + 2 H + Cr 2 O 7 2 + H 2 O is 4.2 × 1014. The molar absorptivities for the two principal species in a solution of K2 Cr2 O2 are

Chapter 13, Problem 13.12QAP, The equilibrium constant for the reaction 2CrO42+2H+Cr2O72+H2O is 4.2  1014. The molar

Four solutions were preparedby dissolving 4.00 × 10-4, 3.00 × 10-4, 2.00 × 10-4,and 1.00 × 10-4 moles of K2 Cr2 O7 in water and diluting to 1.00 L with a pH 5.60 buffer. Derive theoretical absorbance values (1.00-cm cells) for each solution and plot the data for (a) 345 nm, (b) 370 nm, and (c) 400 nm.

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The theoretical absorbance value for 345 nm should be derived and the data should be plotted.

Concept introduction:

The relationship between absorbance and concentration of absorbance is linear. If the incident light Io falls on the sample solutions then only I amount of light gets transmitted through the solution. And the relationship is defined as below:

Absorbance(A)=IoI

Transmission(T)=IIo

In the case of no absorbing sample, all the light gets passed and the value of T=100%. But in the case of absorbing sample, the absorbance is given by A=log(1T).

Explanation of Solution

Given:

The equilibrium constant is 4.2×1014 and the pH is 5.6. λ=345,ε1(CrO42)=1.84×103,ε2(CrO72)=10.7×102.

The given reaction is 2CrO42+2H+Cr2O72+H2O

The equilibrium constant for the given reaction is 4.2×1014 and the pH is 5.6.

The formula to determine pH is: pH=log[H+].

Therefore

5.6=log[H+][H+]=2.51×106M

For the given reaction the expression for the equilibrium constant can be written as

K=[Cr2O72][Cr2O42]2[H+]2

The equilibrium concentration of dichromate is 4.00×104. Now, substitute the values and find out the equilibrium concentration of Cr2O42 at the given pH.

4.2×1014=[4.00×104][CrO42]2[2.51×106]2[CrO42]2=[4.00×104][4.2×1014][2.51×106]2[CrO42]2=3.88×104

The theoretical absorbance value of the first solution can be calculated as below

A1 = [(1.84×103 M1cm1)×(3.88×104 M) + (10.7×102 M1cm1)×(4.00×104 M)]×(1.00 cm)]A1 = 1.14192  1.14

For the second solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 3.00×104.

So, the new required concentration will be as

4.2×1014=[3.00×104][CrO42]2[2.51×106]2[CrO42]2=[3.00×104][4.2×1014][2.51×106]2[CrO42]2=2.91×104

The theoretical absorbance value of the first solution can be calculated as below

A1 = [(1.84×103 M1cm1)×(2.91×104 M) + (10.7×102 M1cm1)×(3.00×104 M)]×(1.00 cm)]A1 = 0.8564  0.85

For the third solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 2.00×104.

So, the new required concentration will be as

4.2×1014=[2.00×104][CrO42]2[2.51×106]2[CrO42]2=[2.00×104][4.2×1014][2.51×106]2[CrO42]2=1.94×104

The theoretical absorbance value of the first solution can be calculated as below

A1 = [(1.84×103 M1cm1)×(1.94×104 M) + (10.7×102 M1cm1)×(2.00×104 M)]×(1.00 cm)]A1   0.57

For the fourth solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 1.00×104.

So, the new required concentration will be as

4.2×1014=[1.00×104][CrO42]2[2.51×106]2[CrO42]2=[1.00×104][4.2×1014][2.51×106]2[CrO42]2=0.97×104

The theoretical absorbance value of the first solution can be calculated as below:

A1 = [(1.84×103 M1cm1)×(0.97×104 M) + (10.7×102 M1cm1)×(1.00×104 M)]×(1.00 cm)]A1   0.28

Now take the values in excel and plot to get the graph:

Principles of Instrumental Analysis, Chapter 13, Problem 13.12QAP , additional homework tip  1

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The theoretical absorbance value for 370 nm should be derived and the data should be plotted.

Concept introduction:

The relationship between absorbance and concentration of absorbance is linear. If the incident light Io falls on the sample solutions then only I amount of light gets transmitted through the solution. And the relationship is defined as below:

Absorbance(A)=IoI

Transmission(T)=IIo

In the case of no absorbing sample, all the light gets passed and the value of T=100%. But in the case of absorbing sample, the absorbance is given by A=log(1T).

Explanation of Solution

Given:

The equilibrium constant is 4.2×1014 and the pH is 5.6. λ=370,ε1(CrO42)=4.81×103,ε2(CrO72)=7.28×102.

The given reaction is 2CrO42+2H+Cr2O72+H2O

The equilibrium constant for the given reaction is 4.2×1014 and the pH is 5.6.

The formula to determine pH is: pH=log[H+].

Therefore

5.6=log[H+][H+]=2.51×106M

For the given reaction the expression for the equilibrium constant can be written as

K=[Cr2O72][Cr2O42]2[H+]2

The equilibrium concentration of dichromate is 4.00×104. Now substitute the values and find out the equilibrium concentration of Cr2O42 at the given pH.

4.2×1014=[4.00×104][CrO42]2[2.51×106]2[CrO42]2=[4.00×104][4.2×1014][2.51×106]2[CrO42]2=3.88×104

The theoretical absorbance value of the first solution can be calculated as below

A1 = [(4.81×103 M1cm1)×(3.88×104 M) + (7.28×102 M1cm1)×(4.00×104 M)]×(1.00 cm)]A1   1.93

For the second solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 3.00×104.

So, the new required concentration will be as

4.2×1014=[3.00×104][CrO42]2[2.51×106]2[CrO42]2=[3.00×104][4.2×1014][2.51×106]2[CrO42]2=2.91×104

The theoretical absorbance value of the first solution can be calculated as below

A1 = [(4.81×103 M1cm1)×(2.91×104 M) + (7.28×102 M1cm1)×(4.00×104 M)]×(1.00 cm)]A1   1.47

For the third solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 2.00×104.

So, the new required concentration will be as

4.2×1014=[2.00×104][CrO42]2[2.51×106]2[CrO42]2=[2.00×104][4.2×1014][2.51×106]2[CrO42]2=1.94×104

The theoretical absorbance value of the first solution can be calculated as below

A1 = [(4.81×103 M1cm1)×(1.94×104 M) + (7.28×102 M1cm1)×(4.00×104 M)]×(1.00 cm)]A1   1.00

For the fourth solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 1.00×104.

So, the new required concentration will be as

4.2×1014=[1.00×104][CrO42]2[2.51×106]2[CrO42]2=[1.00×104][4.2×1014][2.51×106]2[CrO42]2=0.97×104

The theoretical absorbance value of the first solution can be calculated as below

A1 = [(4.81×103 M1cm1)×(0.97×104 M) + (7.28×102 M1cm1)×(4.00×104 M)]×(1.00 cm)]A1   0.54

Now take the values in excel and plot to get the graph:

Principles of Instrumental Analysis, Chapter 13, Problem 13.12QAP , additional homework tip  2

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The theoretical absorbance value for 400 nm should be derived and the data should be plotted.

Concept introduction:

The relationship between absorbance and concentration of absorbance is linear. If the incident light Io falls on the sample solutions then only I amount of light gets transmitted through the solution. And the relationship is defined as below:

Absorbance(A)=IoI

Transmission(T)=IIo

In the case of no absorbing sample, all the light gets passed and the value of T=100%. But in the case of absorbing sample, the absorbance is given by A=log(1T).

Explanation of Solution

Given:

The equilibrium constant is 4.2×1014 and the pH is 5.6. λ=400,ε1(CrO42)=1.88×103,ε2(CrO72)=1.89×102.

The given reaction is 2CrO42+2H+Cr2O72+H2O

The equilibrium constant for the given reaction is 4.2×1014 and the pH is 5.6.

The formula to determine pH is: pH=log[H+].

Therefore

5.6=log[H+][H+]=2.51×106M

For the given reaction the expression for the equilibrium constant can be written as

K=[Cr2O72][Cr2O42]2[H+]2

The equilibrium concentration of dichromate is 4.00×104. Now substitute the values and find out the equilibrium concentration of Cr2O42 at the given pH.

4.2×1014=[4.00×104][CrO42]2[2.51×106]2[CrO42]2=[4.00×104][4.2×1014][2.51×106]2[CrO42]2=3.88×104

The theoretical absorbance value of the first solution can be calculated as below

A1 = [(1.88×103 M1cm1)×(3.88×104 M) + (1.89×102 M1cm1)×(4.00×104 M)]×(1.00 cm)]A1   0.74

For the second solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 3.00×104.

So, the new required concentration will be as

4.2×1014=[3.00×104][CrO42]2[2.51×106]2[CrO42]2=[3.00×104][4.2×1014][2.51×106]2[CrO42]2=2.91×104

The theoretical absorbance value of the first solution can be calculated as below

A1 = [(1.88×103 M1cm1)×(2.91×104 M) + (1.89×102 M1cm1)×(4.00×104 M)]×(1.00 cm)]A1   0.57

For the third solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 2.00×104.

So, the new required concentration will be as

4.2×1014=[2.00×104][CrO42]2[2.51×106]2[CrO42]2=[2.00×104][4.2×1014][2.51×106]2[CrO42]2=1.94×104

The theoretical absorbance value of the first solution can be calculated as below

A1 = [(1.88×103 M1cm1)×(1.94×104 M) + (1.89×102 M1cm1)×(4.00×104 M)]×(1.00 cm)]A1   0.38

For the fourth solution, everything will remain the same, just the equilibrium concentration of dichromate will change to 1.00×104.

So, the new required concentration will be as

4.2×1014=[1.00×104][CrO42]2[2.51×106]2[CrO42]2=[1.00×104][4.2×1014][2.51×106]2[CrO42]2=0.97×104

Theoretical absorbance value of first solution can be calculated as below

A1 = [(1.88×103 M1cm1)×(0.97×104 M) + (1.89×102 M1cm1)×(4.00×104 M)]×(1.00 cm)]A1   0.20

Now take the values in excel and plot to get the graph:

Principles of Instrumental Analysis, Chapter 13, Problem 13.12QAP , additional homework tip  3

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