Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
Question
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Chapter 23, Problem 23.FE

(a)

Interpretation Introduction

Interpretation:

The retention factor of compound 2 has to be calculated.

Concept introduction:

Retention factor:

The retention factor (k) is given by

k=tr-tmtmWhere,k-retentionfactortr-timerequiredtoelutethepeaktm-timerequiredformobilephasetopassthroughthecolumn.

(a)

Expert Solution
Check Mark

Answer to Problem 23.FE

  • The retention factor of compound 2 is 5.511.

Explanation of Solution

Given information,

The value of relative retention (α) is 1.068 .

The value of retention factor for compound 1 (k1) is 5.16 .

Calculate the value of k2 by using the relationship between αandk

α=k2k1k2=αk1=(1.068)(5.16)=5.511

The retention factor of compound 2 is 5.511.

(b)

Interpretation Introduction

Interpretation:

The length (L) of the column has to be calculated.

Concept introduction:

Purnell equation:

The relationship between Handα is given by Purnell equation,

Resolution=N4(α-1)α(k21+k2)Where,N-numberoftheoreticalplatesα-relativeretentionofthetwopeaksk2-retentionfactorofthemoreretainedcomponent

(b)

Expert Solution
Check Mark

Answer to Problem 23.FE

  • The length of the column is 6.45m.

Explanation of Solution

Given information,

The value of α is 1.068 .

The value of k2 is 5.511 .

Height of the plate is 0.520mm.

Calculate the number (N) of plates on column by Purnell equation

Resolution=N4(α-1)α(k21+k2)1.5=N4(1.068-1)1.068(5.5111+5.511)N=4×1.5(0.8464)(0.0637)=60.0539=111.32N=12392=1.24×104plates

Calculate the length (L) of the column

H=LengthofthecolumnNumberofplatesoncolumn=LNL=H×N=0.520mm×1.24×104=6448mm=6.45m.

The length of the column is 6.45m.

(c)

Interpretation Introduction

Interpretation:

The values of trandw1/2 has to be calculated for the given peaks.

Concept introduction:

Plates of column:

The relationship between Nandw1/2

N=5.55tr2w1/22Where,N-numberofplatesoncolumntr-retentiontimew1/2-half-width

(c)

Expert Solution
Check Mark

Answer to Problem 23.FE

  • The value of tr for the compound 1 is 12.32min.
  • The value of tr for the compound 2 is 13.02min.
  • The value of w1/2 for the compound 1 is 0.261min.
  • The value of w1/2 for the compound 2 is 0.275min.

Explanation of Solution

Given information,

The value of α is 1.068 .

The value of k1 is 5.16 .

The value of k2 is 5.511 .

The value of N is 1.24×104 .

Calculate the value of tr for the compound 1 by using the relation between retention factor and retention time

k1=(t1-tm)tm=t1tm-1t1=tm(k1+1)=tm(6.16)=(2.00min)(6.16)=12.32min

The value of tr for the compound 1 is 12.32min.

Calculate the value of tr for the compound 2 by using the relation between retention factor and retention time

k2=(t2-tm)tm=t2tm-1t2=tm(k2+1)=tm(6.511)=(2.00min)(6.511)=13.02min

The value of tr for the compound 2 is 13.02min.

Calculate the value of w1/2 for the compound 1 by using the relationship between Nandw1/2

N=5.55tr2w1/22w1/2=5.55N×tr=5.551.24×104×(12.32min)=0.261min.

The value of w1/2 for the compound 1 is 0.261min.

Calculate the value of w1/2 for the compound 2 by using the relationship between Nandw1/2

N=5.55tr2w1/22w1/2=5.55N×tr=5.551.24×104×(13.02min)=0.275min.

The value of w1/2 for the compound 2 is 0.275min.

(d)

Interpretation Introduction

Interpretation:

The partition coefficient value for compound 1 has to be calculated.

Concept introduction:

Partition coefficient:

The relationship between retention factor and partition coefficient is given by

k=KVsVmWhere,k-retentionfactorK-partitioncoefficientVs-volumeofstationaryphaseVm-volumeofmobilephase

(d)

Expert Solution
Check Mark

Answer to Problem 23.FE

  • The partition coefficient value for compound 1 is 17.2.

Explanation of Solution

Given information,

The value of Vs/Vm is 0.30 .

The value of k1 is 5.16 .

Calculate the value of K1 for the compound 1 by using the relation between retention factor and partition coefficient

K1=k1VmVs=(5.16)(1/0.30)=17.2

The partition coefficient value for compound 1 is 17.2.

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