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Chapter 25: Q35P (page 710)

The 鈥渞ule of three鈥� states that the retention factor for a given solute decreases approximately threefold when the organic phase increases by10%.In figure 25-12,tm=2.7min. Find k for peak 5at 40%B .predict the retention time for peak at B and compare the observed and predicted times.

Short Answer

Expert verified

The rule of three states that the retention factor for a given solute decreases.

K=tr-tmtmk=11-27min2.7mintr=27.81min

Step by step solution

01

Definition of retention

Retention implies the maintenance, possession, control, or retention of moisture.

According to the figure 25-12, we see that retention time for peak 5 is 11 min (for 50% B). So, the retention factor is

K=tr-tmtmk=11-27min2.7mink=3.1

02

Rule of three

Then, when B is 40%, k is (according to the rule of three) given by

k=3鈰�3.1=9.3

Retention time is

tr=k.tm+tmtr=9.3.2.7min+2.7mintr=27.81min

According to the figure 25-12, the observed retention time is 20.2 min.

Hencetr=27.81min.

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Most popular questions from this chapter

Two peaks emerge from a reversed-phase chromatography column as sketched in the illustration.

According to Equation 23-33, resolution is given by

Resolution=N4(伪-1)伪(k21+k2)

where Nis plate number, 伪is relative retention (Equation23-20), and k2 is the retention factor for the more retained component (Equation 23-16).

(a) If you decrease the amount of organic solvent in the mobile phase, you will increase retention. Sketch the chromatogram if retention factors increase but Nand 伪are constant.

(b) If you change the solvent type or the stationary phase, you will change the relative retention. Sketch the chromatogram if伪increases but Nandk1are constant.

(c) If you decrease particle size or increase column length, you can increase the plate number. Sketch the chromatogram if Nincreases by (i) decreasing particle size and (ii) increasing column length. Assume 伪and k2are constant.

A reversed-phase separation of a reaction mixture calls for isocratic elution with 48%methanol 52%water. If you want to change the procedure to use acetonitrile/water, what is a good starting percentage of acetonitrile to try?

HPLC peak should generally not have an asymmetry factor, B/A in figure 23-14,outside the range0.9-1.5

  1. Sketch the shape of a peak with an asymmetry of 1.8
  2. What might you do to correct the asymmetry?

(a) Nonpolar aromatic compounds were separated by HPLC on an octadecyl(C18)bonded phase. The eluent was 65 vol% methanol in water. How would the retention times be affected if 90% methanol were used instead?

(b) Octanoic acid and 1-aminooctane were passed through the same column described in (a), using an eluent of 20% methanol/80% buffer (pH 3.0). State which compound is expected to be eluted first and why.

role="math" localid="1656416023291" CH3CH2CH2CH2CH2CH2CH2CO2HOctanoicacidCH3CH2CH2CH2CH2CH2CH2CH2NH21鈥�Aminooctane

(c) Polar solutes were separated by hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase. How would retention times be affected if eluent were changed from 80 vol% to 90 vol% acetonitrile in water?

(d) Polar solutes were separated by normal-phase chromatographyon bare silica using methyl t-butyl ether and 2-propanol solvent. How would retention times be affected if eluent were changed from 40 vol% to 60 vol% 2-propanol? (Hint: See Table 25-4.)

If along 15cmHPCL column has a place height of 5.0 what will be the half-width (in seconds) of a peak eluted at 10.0min? if plate height5渭尘,what will bew1/2?
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