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Chapter 22: Problem 25

The retention volume of a solute is \(76.2 \mathrm{~mL}\) for a column with \(V_{\mathrm{m}}=16.6 \mathrm{~mL}\) and \(V_{\mathrm{s}}=12.7 \mathrm{~mL}\). Calculate the retention factor and the partition coefficient for this solute.

Short Answer

Expert verified
Retention factor is approximately 3.59, and the partition coefficient is approximately 4.69.

Step by step solution

01

Understand the Problem

We are given the retention volume of a solute, the mobile phase volume, and the stationary phase volume for a column. We need to find the retention factor (k') and the partition coefficient (K).
02

Retention Factor Formula

The retention factor (k') is calculated using the formula: \[ k' = \frac{V_r - V_m}{V_m} \]where \(V_r\) is the retention volume, and \(V_m\) is the volume of the mobile phase.
03

Calculate the Retention Factor

Substitute the given values into the formula for the retention factor:\[ V_r = 76.2 \text{ mL}, \quad V_m = 16.6 \text{ mL} \]\[ k' = \frac{76.2 - 16.6}{16.6} = \frac{59.6}{16.6} \]Calculate to get:\[ k' \approx 3.59 \]
04

Partition Coefficient Formula

The partition coefficient (K) is given by the formula:\[ K = \frac{V_r - V_m}{V_s} \]where \(V_s\) is the volume of the stationary phase.
05

Calculate the Partition Coefficient

Substitute the given values into the formula for the partition coefficient:\[ V_s = 12.7 \text{ mL} \]\[ K = \frac{76.2 - 16.6}{12.7} = \frac{59.6}{12.7} \]Calculate to get:\[ K \approx 4.69 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partition Coefficient
The partition coefficient, often symbolized as "K," is a fundamental concept in chromatography. It represents the ratio of a solute's concentration in the stationary phase to its concentration in the mobile phase. This concept helps us understand how a compound distributes itself between the two phases. The formula for the partition coefficient is:\[ K = \frac{(V_r - V_m)}{V_s} \]Where:
  • \(V_r\) is the retention volume
  • \(V_m\) is the mobile phase volume
  • \(V_s\) is the stationary phase volume
Essentially, "K" gives an idea of how much a compound prefers the stationary phase over the mobile phase. A higher "K" value indicates that the compound spends more time in the stationary phase.
Chromatography
Chromatography is a technique used to separate mixtures. It relies on the different affinities substances have for the stationary and mobile phases. This process can separate complex mixtures, allowing each component to be analyzed or used independently. The technique works by allowing a mixture to pass through two phases:
  • **Mobile phase**: This is the liquid or gas that carries the mixture through the stationary phase.
  • **Stationary phase**: This is a solid or liquid stationary in the system that the mobile phase flows through.
Chromatography is widely used in labs for purifying substances, testing sample purity, and analyzing multiple components in a mixture. It plays a significant role in the pharmaceutical industry and environmental testing.
Retention Volume
Retention volume, denoted as \( V_r \), is the volume of the mobile phase required to elute a solute from the column. It describes how long a solute interacts with the stationary phase before being washed out.The equation can be expressed as follows:\[ V_r = V_m + (k' \times V_m) \]Where:
  • \( V_m \) is the mobile phase volume.
  • \( k' \) is the retention factor.
Retention volume is an important parameter since it provides insight into how a solute behaves within the chromatographic system. A larger retention volume means the solute interacts more with the stationary phase, which might indicate stronger binding or interaction.
Mobile Phase
The mobile phase in chromatography, which can be a liquid or gas, carries the sample mixture through the stationary phase. Its primary role is to transport solutes along the column. The choice of the mobile phase directly impacts the efficiency and resolution of the separation. Key factors about the mobile phase include:
  • **Polarity**: The mobile phase's polarity can affect how strongly compounds interact with the stationary phase.
  • **Flow rate**: The speed at which the mobile phase moves influences how quickly the separation process occurs.
  • **Viscosity**: Mobile phase viscosity affects the pressure needed to push it through the system.
Selecting the right mobile phase can enhance separation quality, making it a crucial consideration in chromatographic experiments.

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Most popular questions from this chapter

An open tubular column has an inner diameter of \(207 \mu \mathrm{m}\) and the thickness of the stationary phase on the inner wall is \(0.50 \mu \mathrm{m}\). Unretained solute passes through in \(63 \mathrm{~s}\) and a particular solute emerges in \(433 \mathrm{~s}\). Find the partition coefficient for this solute and find the fraction of time spent in the stationary phase.

A chromatographic band has a width, \(w\), of \(4.0 \mathrm{~mL}\) and a retention volume of \(49 \mathrm{~mL}\). What width is expected for a band with a retention volume of \(127 \mathrm{~mL}\) ? Assume that the only band spreading occurs on the column itself.

(ai) Write the meaning of the retention factor, \(k\), in terms of time spent by solute in each phase. (b) Write an expression in terms of \(k\) for the fraction of time spent by a solute molecule in the mobile phase. (c) The retention ratio in chromatography is defincd as $$ R=\frac{\text { time for solvent to pass through column }}{\text { time for solute to pass through column }}=\frac{t_{\mathrm{m}}}{t_{\mathrm{r}}} $$ Show that \(R\) is related to the retention factor by the cquation \(R=1 /(k+1)\).

Solvent passes through a column in \(3.0 \mathrm{~min}\) but solute requires \(9.0 \mathrm{~min}\). (a) Calculate the retention factor, \(k\). (b) What fraction of time is the solute in the mobile phase in the column? (c) The volume of stationary phase is \(1 / 10\) of the volume of the mobile phase in the column \(\left(V_{\mathrm{s}}=0.10 V_{\mathrm{m}}\right)\). Find the partition coefficient, \(K\), for this system.

1 tut The theoretical limit for extracting solute \(S\) from phase 1 (volume \(V_{1}\) ) into phase 2 (volume \(V_{2}\) ) is attained by dividing \(V_{2}\) into an infinite number of infinitesimally small portions and conducting an infinite number of extractions. With a partition coefficient \(K=[S]_{2} /[S]_{1}\), the limiting fraction of solute remaining in phase 1 is \(^{2} \varphi_{\text {imit }}=\mathrm{e}^{-\left(V_{2} / V_{1}\right) K}\). Let \(V_{1}=V_{2}=50 \mathrm{~mL}\) and let \(K=2\). Let volume \(V_{2}\) be divided into \(n\) equal portions to conduct \(n\) extractions. Find the fraction of \(\mathrm{S}\) extracted into phase 2 for \(n=1,2,10\) extractions. How theoretical limit? A Plumber's View of Chromatography
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