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Chapter 22: Problem 41

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.

Short Answer

Expert verified
The expected width of the band is approximately 6.44 mL.

Step by step solution

01

Understanding the Problem

We are given the width of a chromatographic band as 4.0 mL corresponding to a retention volume of 49 mL. We need to find the expected width ("w") of a band with a retention volume of 127 mL, assuming band spreading is proportional and occurs only in the column.
02

Proportionality Concept

Assume that band spreading is proportional to the square root of the retention volume: \( w_1^2 \propto V_1 \), \( w_2^2 \propto V_2 \). Therefore, \( \frac{w_1^2}{V_1} = \frac{w_2^2}{V_2} \).
03

Setting Up the Proportionality Equation

Using the proportionality relation, express widths in terms of retention volumes:\[ \frac{w_1^2}{49} = \frac{w_2^2}{127} \] where \( w_1 = 4.0 \) mL.
04

Solving for the New Width

Substitute \( w_1 \) into the equation:\[ \frac{4.0^2}{49} = \frac{w_2^2}{127} \] \[ \frac{16}{49} = \frac{w_2^2}{127} \] Multiply both sides by 127 to solve for \( w_2^2 \):\[ w_2^2 = \frac{16}{49} \times 127 \]
05

Calculating the New Width

Perform the multiplication and division:\[ w_2^2 = \frac{2032}{49} = 41.47 \]Take the square root to find \( w_2 \):\[ w_2 = \sqrt{41.47} \approx 6.44 \text{ mL} \]

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Key Concepts

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

Band Spreading
Chromatography is a powerful analytical tool, but one of its challenges is band spreading. Band spreading occurs as solute molecules travel through the column, leading to a broader peak on the chromatogram. This widening can affect the accuracy and resolution of the analysis.
Key factors influencing band spreading include:
  • The type of stationary phase used
  • The flow rate of the mobile phase
  • The temperature of the environment
In our exercise, the focus is specifically on band spreading occurring within the column. It is assumed to be the only factor affecting the broadening of the chromatographic band. As such, this spreading is predictable and can be accounted for, allowing us to anticipate how broad a peak will become as it travels down the column. Understanding this behavior helps us adjust our experimental design to achieve better separation and results.
Retention Volume
Retention volume ( V ) is a critical parameter in chromatography. It represents the volume of mobile phase required to elute a particular solute from the column. Factors that can affect retention volume include the interaction between the solute and the stationary phase, as well as the flow rate and solvent polarity.
Retention volume is used in the solution of our exercise to predict the width of a chromatographic band. By knowing the retention volume of a solute, we can anticipate its elution time and plan the analysis to minimize overlap with other peaks. This is crucial for obtaining accurate measurements in complex mixtures. Simply put, a greater retention volume means more time and volume are needed for the solute to exit the column, which can be linked through mathematical models to how broad the peak will be at the detector.
Proportionality
The concept of proportionality is essential in understanding the relationship between band width and retention volume in chromatography. In our problem, we assume that band spreading is proportional to the retention volume. More specifically, the square of the band width ( w ) is proportional to the retention volume ( V ).
This can be expressed mathematically as:
  • For an initial band with retention volume V1 and width w1 : \( w1^2 \propto V1 \)
  • For a second band with retention volume V2 and width w2 : \( w2^2 \propto V2 \)
By equating these proportions, we have:\[ \frac{w1^2}{V1} = \frac{w2^2}{V2} \]This equation helps us solve for unknowns by maintaining the ratio of w^2 and V . It is a straightforward but powerful tool in scenarios where linearity might not apply. In practice, using proportionality allows researchers to predict and adjust chromatographic conditions, ensuring that separations are efficient and peaks remain distinct from one another.

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Most popular questions from this chapter

Solvent passes through a column in \(3.0\) 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 \(\left(V_{x}=0.10 V_{m}\right)\). Find the partition coefficient, \(K\), for this system.

A \(0.25-\mathrm{mm}\)-diameter open tubular gas chromatography column is coated with stationary phase that is \(0.25 \mu \mathrm{m}\) thick. The diffusion coefficient for a compound with a retention factor \(k=10\) is \(D_{\mathrm{m}}=1.0 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) in the gas phase and \(D_{\mathrm{s}}=1.0 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\) in the stationary phase. Consider longitudinal diffusion and finite equilibration time in the mobile and stationary phases as sources of broadening. Prepare a graph showing the plate height from each of these three sources and the total plate height as a function of linear flow rate (from \(2 \mathrm{~cm} / \mathrm{s}\) to \(1 \mathrm{~m} / \mathrm{s}\) ). Then change the stationary phase thickness to \(2 \mu \mathrm{m}\) and repeat the calculations. Explain the difference in the two results.

A separation of \(2.5 \mathrm{mg}\) of an unknown mixture has been optimized on a column of length \(L\) and diameter \(d\). Explain why you might not achieve the same resolution for \(5.0 \mathrm{mg}\) on a column of length \(2 L\) and diameter \(d\).

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

For the extraction of \(\mathrm{Cu}^{2+}\) by dithizone in \(\mathrm{CCl}_{4}, K_{\mathrm{L}}=1.1 \times 10^{4}\), \(K_{\mathrm{M}}=7 \times 10^{4}, K_{\mathrm{a}}=3 \times 10^{-5}, \beta=5 \times 10^{22}\), and \(n=2\). (a) Calculate the distribution coefficient for extraction of \(0.1 \mu \mathrm{M}\) \(\mathrm{Cu}^{2+}\) into \(\mathrm{CCl}_{4}\) by \(0.1 \mathrm{mM}\) dithizone at \(\mathrm{pH} 1.0\) and at \(\mathrm{pH} 4.0\). (b) If \(100 \mathrm{~mL}\) of \(0.1 \mu \mathrm{M}\) aqueous \(\mathrm{Cu}^{2+}\) are extracted once with \(10 \mathrm{~mL}\) of \(0.1 \mathrm{mM}\) dithizone at \(\mathrm{pH} 1.0\), what fraction of \(\mathrm{Cu}^{2+}\) remains in the aqueous phase?
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