Exploring Chemical Analysis
Exploring Chemical Analysis
5th Edition
ISBN: 9781429275033
Author: Daniel C. Harris
Publisher: Macmillan Higher Education
Question
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Chapter 22, Problem 22.5P

(a)

Interpretation Introduction

Interpretation:

Number of plates and plate height and resolution has to be calculated.

Concept Introduction:

Expression to compute number of theoretical plate is as follows:

  N=5.55 tr2w1/22

Here,

tr denotes analyte’s  retention time.

w1/2 denotes peak width at half-height.

N denotes number of theoretical plate.

It has been found that greater the equilibration steps narrower will be band so this suggests N should be high for more resolution or narrower separation.

Expression to compute plate height is as follows:

  H=LN

Here,

N denotes number of theoretical plate.

L denotes length of chromatographic column.

H denotes plate height.

(a)

Expert Solution
Check Mark

Explanation of Solution

Expression to compute number of theoretical plate is as follows:

  N=5.55 tr2w1/22        (1)

Substitute 50 min for tr and 0.263 mm  for w1/2 in equation (1).

  N=5.55 (50 min )2(0.263 mm )2=2×105

Thus, number of theoretical plate is 2×105.

Expression to compute H from number of plates from is as follows:

  H=LN        (2)

Substitute 100 m for L and 2×105 for N in equation (2).

`

  H=(100 m2×105)(1000 mm1 m)=0.50 mm

Thus, plate height is 0.50 mm.

Value of Δtr is calculated as follows:

  Δtr=63.5 min63 min=0.5 min

Expression to compute resolution is as follows:

  Resolution=Δtrwavg        (3)

Substitute 0.5 min for Δtr, 0.384 min for wavg in equation (3).

  Resolution=0.5 min0.384min=1.3

Thus, value of resolution is 1.3.

Ratio of plate height is calculated as follows:

  Plate height in liquid chromatographyPlate height in gas chromatography=(10 μm0.50 mm)(1 mm1000 μm)=150

Reason behind such massive plate height difference is that rate of longitudinal diffusion is much more rapid in gas than in liquid chromatography.

(a)

Interpretation Introduction

Interpretation:

Number of plates and plate height for peak 11 and resolution for peaks 16 and 17 has to be calculated.

Concept Introduction:

Expression to compute number of theoretical plate is as follows:

  N=5.55 tr2w1/22

Here,

tr denotes analyte’s  retention time.

w1/2 denotes peak width at half-height.

N denotes number of theoretical plate.

It has been found that greater the equilibration steps narrower will be band so this suggests N should be high for more resolution or narrower separation.

Expression to compute plate height is as follows:

  H=LN

Here,

N denotes number of theoretical plate.

L denotes length of chromatographic column.

H denotes plate height.

(a)

Expert Solution
Check Mark

Explanation of Solution

Expression to compute number of theoretical plate is as follows:

  N=5.55 tr2w1/22        (1)

Substitute 50 min for tr and 0.263 mm  for w1/2 in equation (1).

  N=5.55 (50 min )2(0.263 mm )2=2×105

Thus, number of theoretical plate is 2×105.

Expression to compute H from number of plates from is as follows:

  H=LN        (2)

Substitute 100 m for L and 2×105 for N in equation (2).

`

  H=(100 m2×105)(1000 mm1 m)=0.50 mm

Thus, plate height is 0.50 mm.

Value of Δtr is calculated as follows:

  Δtr=63.5 min63 min=0.5 min

Expression to compute resolution is as follows:

  Resolution=Δtrwavg        (3)

Substitute 0.5 min for Δtr, 0.384 min for wavg in equation (3).

  Resolution=0.5 min0.384min=1.3

Thus, value of resolution is 1.3.

(b)

Interpretation Introduction

Interpretation:

Extent of larger plate height has to be calculated and reason behind such massive plate height difference has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Ratio of plate height is calculated as follows:

  Plate height in liquid chromatographyPlate height in gas chromatography=(10 μm0.50 mm)(1 mm1000 μm)=150

Reason behind such massive plate height difference is that rate of longitudinal diffusion is much more rapid in gas than in liquid chromatography.

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Students have asked these similar questions
Can you please help mne with this problem. Im a visual person, so can you redraw it, potentislly color code and then as well explain it. I know im given CO2 use that to explain to me, as well as maybe give me a second example just to clarify even more with drawings (visuals) and explanations.
Part 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answer
Part 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00
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