Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 28: Q20P (page 791)

(a) Explain how dispersive liquid-liquid microextraction reduces the use of solvent in comparison with liquid-liquid extraction.

(b) What is the purpose of the disperser solvent, which is used in much greater volume than the extraction solvent?

Short Answer

Expert verified

(a)How dispersive liquid-liquid microextraction reduces the use of solvent in comparison with liquid-liquid extraction is explained.

(b)The purpose of the disperser solvent is explained.

Step by step solution

01

Derivation of liquid-liquid extraction and dispersive liquid-liquid microextraction.

  • The extraction technique can be used to purify compounds or separate compound mixtures, such as isolating a product from a reaction mixture (known as an extractive work-up). It can also be used to isolate natural products, such as caffeine from tea leaves.
  • Dispersive liquid-liquid microextraction (DLLME) is a quick and easy method for extracting and purifying organic compounds found in trace amounts in aqueous samples.
02

Working of liquid-liquid extraction and dispersive liquid-liquid microextraction.

a) Liquid-liquid extraction:

Liquid-liquid extraction would use of organic solvent to extract aqueous phase via continuous distillation.

Dispersive liquid-liquid microextraction:

Whereas, dispersive liquid-liquid microextraction would use of immiscible organic solvent and also of dispersant solvent in order to make cloudy emulsion for extraction

How the dispersive liquid-liquid microextraction reduces the use of solvent in comparison with liquid-liquid extraction is explained.

b) Disperser solvent can be mixed with both aqueous and organic phases, thereby lowering interfacial energy and also it would permit the formation of high-surface-area emulsion * rapid mass transfer

Hence the purpose of the disperser solvent, which is used in much greater volume than the extraction solvent is explained.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question: Consider a random mixture containing \(4.00\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) (density\(2.532g/mL\)) and \(96.00\;{\rm{g}}\)of \({{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}\) (density\(2.428\;{\rm{g}}/{\rm{mL}}\)) with a uniform spherical particle radius of\(0.075\;{\rm{mm}}\).

(a) Calculate the mass of a single particle of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) and the number of particles of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) in the mixture. Do the same for\({{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}\).

(b) What is the expected number of particles in \(0.100\;{\rm{g}}\)of the mixture?

(c) Calculate the relative sampling standard deviation in the number of particles of each type in a \(0.100\;{\rm{g}}\)sample of the mixture.

From their standard reduction potentials, which of the following metals would you expect to dissolve in \({\rm{HCl}}\)by the reaction\({\rm{M}} + n{{\rm{H}}^ + } \to {{\rm{M}}^{n + }} + \frac{n}{2}{{\rm{H}}_2}:{\rm{Zn}},{\rm{Fe}},{\rm{Co}},{\rm{Al}},{\rm{Hg}},{\rm{Cu}},{\rm{Pt}}\),\({\bf{Au}}\)?

(When the potential predicts that the element will not dissolve, it probably will not. If it is expected to dissolve, it may dissolve if some other process does not interfere. Predictions based on standard reduction potentials at \({\bf{2}}{{\bf{5}}^{^{\bf{o}}}}C\) are only tentative, because the potentials and activities in hot, concentrated solutions vary widely from those in the table of standard potentials.)

When you flip a coin, the probability of its landing on each side is \(p = q = \frac{1}{2}\)in Equations 28-2 and 28-3. If you flip it \(n\)times, the expected number of heads equals the expected number of tails \( = np = nq = \frac{1}{2}n.\)The expected standard deviation for \(n\)flips is\({\sigma _n} = \sqrt {npq} \). From Table 4-1, we expect that \(68.3\% \)of the results will lie within \( \pm 1{\sigma _n}\) and \(95.5\% \)of the results will lie within\( \pm 2{\sigma _n}\).

(a) Find the expected standard deviation for the number of heads in \({\bf{1000}}\) coin flips.

(b) By interpolation in Table 4-1, find the value of \(z\)that includes \(90\% \)of the area of the Gaussian curve. We expect that \(90\% \)of the results will lie within this number of standard deviations from the mean.

(c) If you repeat the\({\bf{1000}}\)coin flips many times, what is the expected range for the number of heads that includes\(90\% \) of the results? (For example, your answer might be, "The range \({\bf{490}}\) to \({\bf{510}}\) will be observed \(90\% \)of the time.")

Barbital can be isolated from urine by solid-phase extraction with\({{\bf{C}}_{18}} - \)silica. The barbital is then eluted with\({\bf{1}}:{\bf{1}}\) vol/volacetone: chloroform. Explain how this procedure works.

Barium titanate, a ceramic used in electronics, was analyzed by the following procedure: Into a Pt crucible was placed \(1.2\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) and \(0.8\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}\;{{\rm{B}}_4}{{\rm{O}}_7}\)plus \(0.3146\;{\rm{g}}\)of unknown. After fusion at \({1000^\circ }{\rm{C}}\)in a furnace for\(30\;{\rm{min}}\), the cooled solid was extracted with \(50\;{\rm{mL}}\)of\(6{\rm{MHCl}}\), transferred to a \(100 - {\rm{mL}}\) volumetric flask, and diluted to the mark. A \(25.00 - {\rm{mL}}\)aliquot was treated with \(5\;{\rm{mL}}\)of \(15\% \)tartaric acid (which complexes \({\rm{T}}{{\rm{i}}^{4 + }}\)and keeps it in aqueous solution) and \(25\;{\rm{mL}}\)of ammonia buffer,\({\rm{pH}}9.5\). The solution was treated with organic reagents that complex\({\rm{B}}{{\rm{a}}^{2 + }}\), and the \({\rm{Ba}}\)complex was extracted into \({\rm{CC}}{{\rm{l}}_4}.\)After acidification (to release the \({\rm{B}}{{\rm{a}}^{2 + }}\) from its organic complex), the \({\rm{B}}{{\rm{a}}^{2 + }}\)was backextracted into\(0.1{\rm{MHCl}}\). The final aqueous sample was treated with ammonia buffer and methylthymol blue (a metal ion indicator) and titrated with \(32.49\;{\rm{mL}}\) of \(0.01144{\rm{M}}\)EDTA. Find the weight per cent of Ba in the ceramic.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free