The distribution coefficient for extraction of a metal complex from aqueous to organic solvents is D=[totalmetal]oegf[totalmetal]aqlocalid="1654846486193" D=[totalmetal]oegf[totalmetal]aqGive physical reasons why localid="1654846489683" β and localid="1654846493841" Kαappear in the numerator of Equation 23-13,but localid="1654846498058" KLand localid="1654846502645" [H+]aqappear in the denominator.

Short Answer

Expert verified

For βand Ka, MLnformation is favoured whereas in KLand H+, the formation of MLnis not favoured.

Step by step solution

01

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  • The distribution coefficient for extraction of a metal complex from aqueous to organic solvents is given as D=[totalmetal]org[totalmetal]aq.
  • Here, let us derive physical reasons of why βand Kαwould appear in the numerator of equation 23-13, where KLand H+appear in the denominator.
02

0f 3

  • The equation for distribution of metal-chelate complex between phases is:

D=KMβKanKanHLorgnH+aqn

  • MLnis the form extracted into organic solvent.
03

0f 3

  • For βand Ka, formation is favoured by the increase of these (an increase in Ka would cause an increase in L-formation which necessary for complex formation).
  • For KLand H+, the formation of MLnis not favoured by the increase of these (as they increase while the formation of MLndecreases).
  • Thus, an increase in the concentration of H+would cause a decrease in the concentration of L-from weak acid in the form HL (if there is less L-it becomes harder for complex to form).

RESULT

L-formation is needed for complex formation of βand Ka, whereas MLnis not favoured by the increase of these in KLand H+as they increase while the formation of MLndecreases.

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Most popular questions from this chapter

Which of the following columns will provide:

(a) highest number of plates?

(b) greatest retention?

(c) highest relative retention?

(d) best separation?

Column1:N=1000;k2=3.9;α=1.16;resolution=0.6Column2:N=5000;k2=3.9;α=1.06;resolution=0.8Column3:N=500;k2=4.7;α=1.31;resolution=1.1Column4:N=2000;k2=2.4;α=1.24;resolution=1.5

Consider a chromatography experiment in which two components with retention factors k1=4.00and k2=5.00are injected into column with N=1.00×103theoretical plates. The retention time for the less-retained component is tr1=10.0min.

(a)Calculate tmandt12.Find w12(width at half height) and w (at the base) for each peak.

(b)Using graph paper , sketch the chromatogram analogous to figure 23-7,supposing that two peaks have same amplitude(height). Draw the half widths accurately.

(c)Calculate the resolution of the two peaks and compare this value with those drawn in Figure 23-10.

Consider the extraction of Mntfrom aqueous solution into organic solution by reaction with protonated ligand HL : D=Mn+(aq)+nHL(org)MLn(org)+nH+(aq)Kextraction=[MLn]org[H+]aqn[Mn+]aq[HL]orgn.Rewrite Equation 23 - 13 in terms of role="math" localid="1654863844402" Kextractionand express Kextractionin terms of the constants in Equation 23 - 13 . Give a physical reason why each constant increases or decreases Kextraction

An open tubular column has an inner diameter of 207μmand the thickness of the stationary phase on the inner wall is 0.50μm. Unretained solute passes through in 63s and a particular solute emerges in 433s . Find the partition coefficient for this solute and find the fraction of time spent in the stationary phase.

In chromatography, resolution is governed by (a) the number of plates, (b) relative retention, and (c) retention factor. Construct a set of graphs to show the dependence of resolution on each of these three parameters. Comment on the nature of the dependence of resolution on each of these three parameters.

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