A 100.0 mL aqueous solution contains two compounds, A and B. The partition coeff

Question

A 100.0 mL aqueous solution contains two compounds, A and B. The partition coefficients between water and dichloromethane (DCM) for these compounds are: (K_H20/DCM)_A = 8.5 A H2O: DCM = 8.5:1)K_H2O/DCM)_B = 0.3 B.... = 0.31 (a) How many extractions with 100.0 mL of DCM are required to enrich the aqueous solution in a relative to B, by a factor of at least 100? (That is, the ratio of the concentrations of A and B the aqueous phase is now 100 times larger than what it was initially.) (b) What percentage of A was lost from the aqueous phase during the enrichment?

Explanation / Answer

A)  To determine the minimum number of extractions:

(Qaq)n=(Vaq / DVorg+Vaq)n

  0.001= [100.00mL / {(8.5/0.3) (100.00mL)+100.00mL]n

     0.001 = (0.034)n

Take the log of both sides and solve for n

   log(0.001) = n log(0.034)

   -3 = n * (-1.47)

   n = -3 / -1.47

   n = 2.04

Hence, minimum of 2 extractions is necessary.

B) The fraction of solute remaining in the aqueous phase after the extraction is given by following equation:

(qaq)1=Vaq / (DVorg+Vaq)

= 100.00mL / {(8.5/0.3) (100.00mL)+100.00mL

=0.034

The fraction of solute in the organic phase is 1 – 0.034 = 0.966 (lost fraction)

Extraction efficiency is the percentage of solute moving into the extracting phase. The extraction efficiency, therefore, is 0.966 * 100 = 96.6 %.

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