Question #a02d1

When you extract, the residue will have the same concentration of material as the extract itself. If you extract again, the concentration will be lower, in the extract as well as in the residue.

So with multiple extractions the amount of material left in the residue will be lower, ergo the extraction will be more complete.

The mass of solute #"X"# that can be extracted in a single extraction is determined by the distribution coefficient #K#:

#K = "[X]"_text(o)/"[X]"_text(aq) = (m_o/V_o)/(m_(aq)/V_(aq)) = (m_oV_(aq))/(m_(aq) V_o#

where

#o# and #aq# represent the organic and the aqueous phases
#m# is the mass of solute in each phase.

Let #m = "total mass of solute"#

Let #m_o = xcolor(white)(l) "g"#.

Then #m_(aq) = (m - x)color(white)(l)"g"#

The formula becomes

#K = (xV_(aq))/((m - x)V_o)#

#K(m-x)V_o = xV_(aq)#

#KmV_o - KxV_o = xV_(aq)#

#KmV_o = KxV_o + xV_(aq) = x(KV_o + V_(aq))#

#x = (KmV_o)/(KV_o + V_(aq)) = m/(1 + V_(aq)/(KV_o))#

Example:

If #D = 4.0#, what mass of compound #"X"# can be extracted from a solution of 10.0 g of #"X"# in 100 mL of water (a) with a single portion of 150 mL of ether? (b) using three 50 mL portions?

(a) Single extraction

#x = m/(1 + V_(aq)/(KV_o)) = "10.0 g"/(1 + (100 color(red)(cancel(color(black)("mL"))))/(4.0 × 150 color(red)(cancel(color(black)("mL"))))) = "10.0 g"/1.167 = "8.6 g"#

∴ 8.6 g extracted and 1.4 g remaining.

We have extracted 86 % of the compound.

(b) Three extractions

First extraction

#x = m/(1 + V_(aq)/(KV_o)) = "10.0 g"/(1 + (100 color(red)(cancel(color(black)("mL"))))/(4.0 × 50 color(red)(cancel(color(black)("mL"))))) = "10.0 g"/1.50 = "6.7 g"#

∴ 6.7 g extracted and 3.3 g remaining.

Second Extraction

#x = "3.3 g"/1.50 = "2.2 g"#

∴ 8.9 g extracted and 1.1 g remaining.

Third extraction

#x = "1.1 g"/1.50 = "0.7 g"#

∴ 9.6 g extracted and 0.4 g remaining.

We have extracted 96 % of the compound.

Multiple extractions are more efficient than a single extraction with the same volume of solvent.