Staged Liquid-Liquid Extraction of an Organic Chemical from Aqueous Solution The acetone-water mixture of the previous illustration is to be treated by a two-stage extraction with methyl isobutyl ketone (MIK). In the first stage 1 kg of MIK is contacted with 1 kg of the acetone-water mixture. The

Question

Staged Liquid-Liquid Extraction of an Organic Chemical from Aqueous Solution

The acetone-water mixture of the previous illustration is to be treated by a two-stage extraction with methyl isobutyl ketone (MIK). In the first stage 1 kg of MIK is contacted with 1 kg of the acetone-water mixture. The water-rich phase will go to a second stage, where it will be contacted with another 1 kg of pure MIK. What will be the amounts and compositions of the equilibrium phases at the exit of each stage?

Accepted Answer

As in the previous illustration, this problem is solved using the triangular diagram of Fig. 11.2-8. First the concentration of the combined streams in stage 1 is determined. It consists of 1 kg MIK, 0.6 × 1 kg = 0.6 kg acetone, and 0.4 × 1 kg water. So the overall feed is 50 wt % MIK, 30 wt % acetone, and 20 wt % water. This point is located on the triangular diagram for this system, and found to be in the two-liquid phase region. Following the procedure used in the previous illustration, the compositions of the two phases are as follows.

	MIK (wt %)	Acetone (wt %)	Water (wt %)
MIK-rich	62	32	6
Water-rich	2	23	75

Next, from an overall mass balance,

[latex]L^{ I }+L^{ II }=2 kg \quad ext { so } \quad L^{ II }=2-L^{ I }[/latex]

and a mass balance on water, we have

[latex]0.06 imes L^{ I }+0.75 imes L^{ II }=0.06 imes L^{ I }+0.75 imes\left(2-L^{ I } ight)=0.4 imes 1=0.4 kg[/latex]

which has the solution [latex]L^{ I }=1.59 ext { and } L^{ II }=0.41 kg[/latex].

The feed to the second stage is the following: MIK = 1 + 0.02 × 0.41 = 1.008 kg, acetone = 0.23 × 0.41 = 0.094 kg, and water = 0.75 × 0.41 = 0.308 kg, for a total of 1.41 kg. Therefore, the composition of the combined feed to the second stage is 71.5 wt % MIK, 6.7 wt % acetone, and 21.8 wt % water, which is also in the two-phase region.

Using the tie line through this new feed point, we obtain the following:

	MIK (wt %)	Acetone (wt %)	Water (wt %)
MIK-rich	88.4	7.6	4
Water-rich	2	3	95

From the mass balances [latex]L^{ I }=1.134 ext { and } L^{ II }=0.276 kg[/latex]., the water-rich stream leaving the second stage extraction unit contains 0.263 kg water, 0.0083 kg acetone, and 0.0055 kg MIK.

Comment

Note that by using two liquid extraction stages instead of a single stage, more acetone has been extracted (only 0.0083 kg in the exit water-rich stream, compared with 0.022 kg in the single-step process), and only 2 kg MIK have been used (of which 0.0055 kg is lost in the water stream), compared with 3 kg in the single-step process with a comparable MIK loss.

This example suggests that staging with smaller amounts of solvent (here MIK) will produce greater recovery than a single-stage process. However, this would involve greater costs since more equipment is needed. Clearly, a careful analysis, including costs, would be required to design the economically optimal process. In such a design, other extraction configurations would have to be considered, such as the countercurrent extraction process shown belowand other, more complicated processes. Such designs are considered in a stagewise operations course elsewhere in the chemical engineering curriculum. The purpose of the illustrations here is merely to show the importance of thermodynamic equilibrium in the design of liquid-liquid extraction processes.

Schematic diagram of a staged liquid-liquid extraction system.

Question 11.2.9: Staged Liquid-Liquid Extraction of an Organic Chemical from ...

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