The countercurrent extraction shown in Figure A6.2 is carried out through step 30.

Question

The countercurrent extraction shown in Figure A6.2 is carried out through step 30. Calculate the fraction of solutes A and B in tubes 5, 10, 15, 20, 25, and 30.

Accepted Answer

The fraction, [latex]q[/latex], of each solute that remains in the lower phase is calculated using equation A6.1.

[latex]\left(q_{\mathrm{aq}} ight)_1=\frac{( ext { moles aq })_1}{( ext { moles aq })_0}=\frac{V_{\mathrm{aq}}}{D V_{\mathrm{org}}+V_{\mathrm{aq}}}[/latex] (A6.1)

Since the volumes of the lower and upper phases are equal, we get

[latex]q_{\mathrm{A}}=\frac{1}{D_{\mathrm{A}}+1}=\frac{1}{5+1}=0.167[/latex]

and

[latex]q_{\mathrm{B}}=\frac{1}{D_{\mathrm{B}}+1}=\frac{1}{0.5+1}=0.667[/latex]

Thus, [latex]p_{\mathrm{A}}[/latex] is 0.833 and [latex]p_{\mathrm{B}}[/latex] is 0.333 . For solute [latex]\mathrm{A}[/latex], the fraction present in tubes 5 , [latex]10,15,20,25[/latex], and 30 after step 30 are

[latex] \begin{gathered} f(5,30)=\frac{30 !}{(30-5) ! 5 !}(0.833)^{5}(0.167)^{30-5}=2.1 imes 10^{-15} \approx 0 \ f(10,30)=\frac{30 !}{(30-10) ! 10 !}(0.833)^{10}(0.167)^{30-10}=1.4 imes 10^{-9} \approx 0 \ f(15,30)=\frac{30 !}{(30-15) ! 15 !}(0.833)^{15}(0.167)^{30-15}=2.2 imes 10^{-5} \approx 0 \ f(20,30)=\frac{30 !}{(30-20) ! 20 !}(0.833)^{20}(0.167)^{30-20}=0.013 \ f(25,30)=\frac{30 !}{(30-25) ! 25 !}(0.833)^{25}(0.167)^{30-25}=0.192 \ f(30,30)=\frac{30 !}{(30-30) ! 30 !}(0.833)^{30}(0.167)^{30-30}=0.004 \end{gathered} [/latex]

The fraction of solute B in tubes [latex]5,10,15,20,25[/latex], and 30 are calculated in the same way, yielding respective values of [latex]0.023,0.153,0.025,0,0[/latex], and 0 . A complete histogram of the distribution of solutes A and B is given in Figure A6.3 and shows that the two solutes have been successfully separated.

Question A6.1: The countercurrent extraction shown in Figure A6.2 is carrie...

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