Question A6.1: The countercurrent extraction shown in Figure A6.2 is carrie...
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.
The fraction, q, of each solute that remains in the lower phase is calculated using equation A6.1.
\left(q_{\mathrm{aq}}\right)_1=\frac{(\text { moles aq })_1}{(\text { moles aq })_0}=\frac{V_{\mathrm{aq}}}{D V_{\mathrm{org}}+V_{\mathrm{aq}}} (A6.1)
Since the volumes of the lower and upper phases are equal, we get
q_{\mathrm{A}}=\frac{1}{D_{\mathrm{A}}+1}=\frac{1}{5+1}=0.167
and
q_{\mathrm{B}}=\frac{1}{D_{\mathrm{B}}+1}=\frac{1}{0.5+1}=0.667
Thus, p_{\mathrm{A}} is 0.833 and p_{\mathrm{B}} is 0.333 . For solute \mathrm{A}, the fraction present in tubes 5 , 10,15,20,25, and 30 after step 30 are
\begin{gathered} f(5,30)=\frac{30 !}{(30-5) ! 5 !}(0.833)^{5}(0.167)^{30-5}=2.1 \times 10^{-15} \approx 0 \\ f(10,30)=\frac{30 !}{(30-10) ! 10 !}(0.833)^{10}(0.167)^{30-10}=1.4 \times 10^{-9} \approx 0 \\ f(15,30)=\frac{30 !}{(30-15) ! 15 !}(0.833)^{15}(0.167)^{30-15}=2.2 \times 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}
The fraction of solute B in tubes 5,10,15,20,25, and 30 are calculated in the same way, yielding respective values of 0.023,0.153,0.025,0,0, 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.
