We plan to recover acetic acid from water using 1-butanol as the solvent. Operation is at 26.7C. The feed flow rate is 10.0 kmol/h of an aqueous solution that contains 0.0046 mole frac acetic acid. The entering solvent is pure and flows at 5.0 kmol/h. This operation will be done with three mixer-settlers arranged as a countercurrent cascade. Each mixer-settler can be assumed to be an equilibrium stage. Equilibrium data are available in Table 13-3. Find the exiting raffinate and extract mole fractions.

Table 13-3. Distribution coefficients for immiscible extraction Solute (A) Solvent Diluent TC K = Y/* 25 30 40 50 60 0.0328 0.0984 0.1022 0.0588 0.0637 26.7 1.613 25 7.10 Equilibrium in Weight Fraction Units (Perry and Green, 1984) Acetic acid Benzene Water Acetic acid Benzene Water Acetic acid Benzene Water Acetic acid Benzene Water Acetic acid Benzene Water Acetic acid 1-Butanol Water Furfural Methylisobutyl ketone Water Ethyl benzene B. B'-Thiodipropionitrile n-Hexane m-Xylene B,B'-Thiodipropionitrile n-Hexane 0-Xylene B. B'-Thiodipropionitrile n-Hexane p-Xylene B. B'-Thiodipropionitrile n-Hexane Equilibrium in Mass Ratio Units (Brian, 1972) Linoleic acid Heptane Methylcellosolve (CH,COOH) + 10 vol % water Abietic acid Heptane Methylcellosolve (C, H,,COOH) + 10 vol % water Oleic acid Heptane Methylcellosolve + 10 vol % water 25 25 25 25 0.100 0.050 0.150 0.080 2.17 1.57 4.14