A liquid containing 47.5% acetic acid and 52.5% water is to be separated by solvent extraction using isopropanol. The solvent used is 1.3 kg per kg of feed. The final extract is found to contain 82% acid on solvent free basis. The residue has 14% acid on solvent free basis. Find the percentage extraction of acid from the feed.
Basis: 1 kg of solvent free acid.
Acetic acid in the feed = 0.475 kg
Water in the feed = 0.525 kg
Entering acid has to come out in the extract and residue (raffinate) phases.
Let as denote the streams with the following representation.
And the mass fractions of acetic acid (on solvent free basis) in various streams:
Acetic acid in the feed: xF
Acetic acid in extract: xE
Acetic acid in residue xR
And the mass fraction of acetic acid in the solvent is zero, since the solvent is a pure one.
Balance on acetic acid:
FxF = ExE + RxR
1 x 0.475 = E x 0.82 + R x 0.14
0.475 = 0.82 E + 0.14 R à 1
similarly writing the balance for water,
0.525 = (1 - 0.82) E + (1 - 0.14) R
0.525 = 0.18 E + 0.86 R à 2
solving equations 1 and 2,
E = 0.493 kg, and
R = 0.507 kg.
Amount of acetic acid in the extract = 0.493 x 0.82 = 0.4043 kg.
Percentage extraction of acetic acid= 100 x Acetic acid in the extract / Acetic acid In feed
= 100 x 0.4043/0.475 =85.12