EXPERIMENT 6 - Extraction

CHEM 2423 Extraction of Benzoic Acid11.25 = ( 2.55/V o ) / (0.45 /100) V o ≈ 51 mlDr. Pahlavand) calculate the total amount of acid extracted by a double extraction of 50 ml ether in each extraction inpart ( c).First extraction (first 50 ml ether)11.25 = ( X o1 /50) / (X w1 /100) 11.25 = ( X o1 /50) / (3.00 –X o1 /100) , solve for X o1X o1 = 2.57 g acid extracted in first extractionX w1 = 3.00 g – 2.57 g = 0.43g acid remained in aqueous solution after first extractionSecond extraction (second 50 ml ether)11.25 = ( X o2 /50) / (X w2 /100) 11.25 = ( X o1 /50) / (0.43 –X o2 /100), solve for X o2X o2 = 0.38 g acid extracted in first extractionX o (total) = X o1 + X o2 = 2.57 g + 0.38 g = 2.93 g%E = (2.93 /3.00) x100 = 97.7 %Or use the following formula;%E = [ 1 - (V aq / V aq + K d V o ) n ] x 100 , where n is the number of extraction.%E = [ 1 - ( 100/ 100+11.25x50) 2 ] x100 = 97.7 %Example (2) – Isobutyric acid has a solubility in water (at 25 0 C) about one-third of its solubility in diethylether; thus K d = 3.0. Imagine 4.0 g of isobutyric acid dissolved in a mixture of 35.0 ml of diethyl ether and100.0 ml of water.a) What weight of acid is in each layer at equilibrium after first extraction and second extraction?b) Calculate the percentage extraction after second extraction.Solubility of butyric acid in water = X g/mlSolubility of butyric acid in diethyl ether = 3X g/mlFirst ExtractionX w = V w . X = (100.0 ml)(X g/ml) = (100.0) XX o = V o . 3X = (35.0 ml)( 3X g/ml) = (105.0) XX w + X o = 4.0 g 4.0 = (100.0) X + 105 X 4.0 = 205.0 X X = 0.0195 gX w(1) = (100.0) X = (100.0)(0.0195) = 1.95 g (remained in water after first extraction)X o(1) = (105.0) X = (105.0)(0.0195) = 2.05 g (extracted by diethyl ether after first extraction)3