2.0 g of a substance in 12 mL of a solvent is extracted with three aliquots of 1
ID: 536525 • Letter: 2
Question
2.0 g of a substance in 12 mL of a solvent is extracted with three aliquots of 10 mL of an extracting solvent How much of the substance by mass was extracted out of the original solvent? (Assume K=2) 15 mL solution of benzoic acid in ether is extracted with 40 mL of 0-20 M NaOH. a. Write the reaction for this process b. What is the purpose of neutralizing the base extract layer with MCI? C. How much HCI with a molarity of 0.68 M must be added to neutralize the base extract? In what layer would you find each of the following molecules after a base extraction with 10% NaOH?Explanation / Answer
Q1.
m = 2 g
V = 12 mL
n = 3
V = 10 mL of solvent
K = 2
initially:
(a) The fraction of solute remaining in the aqueous phase after two extractions and three extractions
(Qaq)n = (Vaq / ( K*Vorg + Vaq))^n
where
Qaq = fraciton of remaining sample in aqueous solution in nth extraction
Vaq = aqueous volume, 12 mL
Vorg = organis layuer volume, 10 mL
K = 2, given
(Qaq)3 = (12/ ( 2*10 + 12))^3
(Qaq)3 = 0.052734 is left
initial mass was = 2 g
final amount lef tin aqueous layer = 2*0.052734 = 0.105468 g left in auqeous layer
mass extracted = 2-0.105468 = 1.894532 g of mass extracted
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