2.0 g of a substance in 12 mL of a solvent is extracted with three aliquots of 1

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) 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 HCl? c. How much HCl with a molarity of 0.68 M must be added to neutralize the base extract? 4. In what layer would you find each of the following molecules after a base extraction with 10% NaOH

Explanation / Answer

Apply:

qn = (1/(K*Vr+1)^n

where

q = fraciton of material remainin in solution

Vr = Volume ratio

K partition cofficient

n = no of extractions

so

Vr = V0/Va = 12 mL / 10 mL = 1.2

K = 2

n = 3 (three aliquots)

q = (1/(2*1.2+1)^3 = 0.0254 g

then:

m = 2-0.0254 = 1.9746 g of substance is extracted.

a.

Benzoic acid = C6H5COOH(aq) + NaOH(aq) = NaC6H5COO(aq) + H2O(l)

b.

The main purpose is to form an aqeuous solution of the benzoic acid, so it is NOT molecular but aquous, so the organic layer will not contain any benzoic/benzoate

c.

mol of base = MV = 40*0.2 = 8 mmol of OH-

then

8 mmol of H+ required

vo

V = mol/M = 0.68/8 = 0.085 mL

4.

expect --> Benzoic acid in molecular extract (ether)

expect water + benzoat in aquoeus solution

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