Question

When doing an extraction, a student mixes 50 mL of 0.40 M benzoic acid (MW = 122 g/mole) with 30 mL of dichlormethane. After separating the layers and allowing the dichloromethane to evaporate, the student recovers 1.22 grams of benzoic acid from the dichloromethane layer. How many moles and grams of benzoic acid were in the original benzoic acid solution? What is the Kd of the of Dichloromethane/water for this extraction?

Answer 1

Answer:

A. 2.44g

B. Kd = 3.2

Explanation:

Concentration of benzoic acid = 50mL = 0.05L

Molecular weight = 122 g/mol

Moles of benzoic acid before the extraction = mass/molecular weight of Benzoic acid

= molar concentration * volume

Moles of benzoic acid before the extraction = 0.05 * 0.4

= 0.02moles of benzoic acid before extraction.

Mass of benzoic acid = 0.02 * 122

= 2.44 g

Partition coefficient, Kd is defined as the ratio of concentration of a substance in an organic medium to the concentration of the substance in the aqueous medium.

Since 1.22g of Benzoic acid was in Dicholoroethane, therefore (2.44 - 1.22)g of benzoic acid was in water.

Mass concentration of benzoic acid in water = mass of benzoic in water/volume of water

= 1.22/0.05

= 24.4 g/L

Mass concentration of benzoic acid in Dicholoroethane = mass of benzoic in Dicholoroethane/volume of Dicholoroethane

= 1.22/0.03

= 40.67g/L

Kd = [concentration]organic/[concentration]aqueous

= 40.66/24.4

= 1.67