Question

a.) Calculate the amount of benzoic acid in the methylene chloride using the titration that required 1.35mL NaOH to calculate the weight of benzoic acid, M.W. 122.12g/mol, in aqueous layer, and subtract from the total sample weight.

b.) Determine K
K= concn org/concn aq=(mol org/vol org)/(mol aq/vol aq)= (wt total -wt aq)/wt aq
wt aq= vol base × molarity base × mol benzoic acid
I used 162.0 mg, 25 mL H2O, 25mL CH2Cl2 separated and got top layers titration of 0.10 NaOH to be 1.35mL
c.) suppose you have 300.0mL of an aqueous solution that contains 30.0g of malononitrile. to isolate malononitrile from this aqueous solution, you decide to use ether as the organic solvent in your liquid-liquid extraction. The solubility of malononitrile in with at room temp is 20.0g/mL and in water is 13.3g/100mL. Consider the following.
1. how many frames of malononitrile can be isolated with three extractions using 100.0nL of either each time ?
2. how many frames of malononitrile can be isolated with one extraction using 300.0 mL of either?
d.) the distribution coefficient between hexanes and water for an unknown is 7.5 if a solution containing 10.0 g of this unknown in 100.0 mL water is extracted with 100.0 mL of hexanes, how many frames of the unknown can be isolated? if four extractions using 25mL portions of hexanes were carried out instead, how many frames of the unknowcan be extracted? how many mL of hexanes will be required to remove 98.5% of the unknown from this solution in a signal extraction?

Answer 1

The amount of benzoic acid in the methylene chloride can be calculated as

= 0.1 M (= mmol/mL)* 1.35 mL * 122.12 g/mol (= mg/mmol), i.e. 16.5 mg

The amount of benzoic acid in the aqueous layer = (162 - 16.5), i.e. 145.5 mg

b) The equilibrium constant (K) = 16.5 mg / 145.5 mg, i.e. 0.113

c) The partition coefficient (K) of malononitrile between water and ether can be calculated as shown below.

= Conc. of malononitrile in ether / Conc. of malononitrile in water

= (20 g/mL) / (0.133 g/mL)

~ 150

Formula: (Final mass of solute)_{in water} = {V₂ / (V₂ + V₁K)}ⁿ * (Initial mass of solute)_{in water}.

Here, solute = malononitrile, V₂ = volume of water, V₁ = volume of ether and n = no. of extractions

In this case, V₂ = 300 mL

2. If n = 1, V₁ = 300 mL

i.e. (Final mass of solute)_{in water} = {300 / (300 + 300*150)}¹ * 30, i.e. 0.2 g

i.e. The amount of malononitrile can be isolated with 1 extraction using 300.0 mL of ether = 30 - 0.2 = 28.8 g

1. If n = 3, V₁ = 100 mL

i.e. (Final mass of solute)_{in water} = {300 / (300 + 100*150)}³ * 30, i.e. 2.26*10^-4 g ~ 0 g

i.e. The amount of malononitrile that can be isolated with 3 extractions using 100.0 mL of ether each time = 30 g

Accordingly, you can calculate for the remaining problem.

All the very best!!!!!!!!!!!!!!!!!