# Distribution Coefficient! Fun for all ages!

[a]1. Calculate the distribution coefficient (Concentration of benzoic acid in dichloromethane: Concentration of benzoic acid in aqueous layer[/b]

Using: molecular weight of benzoic acid=122 Concentration 2.4g/L
Average titre with 0.02M NaOH (39.25 x 0.02)/1000=0.000785

C6H5COOH + NaOH---> C6H5COONa + H20
( 1: 1 RATIO)

2. Moles of benzoic acid left in 50ml aqueous solution: 0.000785

Mass of benzoic acid left in 50ml aqueous solution: 0.000785 x 122=0.90577

Mass of benzoic acid extracted by 10ml dichloromethane:0.12(mass of benzoic acid in 50ml solution) - (0.000785x122)=0.0242

So far, so good. Here's where I ran into brain mush: Calculate the distribution coefficient (Concentration of benzoic acid in dichloromethane: Concentration of benzoic acid in aqueous layer!
Concentration: mass of solute/mass of solution.

Right?
So: here's the problem, I'm not sure what the volume of the solvent is. I think it's the dichloromethane + benzoic (60ml) as there was only dichloromethane (10ml) and benzoic acid (60ml) in the flask.
The concentration of C6H5COOH in dichloromethane= 0.0242/0.06= 0.04033g/L

The concentration of C6H5COOH in aqueous layer: 0.09577/0.06= 1.596 (approx 1.6) g/L

Therefore the distribution coefficient Kd is 0.04/1.6= 0.025. Does this seem correct to you?

Also: Double extraction was carried out in part II of the same experiment. The number of moles of benzoic acid in the aqueous phase was 0.0004 (20ml titre of 0.02M NaOH with a 1:1 stoichiometric ratio) compared to 0.095 for the single extraction. I have to comment on the difference- will I say that the lower amount of benzoic acid in the aqueous phase is due to a more effective separation between the organic and aqueous phases?

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What are you doing here? The concentration (aqueous) of benzoic acid is 2.4 g/L before extraction, yes? What does the Titre refer to? Was the entire methylene chloride layer titrated or a fraction (or whole) of the aqueous layer after extraction?

If the aqueous layer (originally at 2.4 g/L) was titrated in its entirety after extraction with 10 mL of DCM, then the titer refers to the number of moles of benzoic acid that are not in the DCM layer. Subtraction of the original concentration from this titre will give you the moles of benzoic acid in the DCM layer. edit: you can't subtract concentrations, you will have to convert to grams...

The ratio (distribution coefficient) is pretty straightforward from there...

The benzoic acid was added to the dichloromethane and then the entire dichloro layer was removed. The remainder was titrated with the NaOH.

I redid my calculations. How does this look to you now?

You've got the mass of the benzoic acid in the aqueous layer. 50 mL of water has a mass of 50 grams. so the concentration of HBz in the water is 0.0242 g HBz/50.0 g H2O = 0.000484 g/g

The mass in the dicholormethane is 0.0958 g/(10mLx1.33g/mL) = .00720 g/g

[HBz]in dichloromethane:[HBz] in water = .00720/0.000484 = 14.9