[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? Thanks for your help!