A solution of Ethanoic acid in a organic solvent called A which is not mixed with water has been shaked in some volume of water and then the mixture has came in to the equilibrium state in 300k.Phynopthelene has been used as the index and the two layers has been titrated by dilute NaOH seperatedly. 10 cm^3 of NaOH solution has been used to completely react with 50 cm^3 of water layer.20 cm^3 of NaOH solution has been used to react with 5.0 cm^3 of organic layer.
Find the Distribution Constant of Ethanoic acid in water and the oraganic solvent A.
Please help me on this.
Water = aq
Organic Solvent = org
When the ethanoic acid the in organic solvent mix with water ethanoic acid will go in to the water phase because ethonoic mix with water. When you titrate with NaOH you will get the ethanoic concentration of each phase water and organic solvent.
`CH_3COOH + NaOH rarr H_2O+CH_3COONa`
Assume the concentration of NaOH = xM
For water layer;
Amount of NaOH consumed `= x/1000xx10 = 0.01x`
But his 0.01x is in 50ml of water.
`[CH_3COOH]_(aq) = (0.01x)/50xx1000 = 0.2x`
For Organic layer;
Using the same calculation above we can get;
`[CH_3COOH]_(org) = (x/1000xx20)/5xx1000 = 4x`
Distribution constant `K_D` is defined as;
`K_D = ([CH_3COOH]_(org))/([CH_3COOH]_(aq))`
`K_D = (4x)/(0.2x)`
`K_D = 20`
So the Distribution Constant of Ethanoic acid in water and the organic solvent is 20.