Distribution coefficient of an organic acid between water and benzene is 4.1 in favour of `C_(6)H_(6)`. If 5g of acid is distributed in between 50mL of benzeen and `100mL` of water, calculate the concentration of acid in two solvents.

Explanation

Let the amount of organic acid in
`C_(6)H_(6)` layer `= ag`
volume of `C_(6)H_(6) = 50 mL`
`:.` concentration of acid in `C_(6)H_(6) = (a)/(50)g mL^(-1)`
since total amount of acid `= 5g`
and volume of `H_(2)O = 100mL`
and volume of `H_(2)O = 100mL`
`:.` conc. of acid in `H_(2)O = ((5-a)/(100)) (g)/(mL)`
Now, `K =("Conc. of acid in"C_(6)H_(6))/("Conc. of acid in"H_(2)O)`
`= (a)/(50) xx (100)/((5-a))`
`:. 4.1 =(a)/(50) xx (100)/((5-a))`
or `a = 3.361g`
`:.` Amount of acid in `50mL`
`C_(6)H_(6) = 3.361 g`
`:.` acid concentration in `C_(6)H_(6) = (3.361)/(50) xx 100 = 67.22 g//L`
Also, amount of acid in `100mL H_(2)O = 5 -a = 5 - 3.361 = 1.639 g`
`:.` acid concentration in `H_(2)O = (1.639)/(100) xx 1000 = 16.39g//L`