Acetic acid tends to form dimer due to formation of intermolcular hydrogen bonding.
`2CH_(2)COOHhArr(CH_(3)COOH)_(2)`
The equilibrium constant for this reaction is `1.5xx10^(2)M^(-1)`in benzene solution and `3.6xx10^(-2)` in water. In benzene, monomer does not dissociate but it water, monomer dissociation simultaneously with acid dissociation constant `2.0xx10^(-5)`M. Dimer does not dissociate in benzene as well as water.
The molar ration of dimer to monomer for 0.1 M acetic acid in benzene is equal to :

Options

A. `150:1`
B. `1:150`
C. `5:2`
D. `2:5`

Answer

C

Explanation

`{:(2CH_(3)COOH hArr,(CH_(3)COOH)_(2)),(0.1-2x,x):}`
`1.5 xx 10^(2) = (x)/((0.1 -2x)^(2))`
`0.01 +4x^(2) - 0.4x = 0.0067 x`
`4x^(2) - 0.4067 x + 0.01 = 0`
`x =(+0.4067 +-sqrt(0.165 -0.16))/(2xx4)=0.0416`
`([CH_(3)COOH]_(2))/([CH_(3)COOH]) = (0.0416)/(0.0167)=2.5 =(5)/(2)`