`0.6 mL` of acetic acid `(CH_(3)COOH)` having density `1.06 g mL^(-1)` is dissolved in `1 L` of water. The depression in freezing point observed for this strength of acid was `0.0205^(@)C`.Calculate the Van't Hoff factor and dissociation constant of the acid. `K_(f)` for `H_(2)O=1.86 K kg ^(-1) "mol"^(-1))`

Explanation

Calculation of Van't Hoff factor
Number of moles of acetic acid `(CH_(3)COOH)= ((0.6 mL)xx(1.06 g mL^(-1)))/((60 g mol^(-1)))=0.0106 mol`
Mass of water= `(100mL)xx(1.0 g mL^(-1))=100 g= 1kg`
`"Molality of solution "(m)=("No. of moles of acetic acid")/("Mass of water in kg")`
= `((0.0106 mol))/(1 kg)=0.0106 "mol kg"^(-1)`
Depression in freezing point `(DeltaT_(f))=K_(f)xxm=(1.86 "K g mol"^(-1))xx(0.0106 mol kg^(-1))`
=0.0197 K
`"Van't Hoff factor"("Observed freezing point")/("Calculated freezing point")`
`=((0.0205 K))/((0.0197 K))=1.041`
Calculation of dissociation constant of acetic acid.
Acetic acid in aqueous solution dissociates as:
`CH_(3)COOHoverset("(aq)")LeftrightarrowCH_(3)COO^(-)(aq)+H^(+)(aq)`
Value of n=2,i=1.041
Degree of dissociation of acetic acie `(alpha)=((i-1))/((n-1))((1.041-1))/(2-1)=0.041`
For acetic acid
`CH_(3)COOH+H_(2)OLeftrightarrowCH_(3)CObarO(aq)+H_(3)overset(+)O(aq)`
`{:(CH_(3)COOH+H_(2)OLeftrightarrowCH_(3)CObarO(aq)+H_(3)overset(+)O(aq)),(c" "0" "0),(c(1-alpha)" "calpha" "aalpha):}`
`K_(a)=((Calpha)(Calpha))/(C(1-alpha))=(Calpha^(2))/(1-alpha)=Calpha^(2)(alphaltltltlt1)`
= `(0.0106)xx(0.041)^(2)=1.78xx10^(-5)`