`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

Number of moles of acetic acid `=(0.6" mL "xx1.06"g "mL^(-1))/(60" g "mol^(-1))`
`=0.0106" mol"=n`
Molality `=(0.0106" mol")/(1000mLxx1" g "mL^(-1))=0.0106" mol "kg^(-1)`
Using equation
`DeltaT_(f)=1.86" K kg "mol^(-1)xx0.0106" mol "kg^(-1)=0.0197K`
van't Hoff factor (i) `=("Observed freezing point")/("calculated freezing point")=(0.0205K)/(0.0197K)=1.041`
Acetic acid is a weak electrolyte and will dissociate into two ions: acetate and hydrogen ions per molecule of acetic acid. if x is the degree of dissociation of acetic acid, then we would have `n(1-x)` moles of undissociated acetic acid, nx moles of `CH_(3)COO^(-)` and nx moles of `H^(+)` ions
`{:(CH_(3)COOH,hArr,H^(+),+,CH_(3)COO^(-)),("n mol",,0,,0),(n(1-x),,"nx mol",,"nx mol"):}`
Thus total moles of particle are : `n(1-x+x+x)=n(1+x)`
`i=(n(1+x))/(n)=1+x=1.041`
thus degree of dissociation of acetic acid `=x=1.041-1.000=0.041`
Then `[CH_(3)COOH]=n(1-x)=0.016(1-0.041)`.
`[CH_(3)C OO^(-)]=nx=0.0106xx0.041,[H^(+)]=nx=0.0106xx0.041`
`K_(a)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])=(0.0106xx0.041xx0.0106xx0.041)/(0.0106(1.00-0.041))`