0.6 mL of acetic acid `(CH_(3)COOH)`, having density `1.06" g mL"^(-1)`, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was `0.0205^(@)C`. Calculate the van't Hoff factor and the dissociation constant of acid.

Explanation

Number of moles of acetic acid `=(0.6" mL"xx1.06"g mL"^(-1))/(60" g mol"^(-1))`
`=0.0106" mol = n"`
Molality `=(0.0106xx" mol")/(1000" mL"xx 1" g mL"^(-1))=0.0106" mol kg"^(-1)`
Using equation `DeltaT_("f")=(K_("f")xx w_(2)xx1000)/(M_(2)xx w_(1))`
`DeltaT_("f")=1.86" K kg mol"^(-1)xx0.0106" mol kg"^(-1)=0.0197" K"`
van't Hoff factor (i) `=("Observed freezing point")/("Calculated freezing point")=(0.0205"K")/(0.0197"K")=1.041`
Acetic acid is a weak electrolyte and will dissocitate 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,iff,H^(+),+,CH_(3)COO^(-)),(" n mol",," "0,," "0),(" "n(1-x),,"nx mol",," nx mol"):}`
Thus total moles of particles are : `n(1-x+x+x)=n(1+x)`
`t=(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)=0.0106(1-0.041)`
`[CH_(3)COO^(-)]="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))=1.86xx10^(-5)`