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.

Answer

(i) = 1.04, `K_(a)=1.756xx10^(-5)`

Explanation

Mass of acetic acid `=d xx V = 1.06 xx 0.6=0.636g`,
`DeltaT_(f)=0.0205^(@)C`, weight of solvent `=1000g, K_(f)=1.86`
`DeltaT_(f)=(i xx K_(f) xx "wt. of acetic acid" xx 1000)/("Mol. wt of acetic acid" xx "weight of water")` or
(i) `=(0.0205 xx 60 xx 1000)/(1.86xx0.636xx1000)=1.04`,
`C=(0.636)/(60)=0.0106" mol L"^(-1)`
`underset(C(1-alpha))(CH_(3)COOH) hArr underset(C alpha)(CH_(3)COO^(-)) +underset(C alpha)(H^(+))`,
`i=(1+alpha)/(1) or 1.04=1+alpha, alpha=0.04`
`K_(alpha)=(C alpha^(2))/((1-alpha))=(0.016 xx(0.04)^(2))/(1-0.04)=(0.0106 xx (0.04)^(2))/(0.96)`
`=1.756xx10^(-5)`