0.400 g of an acid HA (mol. mass = 80) was dissolved in 100 g of water. The solution showed a depression of freezing point of 0.12 K. What will be the dissociation constant (in multiple of `10^(-3)` ) of the acid at about 0°C? Given `K_f` (water) = 1.86 K Kg `"mol"^(-1)` (Assume molarity of solution a molality)

Answer

5.92

Explanation

Molality (m) of solution `=(0.4 xx 1000)/(80 x 100) = 0.05`
`DeltaT_(f)` (normal) `=K_(f) xx m = 1.86 xx 0.05 = 0.093` K
Van.t Hoff factor,
`i=(DeltaT_(f) ("observed"))/(DeltaT_(f)("normal")) = 0.12/0.093 = 1.290`
`HA + H_(2)O `alpha =(i-1)/(n-1) = (1.290-1)/(2-1) = 0.29`
`K_(a) = (Calpha^(2))/(1-alpha) = (0.05 xx (0.29)^(2))/(1-0.29) = 5.92 xx 10^(-3)`