`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 (i)
0.6 mL of acid means solute `(w_(2)) = 0.6 xx 1.6 `
1 Litre of water means solvent `(w_(1)) = 1000 g , (Delta T_(f))_("observed") = 0.0205 ^(@) C`
` :. ` Observed molar mass , `(M_(2)) _("observed") = (1000 xx K_(f) xx w_(2))/(w_(1) xx Delta T _(f))= (1000 " g mol"^(-1) xx 1.86" K kg mol"^(-1) xx 0.636 g )/(1000g xx 0.0205 K)`
` = 57.7 g " mol"^(-1)`
Calculated molar mass of `CH_(3) COOH = "60 g mol"^(-1)`
` :. ` van't Hoff factor (i) = `("Calculated molar mass")/("Observed molar mass ") = (60"g mol"^(-1))/(57.7"g mol"^(-1))= 1.04`
Calculation of degree of dissociation `(alpha)` from van't Hoff factor (i)
If `alpha` is the degree of dissociation of acetic acid, then
` {:(,CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),("Intial moles",1,,,,),("Moles at eqm.",1-alpha,,alpha,,alpha", Total" = 1+alpha):}`
` :. i (1+alpha)/1 = 1 + alpha or alpha = 1 - i = 1.04 - 1.0 = 0.04`
Calculation of dissociation constant , If we start with C mol `L^(-1)` of acetic acid, then
` {:(,CH_(3) COOH,hArr,CH_(3)COO^(-),+,H^(+)),("Intial conc.",C " mol "L^(-1),,,,),("Conc.at eqm.",C(1-alpha),,Calpha,,Calpha):}`
Dissociation constant `(K_(a)) = ([CH_(3)COO^(-1)][H^(+)])/([CH_(3)COOH])=(C alpha. C alpha )/(C(1-alpha)) = (Calpha^(2))/(1-alpha) `
But C = 0.636 g `L^(-1) = (0.636 g L^(-1))/(60 g " mol"^(-1)) = 0.0106 "mol L"^(-1) :. K _(a) = ((0.0106)(0.04)^(2))/(1-0.04) = 1.76 xx 10^(-5)`