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 xx 1.06 " g mL"^(-1))/(60 " g mol"^(-1))`
` = 0.106 mol = n `
Molality = ` = (0.0106 mol)/(1000 mL xx 1 " g mL"^(-1)) = 0.0106 " mol kg"^(-1)`
Using equation (2.35)
`DeltaT_(f) = 1.86 " K kg mol"^(-1) xx 0.0106 " 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 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^(-)),("nmol",0,0),(n(1-x),"nx mol","nx mol"):}`
Thus total moles of particles 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) 0.0106 (1-0.041 )`
` [CH_(3)COO^(-)] = nx = 0.106 xx0.41 , [H^(+)] = nx = 0.0106 xx 0.041 `
`K_(a) = ([CH_(3)COO^(-)][H^(+)] )/([CH_(3)COOH]) = (0.016 xx 0.041 xx 0.0106 xx 0.041 )/(0.0106 (1.00 - 0.041))`
` = 1.86 xx10^(-5)`