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))/(60g mol^(-1))=0.0106 mol = n`
Molality `= (0.0106 mol)/(1000 mL xx 1g mL^(-1))`
`=0.0106 mol kg^(-1)`
`therefore Delta T_(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 freezsing 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^(-)),("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)`
`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=0.0106xx0.041`,
`[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)`