`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

Step 1. Determining van't hoff factor
`i=((DeltaT_(f))_("observed"))/((DeltaT_(f))_("calculated"))`
`(DeltaT_(f))_("observed")=0.0205 K`
`(DeltaT_(f))_("calculated")=K_(f)m`
To calculate molality, we need moles of acetic acid dissolved in water and mass of solvent `(H_(2)O)` in kilograms.
`n_("acetic acid")=(mass)/(molar mass)`
`=(volume xx density)/(molar mass)`
`=((0.6 mL)(1.06 g mL^(-1)))/(60 g mL^(-1))`
`=0.0106 mol`
`mass_(H_(2)O)=volume xx density`
`=(1000 mL)(1 gm mL^(-1))`
`=1000 g`
`=1 kg`
`molarity=n_(acetic acid)/(1000 mL xx 1 g mL^(-1))=0.0106 mol kg^(-1)`
Therefore
`(DeltaT_(alpha))_("calculated")=(1.86 K kg mol^(-1))(0.0106 mol kg^(-1))`
`=0.0197 K`
Now
`i=(0.0205 K)/(0.0197 K)`
`=1.04`
Step 2. Determining degree of dissociation Acetic acid is a week electrolyte and thus ionizes into ions (acetate and hydrogen ions) to a limited extent measured as `alpha` (the degree of dissociation). For dilute aqueous solutions, molality (m) and molarity (c ) are almost equal. Thus
`CH_(3)CO_(2)H(aq) hArr CH_(3)COO^(-) (aq) + H^(+) (aq)`
`{:("Moles per litre,C,0,0),("Moles per litre",C- C alpha, C alpha, C alpha):}`

`i=("Moles per litre after ionization")/("Moles per litre before ionization")=(C-Calpha+Calpha+Calpha)/(Calpha)`
`=(C+Calpha)/(Calpha)=1+alpha`
or `alpha=i-1`
`=1.091-1.000`
`=0.041`
Step 3: Calculating the dissociation constant
`CH_(3)COOH(aq)`
Initial C 0 0
concentration
Concentration `C - C alpha` `C alpha` `C alpha`
Concentration
`K_(a)=(C_(CH_(3)COO^(-))C_(H^(+)))/C_(CH_(3)COOH)`
`=((Calpha)(Calpha))/(C(1-alpha))`
`=(Calpha^(2))/(1-alpha)`
Neglecting `alpha` relative to 1 as `CH_(3)CO_(2)H` is a weak electrolyte, we have
`K_(a)=Calpha^(2)`
`=(0.0106)(0.041)^(2)`
`=0.0000178`
`=1.78xx10^(-5)`
`=1.8xx10^(-5)`