Question

0.6 mL of acetic acid $\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right)$, having density $1.06\,{\rm{g}}\,{\rm{m}}{{\rm{L}}^{ - 1}}$, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was $0.0205^\circ {\rm{C}}$. Calculate the van’t Hoff factor and the dissociation constant of acid.

Answer 1

Hint: Here, first we have to calculate the molality of the solution. Then we have to calculate the van’t Hoff factor using the formula $\Delta {T_f} = i{K_f}m$, where, $\Delta {T_f}$
is freezing point depression, I is van’t Hoff factor and m is molality.

Complete step by step answer:
To calculate molality of the solution we need moles of solute, molar mass of solute and mass of solvent.
The solute is acetic acid $\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right)$.
The molar mass of acetic acid $\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right) = 2 \times 12 + 4 \times 1 + 16 \times 2 = 24 + 4 + 32 = 60\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$
Now, we have to calculate the mass of acetic acid using the formula of density, that is,
Density = $\dfrac{{{\rm{Mass}}}}{{{\rm{Volume}}}}$
The density of acetic acid is given as $1.06\,{\rm{g}}\,{\rm{m}}{{\rm{L}}^{ - 1}}$ and volume is 0.6 mL.
So, the mass of acetic acid is,
Mass = ${\rm{Density}} \times {\rm{Volume}}$
$ \Rightarrow $Mass = $1.06\,{\rm{g}}\,{\rm{m}}{{\rm{L}}^{ - 1}} \times 0.6\,{\rm{mL}} = {\rm{0}}{\rm{.636}}\,{\rm{g}}$
Now, we have to calculate the mass of solvent (water) using the formula of density.
The volume given is 1 L and the density of water is 1000 g/L. So, using formula of density the mass of solvent is,
Mass = $1\,{\rm{L}} \times {\rm{1000}}\,{\rm{g}}\,{{\rm{L}}^{ - 1}} = 1000\,{\rm{g}}$
Now, we have to calculate the molality using the below formula.
$m = \dfrac{{{w_2}}}{{{M_2}}} \times \dfrac{{1000}}{{{w_1}}}$
Here, m is molality, ${w_2}$ is mass of solute, ${M_2}$ is molar mass of solute, ${w_1}$ is the mass of solvent.
As calculated above values of ${w_2} = 0.636\,{\rm{g}}$, ${M_2} = 60\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$ and ${w_1} = 1000\,{\rm{g}}$


Now, we have to put the above values in the formula of molality.
$m = \dfrac{{0.636\,}}{{60}} \times \dfrac{{1000}}{{1000\,}} = 0.0106\,{\rm{m}}$
So, the molality of the solution is 0.0106 m.
Now, we have to calculate the van’t Hoff factor using the formula
$\Delta {T_f} = i{K_f}m$ …… (1)
The freezing point depression is given as $0.0205^\circ {\rm{C}}$, ${K_f} = 1.86\,{\rm{K}}\,{\rm{kg}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$ and m is 0.0106 m.
Now we put the above formula in equation (1).
$\Delta {T_f} = i{K_f}m$
$ \Rightarrow 0.0205 = i \times 1.86 \times 0.0{\rm{1}}0{\rm{6}}$
$ \Rightarrow i = 1.041$
Therefore, van’t Hoff factor is 1.041.
Now, we have to calculate the dissociation constant of the acetic acid. We know that acetic acid is a weak electrolyte and it undergoes dissociation to form two ions namely acetate and hydrogen ions per molecule of acetic acid. We take x as the degree of dissociation of acetic acid, then we would have n(1-x) moles of undissociated acetic acid, nx moles of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }$ and nx moles of ${{\rm{H}}^ + }$ ions.
$CH_3COOH\rightleftharpoons H^{+}+CH_3COO^{-}$ …… (2)
Initial moles;
$\text{n mol 0 0 }$
Equilibrium mole;
$\text{n(1-x) nx nx}$
Therefore, total moles of particles are = $n\left( {1 - x + x + x} \right) = n\left( {1 + x} \right)$
We know that, van’t Hoff factor is equal to the ratio of moles of particles after dissociation and moles of particles before association.

$i = \dfrac{{{\rm{Moles}}\,{\rm{after}}\,{\rm{dissociation}}}}{{{\rm{Moles}}\,{\rm{before}}\,{\rm{dissociation}}}}$
$ \Rightarrow i = \dfrac{{n\left( {1 + x} \right)}}{n} = 1 + x$
From the above calculation the van’t Hoff factor is 1.041. So, the above equation becomes,
$ \Rightarrow 1 + x = 1.041$
$ \Rightarrow x = 1.041 - 1 = 0.041$
So, the degree of dissociation of acetic acid is 0.041.
Now, we have to calculate the dissociation constant $\left( {{K_a}} \right)$.
From equation (2), the value of ${K_a}$ is,
${K_a} = \dfrac{{\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }} \right]\left[ {{{\rm{H}}^ + }} \right]}}{{\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right]}}$ …… (3)
$\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right] = n\left( {1 - x} \right) = 0.0106\left( {1 - 0.041} \right)$
$\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }} \right] = nx = 0.0106 \times 0.041$
$\left[ {{{\rm{H}}^ + }} \right] = nx = 0.0106 \times 0.041$
Now, put the above values in equation (3).
${K_a} = \dfrac{{\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }} \right]\left[ {{{\rm{H}}^ + }} \right]}}{{\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right]}}$
$ \Rightarrow {K_a} = \dfrac{{0.0106 \times 0.041 \times 0.0106 \times 0.041}}{{0.0106\left( {1 - 0.041} \right)}} = 1.86 \times {10^{ - 5}}$

Hence, the dissociation constant is $1.86 \times {10^{ - 5}}$

Note: Remember that van’t Hoff factor (i) is used to account for the extent of association or dissociation. In case of association, the value of i is less than 1 and in case of dissociation the value of i is greater than one.