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Question: 0.6 mL of acetic acid \(\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right)\), having density \(...

0.6 mL of acetic acid (CH3COOH)\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right), having density 1.06 g mL−11.06\,{\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∘C0.0205^\circ {\rm{C}}. Calculate the van’t Hoff factor and the dissociation constant of acid.

Explanation

Solution

Here, first we have to calculate the molality of the solution. Then we have to calculate the van’t Hoff factor using the formula ΔTf=iKfm\Delta {T_f} = i{K_f}m, where, ΔTf\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 (CH3COOH)\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right).
The molar mass of acetic acid (CH3COOH)=2×12+4×1+16×2=24+4+32=60 g mol−1\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 = MassVolume\dfrac{{{\rm{Mass}}}}{{{\rm{Volume}}}}
The density of acetic acid is given as 1.06 g mL−11.06\,{\rm{g}}\,{\rm{m}}{{\rm{L}}^{ - 1}} and volume is 0.6 mL.
So, the mass of acetic acid is,
Mass = Density×Volume{\rm{Density}} \times {\rm{Volume}}
⇒\RightarrowMass = 1.06 g mL−1×0.6 mL=0.636 g1.06\,{\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 L×1000 g L−1=1000 g1\,{\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=w2M2×1000w1m = \dfrac{{{w_2}}}{{{M_2}}} \times \dfrac{{1000}}{{{w_1}}}
Here, m is molality, w2{w_2} is mass of solute, M2{M_2} is molar mass of solute, w1{w_1} is the mass of solvent.
As calculated above values of w2=0.636 g{w_2} = 0.636\,{\rm{g}}, M2=60 g mol−1{M_2} = 60\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}} and w1=1000 g{w_1} = 1000\,{\rm{g}}

Now, we have to put the above values in the formula of molality.
m=0.636 60×10001000 =0.0106 mm = \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
ΔTf=iKfm\Delta {T_f} = i{K_f}m …… (1)
The freezing point depression is given as 0.0205∘C0.0205^\circ {\rm{C}}, Kf=1.86 K kg mol−1{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).
ΔTf=iKfm\Delta {T_f} = i{K_f}m
⇒0.0205=i×1.86×0.0106\Rightarrow 0.0205 = i \times 1.86 \times 0.0{\rm{1}}0{\rm{6}}
⇒i=1.041\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 CH3COO−{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - } and nx moles of H+{{\rm{H}}^ + } ions.
CH3COOH⇌H++CH3COO−CH_3COOH\rightleftharpoons H^{+}+CH_3COO^{-} …… (2)
Initial moles;
n mol 0 0 \text{n mol 0 0 }
Equilibrium mole;
n(1-x) nx nx\text{n(1-x) nx nx}
Therefore, total moles of particles are = n(1−x+x+x)=n(1+x)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=Moles after dissociationMoles before dissociationi = \dfrac{{{\rm{Moles}}\,{\rm{after}}\,{\rm{dissociation}}}}{{{\rm{Moles}}\,{\rm{before}}\,{\rm{dissociation}}}}
⇒i=n(1+x)n=1+x\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,
⇒1+x=1.041\Rightarrow 1 + x = 1.041
⇒x=1.041−1=0.041\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 (Ka)\left( {{K_a}} \right).
From equation (2), the value of Ka{K_a} is,
Ka=[CH3COO−][H+][CH3COOH]{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)
[CH3COOH]=n(1−x)=0.0106(1−0.041)\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}} \right] = n\left( {1 - x} \right) = 0.0106\left( {1 - 0.041} \right)
[CH3COO−]=nx=0.0106×0.041\left[ {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }} \right] = nx = 0.0106 \times 0.041
[H+]=nx=0.0106×0.041\left[ {{{\rm{H}}^ + }} \right] = nx = 0.0106 \times 0.041
Now, put the above values in equation (3).
Ka=[CH3COO−][H+][CH3COOH]{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]}}
⇒Ka=0.0106×0.041×0.0106×0.0410.0106(1−0.041)=1.86×10−5\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 1.86×10−51.86 \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.