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Question: Given,0.6 ml of acetic acid having density \(1.06gm{{l}^{-1}}\), is dissolved in 1 liter of water. T...

Given,0.6 ml of acetic acid having density 1.06gml−11.06gm{{l}^{-1}}, is dissolved in 1 liter of water. The depression of freezing point observed for the strength of acid was 0.0205∘C0.0205{}^\circ C. The dissociation constant of acid is:
A. 1.86×10−51.86\times {{10}^{-5}}
B. 2.21×10−72.21\times {{10}^{-7}}
C. 3.56×10−53.56\times {{10}^{-5}}
D. 4.44×10−74.44\times {{10}^{-7}}

Explanation

Solution

Freezing point is a temperature at which liquid starts to convert itself in a solid state i.e. the intermolecular forces gets stronger as we know that intermolecular forces are lesser in liquids due to which they have tendency to flow but by decreasing the temperature forces some close to each other and convert in solid state.

Complete answer:
Depression in freezing point can be calculated by the following formula:
ΔTf=Kfm\Delta {{T}_{f}}={{K}_{f}}m
Where Kf{{K}_{f}}is freezing point depression constant of solvent and m = molarity
Molarity can be defined as the number of solutes present in 1 liter of the solvent it can be calculated by the formula:
Ka=[CH3COO−][H+][CH3COOH]=(0.0106)2(0.041)2(0.0106)(1−0.041)=1.86×10−5{{K}_{a}}=\dfrac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}=\dfrac{{{(0.0106)}^{2}}{{(0.041)}^{2}}}{(0.0106)(1-0.041)}=1.86\times {{10}^{-5}}
number of moles of CH3COOH=0.6×1.0660=0.0106molnumber\ \text{of moles of C}{{\text{H}}_{\text{3}}}\text{COOH}=\dfrac{0.6\times 1.06}{60}=0.0106mol
molarity=0.01061000ml×1gmol−1=0.0106molkg−1molarity=\dfrac{0.0106}{1000ml\times 1gmo{{l}^{-1}}}=0.0106molk{{g}^{-1}}
By putting all the values in equation: ΔTf=Kfm\Delta {{T}_{f}}={{K}_{f}}m
ΔTf=1.86×0.0106\Delta {{T}_{f}}=1.86\times 0.0106= 0.0197 K
Now we can calculate Von’t hoff factor = Observed freezing pointCalculated freezing point=0.02050.0197=1.041\dfrac{Observed\ freezing\ \text{point}}{Calculated\ freezing\ \text{point}}=\dfrac{0.0205}{0.0197}=1.041
CH3COOH⇌CH3COO−+H+C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}
n 0 0
n(1-x) nx nx
Total moles will be equal to n-nx+2nx = n-nx
Then the value of Van't hoff factor i.e. i=n(1+x)n=1+x=1.041i=\dfrac{n(1+x)}{n}=1+x=1.041
Value of x would be 0.041
This defines degree of dissociation is 0.041
[CH3COOH]=n(1−x)=(0.0106)(1−0.041)[C{{H}_{3}}COOH]=n(1-x)=(0.0106)(1-0.041)
[CH3COO−]=[H+]=nx=(0.0106)(0.041)[C{{H}_{3}}CO{{O}^{-}}]=[{{H}^{+}}]=nx=(0.0106)(0.041)
Ka=[CH3COO−][H+][CH3COOH]=(0.0106)2(0.041)2(0.0106)(1−0.041)=1.86×10−5{{K}_{a}}=\dfrac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}=\dfrac{{{(0.0106)}^{2}}{{(0.041)}^{2}}}{(0.0106)(1-0.041)}=1.86\times {{10}^{-5}}.

Thus we can say that dissociation constant of acid is 1.86×10−51.86\times {{10}^{-5}}, option A is the correct answer.

Note:
Depression of freezing point which can be defined as the changes in temperature between solvent and solution ad this can be shown by the formula
ΔTf=T0−Ts\Delta {{T}_{f}}={{T}^{0}}-{{T}^{s}}
Where ΔTf\Delta {{T}_{f}} is depression in freezing point.
T0{{T}_{0}} = Freezing point of solvent
Ts{{T}_{s}}= Freezing point of solution