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Question: In a \(0.2\) molal aqueous solution of weak acid \(HX\), the degree of ionisation is \(0.3\). Taking...

In a 0.20.2 molal aqueous solution of weak acid HXHX, the degree of ionisation is 0.30.3. Taking Kf{K_f} for water as 1.851.85, the freezing point of the solution will be nearest to:
A. −0.360∘C - 0.360^\circ C
B. −0.260∘C - 0.260^\circ C
C. +0.480∘C + 0.480^\circ C
D. −0.480∘C - 0.480^\circ C

Explanation

Solution

Freezing point of the solution can be defined as the lowering of the freezing point of the solvents on the addition of solutes. It is categorized as one of the colligative properties of a solution. It is directly related to the molality of solute added in the solution. The formula can be written as: ΔTf=i×Kf×m\Delta {T_f} = i \times {K_f} \times m
Here, m represents molality
Kf{K_f} represent cryoscopic constant
i represents Van’t Hoff factor
ΔTf\Delta {T_f} represent freezing point depression in the solution

Complete step by step answer:
In the question, we are given with the 0.20.2molal aqueous solution of a weak acid and the value of the degree of ionisation is 0.30.3
If we ionise a weak acid, it can be represented as follows:
HX⇌H++X−HX \rightleftharpoons {H^ + } + {X^ - }
(1−0.3)(1 - 0.3) (0.3)(0.3) (0.3)(0.3)
For this reaction, i can be written as:
i=1−0.3+0.3+0.31i = \dfrac{{1 - 0.3 + 0.3 + 0.3}}{1}
i=1.3i = 1.3
Now, we have the value of Van’t Hoff factor i.e. 1.31.3
As mentioned the formula used to determine the freezing point depression of the solution is
ΔTf=i×Kf×m\Delta {T_f} = i \times {K_f} \times m
So, we are given Kf{K_f} for water is 1.851.85
After substituting the values in the formula we have
ΔTf=1.3×1.85×0.2\Delta {T_f} = 1.3 \times 1.85 \times 0.2
ΔTf=0.480∘C\Delta {T_f} = {0.480^\circ }C
Here, we have to calculate the freezing point of the solution that will be as follows:
Tf=0∘C−0.480∘C{T_f} = 0^\circ C - 0.480^\circ C
Tf=−0.480∘C{T_f} = - 0.480^\circ C
Thus, we can conclude that the freezing point of the solution will be nearest to the −0.480∘C - 0.480^\circ C

So, the correct answer is Option D.

Note: There are several reasons for the occurrence of freezing point depression mentioned as follows:
There will be an equilibrium condition between the liquid state and the solid- state at the freezing point, so we can say that vapour pressure of solid and liquid are equal.
When there is the addition of non-volatile solute, the vapour pressure of the solution is determined to be lower than the vapour pressure of the solvent present in the solution. So, the equilibrium point will be attained at a lower temperature between the solid and the solution.