Question

Debye-Huckel-Onsager equation is represented as ${{\Lambda }_{c}}={{\Lambda }_{0}}-b\sqrt{C}$. From the given options, identify the equation which fits the above equation.
A)$\dfrac{82.4}{{{(DT)}^{{1}/{2}\;}}\eta }+\dfrac{8.20\times {{10}^{5}}}{{{(DT)}^{{3}/{2}\;}}}{{\Lambda }_{0}}$
B) $\dfrac{82.4}{{{(DT)}^{{1}/{2}\;}}\eta }+\dfrac{8.20\times {{10}^{5}}}{{{(DT)}^{{1}/{2}\;}}}{{\Lambda }_{0}}$
C) $\dfrac{82.4}{{{(DT)}^{{1}/{2}\;}}\eta }+\dfrac{8.20\times {{10}^{5}}}{{{(DT)}^{{3}/{2}\;}}}$
D) $\dfrac{8.24}{{{(DT)}^{{1}/{2}\;}}\eta }+\dfrac{8.20\times {{10}^{5}}}{{{(DT)}^{{3}/{2}\;}}}{{\Lambda }_{0}}$

Answer 1

Here, the answer includes mainly the substitution of values of the constant ‘b’ which is assigned a value based on the derivation of Debye-Huckel-Onsager equation.

Complete step by step answer:
In our previous concepts of physical chemistry we have learnt about the Debye-Huckel theory which is a theoretical explanation for the non – ideality of the electrolytic solutions.

Let us now have a look at this theory which makes us understand the base for the Debye-Huckel-Onsager equation.
The drawback or limitation for Debye-Huckel theory was that the equation for this theory was suitable for the solutions of low electrolyte concentration. This theory was not applicable for the solutions of higher electrolyte concentrations and also with the electrolyte which produces ions of higher charges which are particularly unsymmetrical electrolytes.
Thus, Debye and Huckel modified the theory and which was further modified by Onsager with the retention of all the original postulates.
This equation is called as Debye-Huckel-Onsager equation which is given by ${{\Lambda }_{c}}={{\Lambda }_{0}}-b\sqrt{C}$
Where ${{\Lambda }_{0}}$is the limiting molar conductivity
b is the empirical constant and C is the electrolyte concentration
Thus, the derivation of this law includes the value for the empirical constant which is given by,
${{\Lambda }_{c}}={{\Lambda }_{0}}-\left[ \dfrac{82.4}{{{(DT)}^{{1}/{2}\;}}\eta }+\dfrac{8.20\times {{10}^{5}}}{{{(DT)}^{{3}/{2}\;}}}{{\Lambda }_{0}} \right]\sqrt{C}$
Therefore, by comparing this equation to the original equation we have ,
$b=\dfrac{82.4}{{{(DT)}^{{1}/{2}\;}}\eta }+\dfrac{8.20\times {{10}^{5}}}{{{(DT)}^{{3}/{2}\;}}}{{\Lambda }_{0}}$
Thus, the correct answer to this question is option A) $\dfrac{82.4}{{{(DT)}^{{1}/{2}\;}}\eta }+\dfrac{8.20\times {{10}^{5}}}{{{(DT)}^{{3}/{2}\;}}}{{\Lambda }_{0}}$
So, the correct answer is “Option A”.

Note: Here, it is important to make note about the fact that in competitive exams you cannot derive the whole equation and therefore the values of several constants are to be memorised so that answers can be approached directly.