Animated Solution for Chemistry - Electrochemistry: The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.1 M). If ${\lambda }_{X^{-}}^0\approx {\lambda }_{Y^{-}}^0$, the difference in their $p{K}_a$ values, $p{K}_a(HX)-p{K}_a(HY)$, is (consider degree of ionization of both acids to be <<1).

Welcome, future engineers and scientists! Today, we are going to unravel a beautiful problem from JEE Advanced 2015. This question is a perfect blend of two fundamental concepts in physical chemistry: Electrochemistry and Ionic Equilibrium. It tests your ability to connect the macroscopic observable of molar conductivity with the microscopic reality of acid dissociation.

Let's take a deep breath and look at what the problem is telling us. We have two weak acids, HX and HY. The concentration of HX is $0.01\text{ M}$, and the concentration of HY is $0.1\text{ M}$. We are given a crucial piece of information: the molar conductivity of the HX solution is 10 times smaller than that of the HY solution.

Mathematically, we can write this as:

This is our first anchor. But conductivity alone doesn't tell us about the strength of the acid. For that, we need to look at the degree of dissociation, denoted by $\alpha$.

The problem also gives us a subtle hint: ${\lambda }_{X^{-}}^0\approx {\lambda }_{Y^{-}}^0$. This is where Kohlrausch's Law of Independent Migration of Ions comes into play. According to this law, the limiting molar conductivity of an electrolyte (the conductivity at infinite dilution where the acid is 100% dissociated) is the sum of the limiting molar conductivities of its constituent ions.

For our acids, we can write:

Since the problem states that the limiting molar conductivities of the anions $X^{-}$ and $Y^{-}$ are approximately equal, and both acids share the same cation $H^{+}$, we can confidently conclude that their limiting molar conductivities are practically identical:

This is a massive simplification! It means that when we compare the two acids, we don't have to worry about their inherent maximum conducting abilities being different.

Now, how do we connect conductivity to the degree of dissociation? The relationship is beautifully simple. The degree of dissociation $\alpha$ at any given concentration is the ratio of the molar conductivity at that concentration to the limiting molar conductivity.

Let's find the ratio of the degrees of dissociation for our two acids, ${\alpha }_1$ for HX and ${\alpha }_2$ for HY.

Because we established that ${\Lambda }_m^0(HX)\approx {\Lambda }_m^0(HY)$, these terms cancel out perfectly! We are left with:

Substituting our very first anchor equation into this, we get:

This tells us that the degree of dissociation of HX is one-tenth that of HY.

Now we shift gears from electrochemistry to ionic equilibrium. We need to find the acid dissociation constants, $K_a$, for both acids.

For a generic weak acid HA dissociating into $H^{+}$ and $A^{-}$, the equilibrium expression is:

If the initial concentration is $C$ and the degree of dissociation is $\alpha$, the equilibrium concentrations are $[H^{+}]=C\alpha$, $[A^{-}]=C\alpha$, and $[HA]=C(1-\alpha )$.

Substituting these in, we get:

The problem explicitly tells us to consider the degree of ionization of both acids to be very small ($\alpha \ll 1$). This means $1-\alpha \approx 1$. This approximation is a lifesaver in competitive exams! It simplifies our $K_a$ expression to:

Let's apply this to our specific acids. For HX ($C_1=0.01\text{ M}$):

For HY ($C_2=0.1\text{ M}$):

We are almost there! We need to find the difference in their $p{K}_a$ values. To do that, let's first find the ratio of their $K_a$ values.

We can separate the concentration ratio and the $\alpha$ ratio:

We know that $0.01/0.1=1/10$. And we previously found that ${\alpha }_1/{\alpha }_2=1/10$. Let's plug these in:

Now, we bring in the logarithms. Remember the definition of $p{K}_a$:

Let's take the base-10 logarithm of both sides of our ratio equation:

Using the properties of logarithms ($\log (a/b)=\log (a)-\log (b)$), we get:

Now, we multiply the entire equation by -1 to introduce the $p{K}_a$ terms:

Substitute the definition of $p{K}_a$:

And there we have it! The difference in their $p{K}_a$ values, $p{K}_a(HX)-p{K}_a(HY)$, is exactly 3.

This problem is a fantastic example of how JEE Advanced tests your ability to weave through different concepts. We started with conductivity, used Kohlrausch's law to find a relationship for the degree of dissociation, applied ionic equilibrium approximations to find the acid dissociation constants, and finally used logarithms to find the $p{K}_a$ difference. Keep practicing these multi-concept problems, and you'll develop a strong intuition for physical chemistry!