Step-by-step Derivation

1. Define Variables and Given Information Let's denote the properties of the two weak acids HX and HY with subscripts 1 and 2, respectively.

   • For weak acid HX:

     • Concentration, $C_1 = 0.01 \text{ M}$
     • Molar conductivity, $Λ_{m,1}$
     • Degree of ionization, $α_1$
     • Acid dissociation constant, $K_{a,1}$
     • Limiting molar conductivity, $Λ_{m,1}^0$

   • For weak acid HY:

     • Concentration, $C_2 = 0.10 \text{ M}$
     • Molar conductivity, $Λ_{m,2}$
     • Degree of ionization, $α_2$
     • Acid dissociation constant, $K_{a,2}$
     • Limiting molar conductivity, $Λ_{m,2}^0$

   • Given relations:

     • $Λ_{m,1} = \frac{1}{10} Λ_{m,2}$
     • Limiting molar ionic conductivity of anions: $λ_{x^−}^0 ≈ λ_{y^−}^0$
     • Degree of ionization for both acids is much less than 1, i.e., $α_1 \ll 1$ and $α_2 \ll 1$.

2. Relate Limiting Molar Conductivities According to Kohlrausch's law, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.

   • For HX: $Λ_{m,1}^0 = λ_{H^+}^0 + λ_{x^−}^0$
   • For HY: $Λ_{m,2}^0 = λ_{H^+}^0 + λ_{y^−}^0$ Since we are given $λ_{x^−}^0 ≈ λ_{y^−}^0$, it follows that the limiting molar conductivities of the two acids are approximately equal: $Λ_{m,1}^0 ≈ Λ_{m,2}^0$

3. Relate Degree of Ionization and Molar Conductivity The degree of ionization ($α$) for a weak electrolyte is given by the ratio of its molar conductivity at a given concentration ($Λ_m$) to its limiting molar conductivity ($Λ_m^0$).

   • $α_1 = \frac{Λ_{m,1}}{Λ_{m,1}^0}$
   • $α_2 = \frac{Λ_{m,2}}{Λ_{m,2}^0}$

4. Find the Ratio of Degrees of Ionization Let's find the ratio $\frac{α_1}{α_2}$: $\frac{α_1}{α_2} = \frac{Λ_{m,1} / Λ_{m,1}^0}{Λ_{m,2} / Λ_{m,2}^0} = \left( \frac{Λ_{m,1}}{Λ_{m,2}} \right) \times \left( \frac{Λ_{m,2}^0}{Λ_{m,1}^0} \right)$ Substituting the given information and the result from Step 2: $\frac{α_1}{α_2} ≈ \left( \frac{1}{10} \right) \times (1) = \frac{1}{10}$

5. Relate Acid Dissociation Constant ($K_a$) to Degree of Ionization ($α$) For a weak monoprotic acid HA with initial concentration C and degree of ionization α, the equilibrium is HA <=> H+ + A-. The acid dissociation constant is given by: $K_a = \frac{[H^+][A^−]}{[HA]} = \frac{(Cα)(Cα)}{C(1-α)} = \frac{Cα^2}{1-α}$ Since it is given that $α \ll 1$, we can approximate $1-α ≈ 1$. Therefore: $K_a ≈ Cα^2$ Applying this to our two acids:

   • $K_{a,1} ≈ C_1 α_1^2$
   • $K_{a,2} ≈ C_2 α_2^2$

6. Find the Ratio of Acid Dissociation Constants Now, we find the ratio $\frac{K_{a,1}}{K_{a,2}}$: $\frac{K_{a,1}}{K_{a,2}} = \frac{C_1 α_1^2}{C_2 α_2^2} = \left( \frac{C_1}{C_2} \right) \left( \frac{α_1}{α_2} \right)^2$ Substitute the known values:

   • $\frac{C_1}{C_2} = \frac{0.01 \text{ M}}{0.10 \text{ M}} = \frac{1}{10}$
   • $\frac{α_1}{α_2} = \frac{1}{10}$ So, $\frac{K_{a,1}}{K_{a,2}} = \left( \frac{1}{10} \right) \left( \frac{1}{10} \right)^2 = \frac{1}{10} \times \frac{1}{100} = \frac{1}{1000} = 10^{-3}$

7. Calculate the Difference in pKa Values The pKa of an acid is defined as $pKa = -\log_{10}(K_a)$. We need to find $pKa(HX) - pKa(HY)$, which is $pK_{a,1} - pK_{a,2}$. $pK_{a,1} - pK_{a,2} = (-\log K_{a,1}) - (-\log K_{a,2})$ $pK_{a,1} - pK_{a,2} = \log K_{a,2} - \log K_{a,1} = \log\left( \frac{K_{a,2}}{K_{a,1}} \right)$ From Step 6, we found $\frac{K_{a,1}}{K_{a,2}} = 10^{-3}$, which means $\frac{K_{a,2}}{K_{a,1}} = \frac{1}{10^{-3}} = 10^3$. $pK_{a,1} - pK_{a,2} = \log(10^3) = 3$

Thus, the difference in their pKa values, pKa(HX) - pKa(HY), is 3.