Problem 27 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The molar conductivity of a solution of a weak acid \(\mathrm{H} X(0.01 \mathrm{M})\) is 10 times smaller than the molar conductivity of a solution of a weak acid \(\mathrm{H} Y(0.10 \mathrm{M})\). If \(\lambda_{\mathrm{X}^{-}}^{0} \approx \lambda_{\mathrm{Y}}^{0}\) the difference in their \(\mathrm{p} K_{a}\) values, \(\mathrm{p} K_{a}(\mathrm{H} X)\) \(-\mathrm{p} K_{a}(\mathrm{H} Y)\), is (consider degree of ionization of both acids to be \(<<1\) )

Step-by-Step Solution

Answer

The difference in \( pKa \) values is 3.

Step 1: Understanding Molar Conductivity

Molar conductivity is defined as the conductivity of a solution per unit concentration. For a weak electrolyte like a weak acid, it is calculated using the formula: \[\lambda_m = \frac{1000 \times \kappa}{C}\]where \(\lambda_m\) is molar conductivity, \(\kappa\) is the specific conductivity, and \(C\) is the concentration (in mol/L). For two solutions where \(\lambda_m(HX) = \frac{1}{10} \lambda_m(HY)\), given \(C(HX) = 0.01 \text{ M}\) and \(C(HY) = 0.10 \text{ M}\).

Step 2: Relating Concentration and Conductivity

Since \(\lambda_m(HX) = \frac{1}{10} \lambda_m(HY)\) and \(\lambda = \lambda^{0} \alpha\), where \(\alpha\) is the degree of ionization, and both acids are weak with \(\alpha << 1\), we can write:\[\alpha(HX) \times \lambda_0 \approx \frac{1}{10} (\alpha(HY) \times \lambda_0)\]Thus, \(\alpha(HX) \approx \frac{1}{10} \alpha(HY)\).

Step 3: Calculating the Degree of Ionization

For weak acids \(H\), \(\alpha = \sqrt{\frac{K_a}{C}}\), thus we have:\[\alpha(HX) = \sqrt{\frac{K_{a}(HX)}{0.01}}\]and\[\alpha(HY) = \sqrt{\frac{K_{a}(HY)}{0.10}}\]From \(\alpha(HX) \approx \frac{1}{10} \alpha(HY)\), equate and solve for \(K_a\) differences:

Step 4: Equating and Solving for \(pK_a\) Difference

Equating the expressions from the previous step, we get:\[\sqrt{\frac{K_{a}(HX)}{0.01}} \approx \frac{1}{10} \sqrt{\frac{K_{a}(HY)}{0.10}}\]Squaring both sides:\[\frac{K_{a}(HX)}{0.01} = \frac{1}{100} \times \frac{K_{a}(HY)}{0.10}\]Solve for \(K_a(HX)\):\[K_{a}(HX) = \frac{1}{1000} K_{a}(HY)\]

Step 5: Determining \(pK_a\) Difference

Since \(pK_a = -\log_{10}(K_a)\), applying logarithmic properties:\[pK_a(HX) = -\log_{10}(\frac{1}{1000} K_{a}(HY)) = 3 + pK_a(HY)\]Thus:\[pK_a(HX) - pK_a(HY) = 3\]

Key Concepts

Degree of Ionization

The degree of ionization is an important concept when dealing with weak acids and their solutions. It represents the fraction of the acid that dissociates into ions. In other words, it tells us how much of the acid has turned into ions in the solution. For weak acids, this degree is usually quite small because these acids do not completely dissociate in solution. We represent the degree of ionization by the symbol \( \alpha \). For example, if you have a weak acid solution, and only a small portion of it ionizes, its \( \alpha \) value will be much less than 1. This low value reflects the minimal ionization. Understanding \( \alpha \) helps us to comprehend how the molar conductivity of the solution relates to the acid's nature and concentration. Since most of the acid remains un-ionized, the conductivity — which depends on the presence of ions — will be relatively low. This concept becomes even more relevant when assessing different weak acids, as seen in the original exercise. Here, the degree of ionization helps us compare the conductivities and thus the strengths and behaviors of the acids at different concentrations.

Weak Acids

Weak acids are acids that do not completely dissociate in water. This means that in a solution, there are always both the non-ionized and ionized forms of the acid. Because only a small fraction of the acid releases hydrogen ions, weak acids have lower molar conductivity compared to strong acids, which fully dissociate. Weak acids are essential to study because they offer a deeper understanding of equilibrium in chemical reactions. The balance between the ionized and non-ionized forms depends on factors such as concentration and the intrinsic properties of the acid itself. Examples of weak acids include acetic acid, formic acid, and many organic acids. In terms of chemical reactions and calculations, the weak acid behavior is crucial for calculating the equilibrium positions, pH of solutions, and also leads to concepts like buffer solutions. The understanding of weak acids includes appreciating how their low degree of ionization affects their behavior and doesn’t allow for a complete change in properties upon dissolution. This idea directly contrasts that of strong acids, which fully dissociate under similar conditions.

Dissociation Constant

The dissociation constant, represented as \( K_a \), is a measure of how much a weak acid dissociates into its ions in solution. It is an equilibrium constant specific for acids, indicating the extent of the reaction where a weak acid breaks into its respective ions. Mathematically, for a weak acid HA dissociating into \( H^+ \) and \( A^- \), the equation is:\[ K_a = \frac{[H^+][A^-]}{[HA]} \]This equation shows that the dissociation constant gives us insight into the acid's strength — the larger the \( K_a \), the stronger the acid. In our exercise, the difference in \( pK_a \) values between two weak acids, where \( pK_a = -\log_{10}(K_a) \), illustrates how low \( K_a \) values reflect greater differences in acid strengths and a greater tendency for the acid to remain undissociated. Understanding \( K_a \) is fundamental in chemistry, as it helps predict how acids will react in different environments and within various chemical reactions. By comparing \( K_a \) values across different acids, you can learn their relative strengths and behaviors in solutions, aiding in the design of chemical processes or understanding natural phenomena.