JEE Challenger
More from Electrochemistry
ChemMedAdv 2015 • Paper 2

Difference in pKa Values of Two Weak Monoprotic Acids from Conductivity

The molar conductivity of a 0.01 M0.01\text{ M} solution of a weak acid HX\text{HX} is found to be 1010 times smaller than that of a 0.10 M0.10\text{ M} solution of another weak acid HY\text{HY}. Assuming λX−0≈λY−0\lambda^0_{\text{X}^-} \approx \lambda^0_{\text{Y}^-} and that the degree of ionization for both acids is ≪1\ll 1, find the value of the difference pKa(HX)−pKa(HY)\text{p}K_a(\text{HX}) - \text{p}K_a(\text{HY}).

Official Numerical Answer3

Topics & Concepts

ElectrochemistryAcid-Base Equilibrium

Step-by-Step Solution

To find the difference pKa(HX)−pKa(HY)\text{p}K_a(\text{HX}) - \text{p}K_a(\text{HY}), we use the relation between molar conductivity, the degree of ionization (α\alpha), and the acid dissociation constant (KaK_a).

Step 1: Limiting Molar Conductivities The limiting molar conductivity of a weak monoprotic acid HA\text{HA} is given by Kohlrausch's law: Λm∘=λH+∘+λA−∘\Lambda_m^\circ = \lambda^\circ_{\text{H}^+} + \lambda^\circ_{\text{A}^-}

Given that λX−0≈λY−0\lambda^0_{\text{X}^-} \approx \lambda^0_{\text{Y}^-}, it follows that: Λm∘(HX)≈Λm∘(HY)=Λm∘\Lambda_m^\circ(\text{HX}) \approx \Lambda_m^\circ(\text{HY}) = \Lambda_m^\circ

Step 2: Degree of Ionization The degree of ionization α\alpha is related to the molar conductivity Λm\Lambda_m by: α=ΛmΛm∘\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}

Given that the molar conductivity of HX\text{HX} is 1010 times smaller than that of HY\text{HY}: Λm(HX)=110Λm(HY)\Lambda_m(\text{HX}) = \frac{1}{10}\Lambda_m(\text{HY})

Therefore, the ratio of their degrees of ionization is: αHXαHY=Λm(HX)Λm(HY)=110=10−1\frac{\alpha_{\text{HX}}}{\alpha_{\text{HY}}} = \frac{\Lambda_m(\text{HX})}{\Lambda_m(\text{HY})} = \frac{1}{10} = 10^{-1}

Step 3: Acid Dissociation Constant (KaK_a) For a weak acid with degree of ionization α≪1\alpha \ll 1: Ka=Cα21−α≈Cα2K_a = \frac{C\alpha^2}{1 - \alpha} \approx C\alpha^2

For HX\text{HX} with concentration C1=0.01 MC_1 = 0.01\text{ M}: Ka(HX)≈C1αHX2K_a(\text{HX}) \approx C_1 \alpha_{\text{HX}}^2

For HY\text{HY} with concentration C2=0.10 MC_2 = 0.10\text{ M}: Ka(HY)≈C2αHY2K_a(\text{HY}) \approx C_2 \alpha_{\text{HY}}^2

Taking the ratio of their dissociation constants: Ka(HX)Ka(HY)=(C1C2)(αHXαHY)2\frac{K_a(\text{HX})}{K_a(\text{HY})} = \left(\frac{C_1}{C_2}\right) \left(\frac{\alpha_{\text{HX}}}{\alpha_{\text{HY}}}\right)^2

Substitute the given values: C1C2=0.01 M0.10 M=10−1\frac{C_1}{C_2} = \frac{0.01\text{ M}}{0.10\text{ M}} = 10^{-1} (αHXαHY)2=(10−1)2=10−2\left(\frac{\alpha_{\text{HX}}}{\alpha_{\text{HY}}}\right)^2 = (10^{-1})^2 = 10^{-2}

Thus: Ka(HX)Ka(HY)=10−1×10−2=10−3\frac{K_a(\text{HX})}{K_a(\text{HY})} = 10^{-1} \times 10^{-2} = 10^{-3}

Step 4: Calculation of pKa(HX)−pKa(HY)\text{p}K_a(\text{HX}) - \text{p}K_a(\text{HY}) pKa(HX)−pKa(HY)=−log⁡10Ka(HX)−(−log⁡10Ka(HY))\text{p}K_a(\text{HX}) - \text{p}K_a(\text{HY}) = -\log_{10} K_a(\text{HX}) - \left(-\log_{10} K_a(\text{HY})\right) pKa(HX)−pKa(HY)=−log⁡10(Ka(HX)Ka(HY))\text{p}K_a(\text{HX}) - \text{p}K_a(\text{HY}) = -\log_{10}\left(\frac{K_a(\text{HX})}{K_a(\text{HY})}\right) pKa(HX)−pKa(HY)=−log⁡10(10−3)=3\text{p}K_a(\text{HX}) - \text{p}K_a(\text{HY}) = -\log_{10}(10^{-3}) = 3

PreviousNext