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JEE Advanced 2015Paper 2
Exam
JEE Advanced
Year
2015
Paper
Paper 2
Subject
Chemistry
Chapter
Electrochemistry

JEE Advanced 2015 · Paper 2 · Chemistry · Electrochemistry

The molar conductivity of a solution of a weak acid HX (0.01 M) is...

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 (0.10 M). If λX−0≈λY−0{\lambda}_{{X}^{-}}^{0} \approx {\lambda}_{{Y}^{-}}^{0} , the difference in their pKap{K}_{a} values, pKa(HX)−pKa(HY)p{K}_{a}\left(HX\right) - p{K}_{a} ( HY ) , is (Consider degree of ionization of both acids to << 1)

Correct answer

3

Step-by-step solution

λX−0≈λY−0{\lambda}_{{X}^{-}}^{0} \approx {\lambda}_{{Y}^{-}}^{0}
⇒λH+0+λX−0≈λH+0+λY−0⇒ {\lambda}_{{H}^{+}}^{0} + {\lambda}_{{X}^{-}}^{0} \approx {\lambda}_{{H}^{+}}^{0} + {\lambda}_{{Y}^{-}}^{0}
⇒λHX0≈λHY0⇒ {\lambda}_{HX}^{0} \approx {\lambda}_{HY}^{0} ...... (i)
Also, λmλm0=α\dfrac{{\lambda}_{m}}{{\lambda}_{m}^{0}} = \alpha ,
So λm(HX)=λm0α1{\lambda}_{m}\left(HX\right) = {\lambda}_{m}^{0}{\alpha}_{1} and λm(HY)=λm0α2{\lambda}_{m}\left(HY\right) = {\lambda}_{m}^{0}{\alpha}_{2}
(Where α1{\alpha}_{1} and α2{\alpha}_{2} are degree of dissociation of HX and HY respectively)
Now, given that
λm(HY)=10λm(HX){\lambda}_{m}\left(HY\right) = 10{\lambda}_{m}\left(HX\right)
⇒λm0α2=10×λm0α1⇒ {\lambda}_{m}^{0}{\alpha}_{2} = 10 \times {\lambda}_{m}^{0}{\alpha}_{1}
α2=10α1{\alpha}_{2} = 10{\alpha}_{1} ...... (ii)
Ka=Cα21−α{K}_{a} = \dfrac{C{\alpha}^{2}}{1 - \alpha} , But α≪1\alpha ≪ 1 , therefore Ka=Cα2{K}_{a} = C{\alpha}^{2}
⇒Ka(HX)Ka(HY)=0.01α120.1α22=0.010.1×(110)2=11000⇒ \dfrac{{K}_{a} ( HX )}{{K}_{a} ( HY )} = \dfrac{0.01{\alpha}_{1}^{2}}{0.1{\alpha}_{2}^{2}} = \dfrac{0.01}{0.1} \times {\left(\dfrac{1}{10}\right)}^{2} = \dfrac{1}{1000}
⇒[log⁡(Ka(HX)−log⁡(Ka(HY)]=−3⇒ [ \log ( {K}_{a}\left(HX\right) - \log \left({K}_{a}\left(HY\right)\right] = - 3
⇒pKa(HX)−pKa(HY)=3⇒ p{K}_{a}\left(HX\right) - p{K}_{a}\left(HY\right) = 3
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