Verified Solution • 2025–26 Session moderate

QJEEAdv23Ch15f0: JEE Advanced 2023 Chemistry - Class 12 JEEAdv23Ch15f0 Solution: Plotting against for aqueous solutions of a monobasic weak acid …

Plotting 1/Λm1 / \Lambda_{\mathrm{m}} against cΛm\mathrm{c} \Lambda_{\mathrm{m}} for aqueous solutions of a monobasic weak acid (HX)(\mathrm{HX}) resulted in a straight line with y\mathrm{y} -axis intercept of P\mathrm{P} and slope of S\mathrm{S} . The ratio P/S\mathrm{P} / \mathrm{S} is

amp;[Λm= molar conductivity amp;Λmo= limiting molar conductivity amp;c= molar concentration amp;Ka= dissociation constant of HX]\begin{aligned}& {\left[\Lambda_{\mathrm{m}}=\right.\text { molar conductivity }} \\\\& \Lambda_{\mathrm{m}}^{\mathrm{o}}=\text { limiting molar conductivity } \\\\& \mathrm{c}=\text { molar concentration } \\\\& \left.\mathrm{K}_{\mathrm{a}}=\text { dissociation constant of } \mathrm{HX}\right]\end{aligned}

Multiple Choice Options

  • A KaΛmo\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}}
  • B KaΛmo/2\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}} / 2
  • C 2 KaΛmo2 \mathrm{~K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}}
  • D 1/(KaΛmo)1 /\left(\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\mathrm{o}}\right)

Step-by-Step Verified Solution

For weak acid, α=ΛmΛ0\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}

amp;Ka=Cα21−α⇒Ka(1−α)=Cα2amp;⇒Ka(1−ΛmΛ0)=C(ΛmΛ0)2amp;⇒Ka−ΛmKaΛ0=CΛm2(Λ0)2\begin{aligned}& \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \Rightarrow \mathrm{K}_{\mathrm{a}}(1-\alpha)=\mathrm{C} \alpha^2 \\\\& \Rightarrow \mathrm{K}_{\mathrm{a}}\left(1-\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}\right)=\mathrm{C}\left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_0}\right)^2 \\\\& \Rightarrow \mathrm{K}_{\mathrm{a}}-\frac{\Lambda_{\mathrm{m}} \mathrm{K}_{\mathrm{a}}}{\Lambda_0}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}^2}{\left(\Lambda_0\right)^2}\end{aligned}

Divide by ' Λm\Lambda_{\mathrm{m}} '

amp;⇒KaΛm=CΛm(Λ0)2+KaΛ0amp;⇒1Λm=CΛmKa(Λ0)2+1Λ0\begin{aligned}& \Rightarrow \frac{\mathrm{K}_{\mathrm{a}}}{\Lambda_{\mathrm{m}}}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}}{\left(\Lambda_0\right)^2}+\frac{\mathrm{K}_{\mathrm{a}}}{\Lambda_0} \\\\& \Rightarrow \frac{1}{\Lambda_{\mathrm{m}}}=\frac{\mathrm{C} \Lambda_{\mathrm{m}}}{\mathrm{K}_{\mathrm{a}}\left(\Lambda_0\right)^2}+\frac{1}{\Lambda_0}\end{aligned}

Plot 1Λm\frac{1}{\Lambda_{\mathrm{m}}} vs CΛm\mathrm{C} \Lambda_{\mathrm{m}} has

Slope =1 Ka(Λ0)2=S=\frac{1}{\mathrm{~K}_{\mathrm{a}}\left(\Lambda_0\right)^2}=\mathrm{S}

y\mathrm{y} -intercept =1Λ0=P=\frac{1}{\Lambda_0}=\mathrm{P}

Then, PS=1Λ01 Ka(Λ0)2=KaΛ0\frac{\mathrm{P}}{\mathrm{S}}=\frac{\frac{1}{\Lambda_0}}{\frac{1}{\mathrm{~K}_{\mathrm{a}}\left(\Lambda_0\right)^2}}=\mathrm{K}_{\mathrm{a}} \Lambda_0

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