JEE Challenger
More from Electrochemistry
ChemMedAdv 2023 • Paper 1

Ratio of Intercept to Slope for Weak Acid Molar Conductivity Plot

Plotting 1/Λm1/\Lambda_m against cΛmc\Lambda_m for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio P/S is

[Λm=molar conductivity[\Lambda_m = \text{molar conductivity} Λmo=limiting molar conductivity\Lambda_m^o = \text{limiting molar conductivity} c=molar concentrationc = \text{molar concentration} Ka=dissociation constant of HX]K_a = \text{dissociation constant of HX}]

Options

A

KaΛmoK_a \Lambda_m^o

Correct
B

KaΛmo/2K_a \Lambda_m^o / 2

C

2KaΛmo2 K_a \Lambda_m^o

D

1/(KaΛmo)1 / (K_a \Lambda_m^o)

Topics & Concepts

ElectrochemistryAcid-Base EquilibriumElectrochemical Cell

Step-by-Step Solution

For a monobasic weak acid (HX\text{HX}), the dissociation equilibrium is given by: HX(aq)⇌H+(aq)+X−(aq)\text{HX(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{X}^-\text{(aq)}

The degree of dissociation (α\alpha) of the weak acid at concentration cc is related to its molar conductivity (Λm\Lambda_m) and limiting molar conductivity (Λmo\Lambda_m^o) by Kohlrausch's law: α=ΛmΛmo\alpha = \frac{\Lambda_m}{\Lambda_m^o}

The acid dissociation constant (KaK_a) according to Ostwald's dilution law is expressed as: Ka=cα21−αK_a = \frac{c \alpha^2}{1 - \alpha}

Substituting α=ΛmΛmo\alpha = \frac{\Lambda_m}{\Lambda_m^o} into the KaK_a expression: Ka=c(ΛmΛmo)21−ΛmΛmo=cΛm2Λmo(Λmo−Λm)K_a = \frac{c \left(\frac{\Lambda_m}{\Lambda_m^o}\right)^2}{1 - \frac{\Lambda_m}{\Lambda_m^o}} = \frac{c \Lambda_m^2}{\Lambda_m^o (\Lambda_m^o - \Lambda_m)}

Rearranging the above equation: 1Ka=Λmo(Λmo−Λm)cΛm2=(Λmo)2−ΛmoΛmcΛm2\frac{1}{K_a} = \frac{\Lambda_m^o (\Lambda_m^o - \Lambda_m)}{c \Lambda_m^2} = \frac{(\Lambda_m^o)^2 - \Lambda_m^o \Lambda_m}{c \Lambda_m^2}

Multiplying both sides by cΛmc \Lambda_m: cΛmKa=(Λmo)2Λm−Λmo\frac{c \Lambda_m}{K_a} = \frac{(\Lambda_m^o)^2}{\Lambda_m} - \Lambda_m^o

Dividing both sides by (Λmo)2(\Lambda_m^o)^2: cΛmKa(Λmo)2=1Λm−1Λmo\frac{c \Lambda_m}{K_a (\Lambda_m^o)^2} = \frac{1}{\Lambda_m} - \frac{1}{\Lambda_m^o}

Rearranging to isolate 1Λm\frac{1}{\Lambda_m}: 1Λm=1Ka(Λmo)2(cΛm)+1Λmo\frac{1}{\Lambda_m} = \frac{1}{K_a (\Lambda_m^o)^2} (c \Lambda_m) + \frac{1}{\Lambda_m^o}

This equation is of the linear form y=Sx+Py = S x + P, where:

  • y=1Λmy = \frac{1}{\Lambda_m}
  • x=cΛmx = c \Lambda_m
  • Slope, S=1Ka(Λmo)2S = \frac{1}{K_a (\Lambda_m^o)^2}
  • yy-axis intercept, P=1ΛmoP = \frac{1}{\Lambda_m^o}

Now, evaluating the ratio of the intercept PP to the slope SS: PS=1Λmo1Ka(Λmo)2=Ka(Λmo)2Λmo=KaΛmo\frac{P}{S} = \frac{\frac{1}{\Lambda_m^o}}{\frac{1}{K_a (\Lambda_m^o)^2}} = \frac{K_a (\Lambda_m^o)^2}{\Lambda_m^o} = K_a \Lambda_m^o

Thus, the correct option is (A).

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