Plotting 1/Λm1/\Lambda_\mathrm{m}1/Λm against cΛmc\Lambda_\mathrm{m}cΛm 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\mathrm{P}/\mathrm{S}P/S is
[Λm\Lambda_\mathrm{m}Λm = molar conductivity; Λm∘\Lambda^\circ_\mathrm{m}Λm∘ = limiting molar conductivity; c = molar concentration; KaK_\mathrm{a}Ka = dissociation constant of HX]
Official answer: A
Ostwald's dilution law with α=Λm/Λm∘\alpha=\Lambda_m/\Lambda_m^\circα=Λm/Λm∘:Ka=cα21−α=cΛm2Λm∘(Λm∘−Λm).K_a=\frac{c\alpha^2}{1-\alpha}=\frac{c\Lambda_m^2}{\Lambda_m^\circ(\Lambda_m^\circ-\Lambda_m)}.Ka=1−αcα2=Λm∘(Λm∘−Λm)cΛm2.
Rearrange. Divide KaΛm∘2−KaΛm∘Λm=cΛm2K_a\Lambda_m^{\circ2}-K_a\Lambda_m^\circ\Lambda_m=c\Lambda_m^2KaΛm∘2−KaΛm∘Λm=cΛm2 by KaΛm∘2ΛmK_a\Lambda_m^{\circ2}\Lambda_mKaΛm∘2Λm:1Λm=1Λm∘+1KaΛm∘2 (cΛm).\frac1{\Lambda_m}=\frac1{\Lambda_m^\circ}+\frac{1}{K_a\Lambda_m^{\circ2}}\,(c\Lambda_m).Λm1=Λm∘1+KaΛm∘21(cΛm).
So P=1Λm∘P=\dfrac1{\Lambda_m^\circ}P=Λm∘1 and S=1KaΛm∘2S=\dfrac1{K_a\Lambda_m^{\circ2}}S=KaΛm∘21, givingPS=KaΛm∘.\frac PS=K_a\Lambda_m^\circ.SP=KaΛm∘.