In a study of the dissociation of n-butyric acid in water at 298K, a graph of $1/{\mathrm{\Lambda }}_{\mathrm{C}}$ vs ${\mathrm{\Lambda }}_c\times C$ at various concentrations yielded a straight line with an intercept equal to 2.62 × 10^-3 and slope equal to 0.455 (concentration is expressed as mol dm^-3 and ${\mathrm{\Lambda }}_{\mathrm{C}}$ as ohm^-1 cm² mol^-1). Identify correct statements among the following

To solve this problem, we have

                                      $\alpha =\frac{{\mathrm{\Lambda }}_C}{{\mathrm{\Lambda }}_{\mathrm{\infty }}}; K_a=\frac{{\alpha }^{2}c}{1-\alpha }$

Rewriting      $\begin{array}{l}{K}_a\text{ as, }{K}_a=\frac{\alpha }{1-\alpha }\times \alpha c\text{ or }\frac{\alpha c}{\frac{1}{\alpha }-1}=K_a\\ \frac{1}{\alpha }-1=\frac{\alpha c}{K_a} \text{ or }\frac{1}{\alpha }=\left(\frac{\alpha c}{K_a}+1\right)\end{array}$

$\text{ Using }\alpha =\frac{{\mathrm{\Lambda }}_C}{{\mathrm{\Lambda }}_{\mathrm{\infty }}}\text{ we have }\frac{{\mathrm{\Lambda }}_{\mathrm{\infty }}}{{\mathrm{\Lambda }}_C}=\left(\frac{\frac{{\mathrm{\Lambda }}_C}{{\mathrm{\Lambda }}_{\mathrm{\infty }}}\times C}{K_a}+1\right)$

or $\frac{1}{{\mathrm{\Lambda }}_C}=\frac{{\mathrm{\Lambda }}_{C}C}{K_a{\left({\mathrm{\Lambda }}_{\mathrm{\infty }}\right)}^2}+\frac{1}{{\mathrm{\Lambda }}_{\mathrm{\infty }}}$

$\text{ A plot of }\frac{1}{{\mathrm{\Lambda }}_C}Vs{\mathrm{\Lambda }}_C\times C\text{ is linear with an intercept of }\frac{1}{{\mathrm{\Lambda }}_{\mathrm{\infty }}}\text{ and slope of }\frac{1}{K_a{\mathrm{\Lambda }}_{\mathrm{\infty }}^2}\text{. }$

(1) From the given Problem,

$\text{ intercept }=\frac{1}{{\mathrm{\Lambda }}_{\mathrm{\infty }}}=2.62\times {10}^{-3};{\mathrm{\Lambda }}_{\mathrm{\infty }}=\frac{1}{2.62\times {10}^{-3}}=381.7\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$

(2) $\text{ Slope }=0.455=\frac{1}{K_a(381.7{)}^2}\text{ or }{K}_a=\frac{1}{0.455\times (381.7{)}^2}=1.51\times {10}^{-5}.$

(3) ${\alpha }^{2}C\approx 1.51\times {10}^{-5}{\alpha }^2=\frac{1.51\times {10}^{-5}}{{10}^{-1}}=1.51\times {10}^{-4}$

$\alpha =\sqrt{1.51\times {10}^{-4}}=1.23\times {10}^{-2}$

(4) $$\begin{gathered} \mathrm{\kappa }(\mathrm{specific} \mathrm{conductance}) \text{ at }0.02\mathrm{M} \\ {\mathrm{\Lambda }}_{\mathrm{C}}=\mathrm{\alpha }\times {\mathrm{\Lambda }}_{\mathrm{\infty }}\text{ or }\mathrm{\alpha }=\sqrt{\frac{1.51\times {10}^{-5}}{2\times {10}^{-2}}}=2.75\times {10}^{-2} \end{gathered}$$

$\begin{array}{l}\therefore {\mathrm{\Lambda }}_C=2.75\times {10}^{-2}\times 381.7=10.55{\mathrm{cm}}^2{\mathrm{eq}}^{-1}\\ k=\frac{C{\mathrm{\Lambda }}_C}{1000}=\frac{0.02\times 10.5}{1000}=2.1\times {10}^{-4}\mathrm{S}{\mathrm{cm}}^{-1}\\ =2.1\times {10}^{-2}{\mathrm{ohm}}^{-1}{\mathrm{m}}^{-1}\end{array}$