Plotting 1 / Lambda_{m} against c Lambda_{m} | vedclass

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(A) For a weak acid,the degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$.The dissociation constant $K_a$ is given by $K_a = \frac

Vedclass · Accepted Answer

$K_{a} \Lambda_{m}^{\circ}$

Vedclass · Answer

(A) For a weak acid,the degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$.The dissociation constant $K_a$ is given by $K_a = \frac{c \alpha^2}{1 - \alpha}$.Substituting $\alpha$,we get $K_a = \frac{c (\Lambda_m / \Lambda_m^{\circ})^2}{1 - (\Lambda_m / \Lambda_m^{\circ})} = \frac{c \Lambda_m^2}{\Lambda_m^{\circ}(\Lambda_m^{\circ} - \Lambda_m)}$.Rearranging gives $K_a \Lambda_m^{\circ} (\Lambda_m^{\circ} - \Lambda_m) = c \Lambda_m^2$.Dividing by $K_a \Lambda_m \Lambda_m^{\circ}$,we get $\frac{\Lambda_m^{\circ} - \Lambda_m}{\Lambda_m} = \frac{c \Lambda_m}{K_a \Lambda_m^{\circ}}$,which simplifies to $\frac{\Lambda_m^{\circ}}{\Lambda_m} - 1 = \frac{c \Lambda_m}{K_a \Lambda_m^{\circ}}$.Dividing by $\Lambda_m^{\circ}$,we get $\frac{1}{\Lambda_m} = \frac{c \Lambda_m}{K_a (\Lambda_m^{\circ})^2} + \frac{1}{\Lambda_m^{\circ}}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = 1 / \Lambda_m$ and $x = c \Lambda_m$:Slope $S = \frac{1}{K_a (\Lambda_m^{\circ})^2}$ and $y$-intercept $P = \frac{1}{\Lambda_m^{\circ}}$.Therefore,the ratio $P / S = \frac{1 / \Lambda_m^{\circ}}{1 / (K_a (\Lambda_m^{\circ})^2)} = K_a \Lambda_m^{\circ}$.

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