m0 for NaCl, HCl and NaA are 126.4, 425.9 and 100.5\ S\ cm2\ mol-1 respectively. If the conductivity of 0.001 M HA is…

Strategy: Use Kohlrausch's Law to find $\Lambda_m^0$ of the weak acid $\ce{HA}$ from its salts. Then calculate the \degree of dissociation $\alpha = \Lambda_m / \Lambda_m^0$.

Step 1: Calculate $\Lambda_m^0(\ce{HA})$ $\Lambda_m^0(\ce{HA}) = \Lambda_m^0(\ce{HCl}) + \Lambda_m^0(\ce{NaA}) - \Lambda_m^0(\ce{NaCl})$ $\Lambda_m^0 = 425.9 + 100.5 - 126.4 = 400.0 \mathrm{S cm^2 mol^{-1}}$

Step 2: Calculate specific $\Lambda_m$ $C = 0.001 \text{ M}$, $\kappa = 5 \times 10^{-5} \text{ S/cm}$. $\Lambda_m = \frac{1000 \times 5 \times 10^{-5}}{0.001} = 50 \mathrm{S cm^2 mol^{-1}}$

Step 3: Solve for $\alpha$ $\alpha = \frac{50}{400} = 0.125$

$\boxed{\text{Answer: (a) 0.125}}$