The molar conductivity of $0.007\text{M}$ acetic acid is $20{\text{S cm}}^2{\text{mol}}^{-1}$. What is the dissociation constant of acetic acid? Choose the correct option. ${\lambda }^{\circ }(\mathrm{H}{}^{+})=350{\text{S cm}}^2{\text{mol}}^{-1}$, ${\lambda }^{\circ }(\mathrm{CH}{}_3\mathrm{COO}{}^{-})=50{\text{S cm}}^2{\text{mol}}^{-1}$

Answer: (A) $1.75\times 10^{-5}{\text{mol L}}^{-1}$. \textbf{Answer:} (A) $K_a=1.75\times 10^{-5}{\text{mol L}}^{-1}$ \textbf{Solution:} Limiting molar conductivity by Kohlrausch's law: ${\Lambda }_m^{\circ }(\mathrm{CH}{}_3\mathrm{COOH})={\lambda }^{\circ }(\mathrm{H}{}^{+})+{\lambda }^{\circ }(\mathrm{CH}{}_3\mathrm{COO}{}^{-})=350+50=400{\text{S cm}}^2{\text{mol}}^{-1}$ Degree of dissociation: $\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{20}{400}=0.05$ Since $\alpha$ is small, $1-\alpha \approx 1$: $K_a=\frac{c{\alpha }^2}{1-\alpha }\approx c{\alpha }^2=0.007\times (0.05{)}^2=1.75\times 10^{-5}{\text{mol L}}^{-1}$