The specific conductance of 0.0025M acetic acid is $$5 \times 10^{-5}$$ S cm$$^{-1}$$ at a certain temperature. The dissociation constant of acetic acid is _______ $$\times 10^{-7}$$. (Nearest integer)
Consider limiting molar conductivity of CH$$_3$$COOH as 400 S cm$$^2$$ mol$$^{-1}$$

We need to find the dissociation constant of acetic acid given its specific conductance and limiting molar conductivity.

Concentration $$C = 0.0025$$ M, Specific conductance $$\kappa = 5 \times 10^{-5}$$ S cm$$^{-1}$$, Limiting molar conductivity $$\Lambda_m^{\circ} = 400$$ S cm$$^2$$ mol$$^{-1}$$.

The molar conductivity is related to specific conductance by:

$$\Lambda_m = \frac{\kappa \times 1000}{C}$$

where $$C$$ is in mol/L and $$\kappa$$ is in S cm$$^{-1}$$. The factor of 1000 converts L to cm$$^3$$.

$$\Lambda_m = \frac{5 \times 10^{-5} \times 1000}{0.0025} = \frac{5 \times 10^{-2}}{2.5 \times 10^{-3}} = 20 \text{ S cm}^2 \text{mol}^{-1}$$

The degree of dissociation is the ratio of molar conductivity to limiting molar conductivity:

$$\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}} = \frac{20}{400} = 0.05$$

For a weak electrolyte $$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$$, the dissociation constant is:

$$K_a = \frac{C\alpha^2}{1 - \alpha}$$

Substituting the values:

$$K_a = \frac{0.0025 \times (0.05)^2}{1 - 0.05} = \frac{0.0025 \times 0.0025}{0.95}$$ $$= \frac{6.25 \times 10^{-6}}{0.95} = 6.579 \times 10^{-6} \approx 65.79 \times 10^{-7}$$

Rounding to the nearest integer: $$K_a \approx 66 \times 10^{-7}$$.

The dissociation constant is 66 $$\times 10^{-7}$$.