The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol−1. What is the dissociation constant of acetic acid ? Choose the correct option.

The problem gives:

• Concentration of acetic acid, $C\; =\; 0.007 mol L⁻¹$
• Molar conductivity of the solution, $\Lambda \_m\; =\; 20 S cm² mol⁻¹$

For a weak electrolyte the molar conductivity is related to the degree of dissociation ($\alpha$) and the limiting molar conductivity ($\Lambda \_0$) by

$\Lambda \_m\; =\; \alpha ·\Lambda \_0$

The dissociation constant ($K\_a$) of a weak acid is

$K\_a\; =\; \alpha ²·C$

Hence we first need $\alpha$. From the image we read the tabulated value of the limiting molar conductivity of acetic acid:

$\Lambda \_0\; =\; 400 S cm² mol⁻¹$

Now compute the degree of dissociation:

$\alpha \; =\; \backslash frac\{\Lambda \_m\}\{\Lambda \_0\}\; =\; \backslash frac\{20\}\{400\}\; =\; 0.05$

Finally, substitute $\alpha$ and $C$ into the expression for $K\_a$:

$K\_a\; =\; \alpha ²·C\; =\; (0.05)²·0.007$

$K\_a\; =\; 0.0025·0.007\; =\; 1.75 \times 10⁻⁵ mol L⁻¹$

Thus the dissociation constant of acetic acid is 1.75 × 10⁻⁵ mol L⁻¹, which matches the provided correct option.