Question

The dissociation constant of a weak monoprotic acid is $1.6 \times 10^{-5}$ and its molar conductance at infinite dilution is $360.5 \times 10^{-4}$ mho m$^2$ mol$^{-1}$. For 0.01 M solution of this acid, the specific conductance is $n \times 10^{-2}$ mho m$^{-1}$. The value of $n$ (rounded off to two decimal places) is ________.

Hint:

For weak electrolytes: $\Lambda_m = \Lambda_m^\infty \sqrt{\frac{K_a}{c}}$ and $\kappa = c\Lambda_m$.

Answer 1

Correct Answer: 1.44

Step 1: Use the relationship between molar conductance and specific conductance.
\[ \kappa = \Lambda_m \times c \] where $\Lambda_m = \alpha \Lambda_m^\infty$ and $\alpha = \sqrt{\frac{K_a}{c}}$.
Step 2: Substitute known values.
\[ \alpha = \sqrt{\frac{1.6 \times 10^{-5}}{0.01}} = 0.04 \] \[ \Lambda_m = 0.04 \times 360.5 \times 10^{-4} = 14.42 \times 10^{-4} \] \[ \kappa = c \times \Lambda_m = 0.01 \times 14.42 \times 10^{-4} = 1.42 \times 10^{-2} \] Step 3: Conclusion.
The value of $n$ = 1.42.