Question

Molar conductance of 0.1 M acetic acid is $7\,\,ohm^{-1} cm^{2} mol^{-1}$. If the molar conductance of acetic acid at infinite dilution is $380.8 ohm^{-1} cm^{2} mol^{-1}$, the value of dissociation constant will be:

• A. $226\times 10^{-5}\, mol \,dm^{-3}$
• B. $1.66 \times 10^{-3}\, mol\, dm^{-1}$
• C. $1.66 \times 10^{-2} \,mol\, dm^{-3}$
• D. $3.38 \times 10^{-5} \,mol\, dm^{-3}$

Answer 1

Answer: (D)

$K_{a}=C\alpha^{2}=0.1 \times\left(\frac{7}{380.8}\right)^{2} =3.38\times10^{-5}$