Question

Equivalent conductance of $0.05 N$ solution of acetic acid is $7.36\, S\, cm^{2}$ at $25^{\circ}C$. What is the value of $K_{a}$ for this acetic acid solution if equivalent conductance of acetic acid at infinite dilution is $390.7 \,S\, cm^{2}$?

• A. $9.4 \times 10^{-4}$
• B. $7.1 \times 10^{-4}$
• C. $1.77 \times 10^{-5}$
• D. $4.7 \times 10^{-5}$

Answer 1

Answer: (C)

Degree of dissociation
$\left(\alpha\right) =\frac{\wedge^{c}_{m}}{\wedge^{\circ}_{m}}$
For monobasic acetic acid, molar conductance = equivalent conductance
$\therefore \alpha = \frac{7.36}{390.7} = 1.88 \times10^{-2}$
For weak acid solution dissociation constant $\left(K\right) = C\alpha^{2}$
$= 0.05\left(1.88 \times 10^{-2}\right)^{2} = 1.77 \times10^{-5}$