Question

The equivalent conductivity of acetic acid at infinite dilution is $370Sc{m}^{2}e{q}^{-1}$. At the same temperature, 0.001 M solution of acetic acid, it is $55.5Sc{m}^{2}e{q}^{-1}$. What is the degree of dissociation of 0.1 N acetic acid? (Assume $1-\alpha \approx 1$)

• A. $0.15\%$
• B. $1.5\%$
  
• C. $0.0015\%$
  
• D. $15\%$

Answer 1

The correct option is B $1.5\%$


$\alpha =\frac{{\lambda }^c}{{\lambda }^{\circ }}=\frac{55.5}{370}=0.15$

$k_a=c{\alpha }^2=0.001\times (0.15{)}^2=0.1\times {\alpha }_1^2$

$\Rightarrow {\alpha }_1=0.015=1.5\%$