Question

The ionization constant $\left(K_a\right)$ of a weak electrolyte is $2.5\times {10}^{-5}$, while ${\Lambda }_{eq}$ of its 0.01 M solution is $19.65Sc{m}^{2}e{q}^{-1},{\Lambda }_{eq}^{\circ }$ is :

• A. $288Sc{m}^{2}e{q}^{-1}$
• B. $393Sc{m}^{2}e{q}^{-1}$
• C. $198Sc{m}^{2}e{q}^{-1}$
• D. None of these

Answer 1

The correct option is B $393Sc{m}^{2}e{q}^{-1}$

We know for weak 1:1 electrolyte,

$K_a=c{\alpha }^2$

$2.5\times {10}^{-5}={10}^{-2}\times {\alpha }^2\Rightarrow \alpha =5\times {10}^{-2}$

$\alpha =\frac{{\Lambda }_{eq}}{{\Lambda }_{eq}^{\circ }}$

$5\times {10}^{-2}=\frac{19.65}{{\Lambda }_{eq}^{\circ }}\Rightarrow {\Lambda }_{eq}^{\circ }=393Sc{m}^{2}e{q}^{-1}$