Question

For $0.0128$ N solution of acetic acid at ${25}^{\circ }C$, equivalent conductance of the solution is ${\lambda }_{eq}=1.4mhoc{m}^{2}e{q}^{-1}$ and ${\lambda }^{\infty }=391mhoc{m}^{2}e{q}^{-1}$. Calculate dissociation constant ($K_a$) of acetic acid.

• A. $1.6\times {10}^{-7}$
• B. $1.6\times {10}^{-8}$
• C. $1.6\times {10}^{-6}$
• D. $3.2\times {10}^{-7}$

Answer 1

The correct option is A $1.6\times {10}^{-7}$

As we know,

$\alpha =\Lambda /{\Lambda }^0=1.4/391$

Now,

Dissociation constant, $K_a=C{\alpha }^2=0.0128\times (1.4/391{)}^2=1.64\times {10}^{-7}$