Question

Molar conductance of $0.1M$ acetic acid is $7oh{m}^{-1}c{m}^{2}mo{l}^{-1}$. If the molar conductance of acetic acid at infinite dilution is $380.80oh{m}^{-1}c{m}^{2}mo{l}^{-1}$, the value of dissociation constant will be:

• A. $2.26\times {10}^{-5}mold{m}^{-3}$
• B. $1.66\times {10}^{-5}mold{m}^{-3}$
• C. $1.66\times {10}^{-2}mold{m}^{-3}$
• D. $3.4\times {10}^{-5}mold{m}^{-3}$

Answer 1

The correct option is D $3.4\times {10}^{-5}mold{m}^{-3}$

Given:
$\text{Molar conductance,}{\Lambda }_m=7oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
$\text{Limiting molar conductance,}{\Lambda }_m^0=380.80oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Degree of dissociation,
$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}$

$\alpha =\frac{7}{380.8}=0.0183$

since, acetic acid is a weak acid

$\text{Dissociation constant,}K=C{\alpha }^2$

$K=0.1\times {\left(0.0183\right)}^2$

$K=3.4\times {10}^{-5}mold{m}^{-3}$