Question

The molar conductivity of 0.0025 mol/lit methanoic acid is 46.1 s cm/mol. Calculate it's degree of dissociation and dissociation constant. Given $\left[H^{+}\right]=349.6$ S-cm/mol and $\left[HCO{O}^{-}\right]=54.6$ S-cm/mol

Answer 1

Given:

$c=0.0025mol/L$

${\Lambda }_m=46.1S{cm}^2{mol}^{-1}$

${\Lambda }^{\circ }\left(H^{+}\right)=349.6S{cm}^2{mol}^{-1}$

${\Lambda }^{\circ }\left({HCOO}^{-}\right)=54.6{cm}^2{mol}^{-1}$

${\Lambda }_m^{\circ }\left(HCOOH\right)$ $={\Lambda }^{\circ }\left(H^{+}\right)+{\Lambda }^{\circ }\left({HCOO}^{-}\right)=349.6+54.6=404.2S{cm}^2{mol}^{-1}$

As we know that,

Degree of dissociation $\left(\alpha \right)=\frac{{\Lambda }^{\circ }\left(HCOOH\right)}{{\Lambda }_m^{\circ }\left(HCOOH\right)}=\frac{46.1}{404.2}\approx 0.114$

Therefore,

Dissociation constant $\left(K\right)=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.0025\times {\left(0.114\right)}^2}{1-0.114}=\frac{3.25\times {10}^{-5}}{0.886}=3.67\times {10}^{-5}mol/L$

Hence the value of degree of dissociation and dissociation constant is $0.114$ and $3.67\times {10}^{-5}mol/L$ respectively.