Question

The molar conductivity of $0.025mol{L}^{-1}$ methanoic acid is $46.1Sc{m}^{2}mo{l}^{-1}$.

Calculate its degree of dissociation and dissociation constant.

Given ${\lambda }_{(H-)}^o=349.6Sc{m}^{2}mo{l}^{-1}$ and ${\lambda }_{\left(HCO{O}^{-}\right)}^o=54.6Sc{m}^{2}mo{l}^{-1}$.

• A. $0.114$ and $3.67\times {10}^{-4}$
• B. $0.124$ and $3.67\times {10}^{-4}$
• C. $0.114$ and $2.67\times {10}^{-4}$
• D. None of these

Answer 1

The correct option is A $0.114$ and $3.67\times {10}^{-4}$

We are given

${\lambda }^0$ for $\left(H^{+}\right)=349.6$ $S$ ${cm}^2/mol$

${\lambda }^0$ for $\left({HCOO}^{-}\right)$ $=$ $54.6$ $S$ ${cm}^2/mol$

${\lambda }_m^0\left(HCOOH\right)={\lambda }^0\left(H^{+}\right)+{\lambda }^0\left({HCOO}^{-}\right)$

$=349.6+54.6=404.2$ $S$ ${cm}^2/mol$

Degree of dissociation $\left(\alpha \right)=\frac{{\Lambda }^{c}m}{{\Lambda }^{0}m}=\frac{46.1}{404.2}=0.114$

Calculation of dissociation constant :-

$HCOOH\rightleftharpoons {HCOO}^{-}+H^{+}$

$C$ $0$ $0$

$C-\alpha$ $C\alpha$ $C\alpha$

$K=\frac{{C\alpha }^2}{1-\alpha }=\frac{0.025\times {\left(0.114\right)}^2}{1-0.114}=3.67\times {10}^{-4}$