Answer

The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

Answer

$ \wedge_m^{o}({H}2 {O})=\lambda{{H}^{+}}^{o}+\lambda^{o} {OH}^{-} $

$ \text {We find out ;} \quad \wedge_m^{o}({HCl}), \wedge_m^{o}({NaOH}) \text { and } \wedge_m^{o}({NaCl}) .$

$\text {Then,} \wedge_m^{o}({H}_2 {O})=\wedge_m^{o}({HCl})+\wedge_m^{o}({NaOH})-\wedge_m^{o}({NaCl})$

Answer

$C=0.025 {~mol} {~L}^{-1}$

$\Lambda_{m}=46.1 {\hspace{0.5mm} S\hspace{0.5mm}cm}^{2} {~mol}^{-1}$

$\lambda^{o}\left({H}^{+}\right)=349.6 {\hspace{0.5mm} S\hspace{0.5mm}cm}^{2} {~mol}^{-1}$

$\lambda^{o}\left({HCOO}^{-}\right)=54.6 {\hspace{0.5mm} S\hspace{0.5mm}cm}^{2} {~mol}^{-1}$

$\Lambda_{m}^{o}({HCOOH})=\lambda^{o}\left({H}^{+}\right)+\lambda^{o}\left({HCOO}^{-}\right)$

$\quad\quad\quad\quad\quad\quad\quad=349.6+54.6$

$\quad\quad\quad\quad\quad\quad\quad=404.2 {\hspace{0.5mm}S\hspace{0.5mm}cm}^{2} {~mol}^{-1}$

Now, degree of dissociation:

$ \begin{aligned} \alpha & =\frac{\Lambda_{m}({HCOOH})}{\Lambda_{m}^{o}({HCOOH})} \\ \\ & =\frac{46.1}{404.2} \\ \\ & =0.114 \text { (approximately) } \end{aligned} $

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

$\text{Initial conc.} \quad\quad\quad\quad\quad c {~mol} {~L}^{-1}$

$\text{Conc. at eqm.} \quad\quad\quad\quad\hspace{0.1cm} c(1-\alpha) \quad\quad c \alpha \quad\quad\quad c \alpha$

Thus, dissociation constant:

$ K_a =\dfrac{c \propto^{2}}{(1-\propto)} $

$K_a =\dfrac{\left(0.025 \right)(0.114)^{2}}{(1-0.114)} $

$K_a =3.67 \times 10^{-4} $