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

Applying Kohlrausch’s law of independent migration of ions, the $\Lambda_{m}^{0}$ value of water can be determined as follows:

$$ \begin{aligned} \Lambda_{m\left(\mathrm{H_2} \mathrm{O}\right)}^{0} & =\lambda_{\mathrm{H}^{+}}^{0}+\lambda_{\mathrm{OH}^{-}}^{0} \\ & =\left(\lambda_{\mathrm{H}^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}\right)+\left(\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{OH}^{-}}^{0}\right)-\left(\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}\right) \\ \Lambda_{m(\mathrm{HCl})}^{0} & +\Lambda_{m(\mathrm{NaOH})}^{0}-\Lambda_{m(\mathrm{NaCl})}^{0} \end{aligned} $$

Hence, by knowing the $\Lambda_{m}^{0}$ values of $\mathrm{HCl}, \mathrm{NaOH}$, and $\mathrm{NaCl}$, the $\Lambda_{m}^{0}$ value of water can be determined.

Answer

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

$\Lambda_{m}=46.1 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

$\lambda^{0}\left(\mathrm{H}^{+}\right)=349.6 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

$\lambda^{0}\left(\mathrm{HCOO}^{-}\right)=54.6 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

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

$=349.6+54.6$

$=404.2 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$

Now, degree of dissociation:

$ \begin{aligned} \alpha & =\frac{\Lambda_{m}(\mathrm{HCOOH})}{\Lambda_{m}^{0}(\mathrm{HCOOH})} \\ \\ & =\frac{46.1}{404.2} \\ \\ & =0.114 \text { (approximately) } \end{aligned} $

Thus, dissociation constant:

$ \begin{aligned} K & =\frac{c \propto^{2}}{(1-\propto)} \\ \\ & =\frac{\left(0.025 \mathrm{~mol} \mathrm{~L}^{-1}\right)(0.114)^{2}}{(1-0.114)} \\ \\ & =3.67 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \end{aligned} $