Question

The Conductivity Of $01\mathrm{N}\mathrm{NaOH}$ Solution Is $0.022\mathrm{s}/\mathrm{cm}$. When The Equal Volume Of $0.1\mathrm{N}\mathrm{HCl}$ Solution Is Added, The Conductivity Of Resultant Solution Decreases To $0.0055\mathrm{s}/\mathrm{cm}.$ The Equivalent Conductivity In s-cm/Equivalent Of $\mathrm{NaCl}$ Solution Is?

Answer 1

Equivalent conductivity:

• The equivalent conductivity of a solution is known as the conducting power of all the ions which are generated by dissolving one gram equivalent of an electrolyte into the solution.
• It is generally expressed as $\lambda$ .

Equivalent conductivity of $\mathrm{NaCl}$ Solution:

Step 1: Given data:
The conductivity Of $01\mathrm{N}\mathrm{NaOH}$ solution is given as = $0.022\mathrm{s}/\mathrm{cm}$

The Volume Of Solution added to the given solution = $0.1\mathrm{N}\mathrm{HCl}$

The specific conductance $\kappa$ is given as= $$0.0055s/cm$$

Step 2: Formula for calculating equivalent conductivity:

We can calculate the equivalent conductivity by using the formula:
$\mathrm{\lambda }=\mathrm{\kappa }\times \mathrm{V}$ ;

where $\kappa$ is the specific conductance

and $$V$$is known as the volume in $$mL$$that contains one gram equivalent of an electrolyte.

Step 3: presenting equivalent conductivity in terms of Normality:

The normality of a solution is given as the ratio of the number of gram equivalents of the solute and the volume of the solvent in litres.

Thus normality can be expressed as:
$\mathrm{N}=\frac{{\mathrm{n}}_{\mathrm{e}}}{\mathrm{V}\left(\mathrm{in}\mathrm{cc}\right)}\times 1000$.

Here the number of equivalents for $$NaCl$$ solution is$$1$$, $$n_e=1$$.
Thus, $V={\displaystyle \frac{1000}{N}}$.
Substituting the value of $$V$$in the formula of equivalent conductivity, we can get :
$\mathrm{\lambda }=\mathrm{\kappa }\times \frac{1000}{\mathrm{N}}$

Step 4: Calculating normality of $\mathrm{NaCl}$ solution:
$\mathrm{NaCl}$ is formed by the mixture of the two solutions i.e $\mathrm{NaOH}$ and $\mathrm{HCl}$

Here the normality of the $\mathrm{NaCl}$ solution can be calculated as = ${\displaystyle \frac{0.1}{2}}N$

This is due to the presence of two ions coming from$\mathrm{NaOH}$ and $\mathrm{HCl}$.

Step 5: Calculating equivalent conductivity:

By inserting the given values in the formula for equivalent conductivity we can get:
$$\begin{gathered} \lambda =0.0055\times \frac{1000}{{\displaystyle \frac{0.1}{2}}} \\ \Rightarrow \lambda =110 \end{gathered}$$
Thus, the equivalent conductivity in ${\mathrm{s}}^{-\mathrm{cm}}/\mathrm{equivalent}$ of $\mathrm{NaCl}$ solution is$110{\mathrm{s}}^{-\mathrm{cm}}/\mathrm{equivalent}$.