Question

The conductance of a 0.0015M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized $Pt$ electrodes. The distance between the electrodes in $120\,cm$ with an area of the cross section of $1\,c{m^2}$. The conductance of this solution was found to be $5\; \times \;{10^{ - 7}}S$. The $pH$ of the solution is $4$. The value of limiting molar conductivity $(\Lambda _m^{\circ})$ of this weak monobasic acid in aqueous solution is $Z \times \;{10^2}\;Sc{m^{ - 1}}mo{l^{ - 1}}$. The value of Z is:

Answer 1

We know that when the solution is diluted at the extreme level than ions will be in extremely less in concentration at that time the solution is said that the conduction of ions will be maximum at zero or infinite dilution, such value of conduction is called limiting molar conductivity $(\Lambda _m^{\circ})$.

Complete step by step solution
The conductivity (k) is given by the formula given by substituting the value of G (cell constant), l (distance between the electrode) and a (cross sectional area of the electrode). Let us write the formula as follows;

$$k = \dfrac{{Gl}}{a}\\\ \Rightarrow k = \dfrac{{5 \times {{10}^{ - 7}\,}s \times 120\,cm}}{{1\,c{m^2}}}$$

$\Rightarrow k = 0.00006\;S/cm$
${\rm{ }}Molar{\rm{ }}\,Conductivity\; = {\Lambda _m}$
conductivity $k = 0.00006\;S/cm$
C (Concentration of electrolyte) $= 0.0015\;mol/L$

$${\Lambda _m}\; = \dfrac{{1000k}}{C}\\\ \Rightarrow {\Lambda _m} = \dfrac{{1000c{m^3}/L \times 0.00006\;S/cm}}{{0.0015\;mol/L}}\\\ \Rightarrow {\Lambda _m} = 40\;S\;c{m^2}/mol$$

Now, the concentration is calculated as shown below.

$$pH = 4\\\ [{H^ + }] = {10^{ - pH}} = {10^{ - 4}}M$$

According to Arrhenius theory of electrolytic dissociation, to the increase in the dissociation of the electrolyte is proportional to the increase in the molar conductivity with dilution, because the dissociation is almost complete the molar conductivity at infinite dilution being maximum.

$$\;\alpha = \dfrac{{{\Lambda _m}}}{{\Lambda _m^{\circ}}} = \dfrac{{number\;of\;dissociated\;ions}}{{total\;number\;of\;ions\;present}}\\\ \Rightarrow \Lambda _m^{\circ} = \dfrac{{{\Lambda _m}}}{\alpha }\\\ \Rightarrow \Lambda _m^{\circ} = \dfrac{{40}}{{0.06667}}Sc{m^2}/mol\\\ \Rightarrow \Lambda _m^{\circ} = 6 \times {10^2}Sc{m^2}/mol\\\ But \Lambda _m^{\circ} = Z \times {10^2}Sc{m^2}/mol$$

Hence Z = 6
The value of limiting molar conductivity $(\Lambda _m^{\circ})$of this weak monobasic acid in aqueous solution is $Z \times \;{10^2}\;Sc{m^{ - 1}}mo{l^{ - 1}}$.

**The value of Z is: $6$

Note: **
Molar conductivity and conductivity is different so do not get confused between those two. The limiting molar value is measured when all the ions will be dissociated completely, otherwise it will be considered as the conductivity.