Question

A 0.1000 mol L^-1 aqueous solution of HCl in a cell with plates having area 7.200 cm² separated by 3.600 cm has a resistance R = 12.553 22 and a cation transference number 0.8314 at 25 °C. Calculate the molar ionic conductivity of hydrogen ions in S cm’ mol"!.

Answer 1

Explanation:

The given data is as follows.

          Molarity = 0.1 M,             Area = 7.2 $cm^{2}$

          Resistance = 12.553 ohm,      Length = 3.6 cm

As it is known that relation between resistance, length and area is as follows.

                    R = $\rho \frac{l}{A}$

and,       $\frac{1}{R} = \frac{1}{\rho} \times \frac{A}{l}$  

                      $k = c \times x$

where,     k = specific conductivity

               c = conductance

                x = cell constant

Therefore, value of c = $\frac{1}{R}$ = $\frac{1}{12.553}$ = 0.0796 per ohm

             x = $\frac{l}{A}$ = $\frac{3.6 cm}{7.2 cm^{2}}$

                           = 0.5 per cm

Hence, calculate the value of specific conductivity as follows.

                 $k = c \times x$

                          = $0.0796 per ohm \times 0.5 per cm$

                          = 0.0398 per ohm per cm

Relation between molar  ionic conductivity and specific conductivity is as follows.

             $\lambda_{m} = \frac{k \times 1000}{M}$

                          = $\frac{0.0398 \text{per ohm per cm} \times 1000}{0.1 M}$

                          = 398 $\Omega^{-1} cm^{2} mol^{-1}$

Also, $\Omega^{-1}$ = Siemen

                $\lambda_{m} = 398 S cm^{2} mol^{-1}$

thus, we can conclude that value of molar ionic conductivity of given hydrogen ions is $398 S cm^{2} mol^{-1}$.