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

Step-by-step 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 (l)/(A)`

and,
`(1)/(R) = (1)/(\rho) * (A)/(l)`

`k = c * x`

where, k = specific conductivity

c = conductance

x = cell constant

Therefore, value of c =
`(1)/(R)` =
`(1)/(12.553)` = 0.0796 per ohm

x =
`(l)/(A)` =
`(3.6 cm)/(7.2 cm^(2))`

= 0.5 per cm

Hence, calculate the value of specific conductivity as follows.

`k = c * x`

=
`0.0796 per ohm * 0.5 per cm`

= 0.0398 per ohm per cm

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

`\lambda_(m) = (k * 1000)/(M)`

=
`\frac{0.0398 \text{per ohm per cm} * 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)`.