Question

When a conductivity cell was filled with 0.02 M KCl, it had a resistance of 82.4 ohm at 25°C and when filled with 0.005 N K₂SO₄ it had a resistance of 326 ohm. Calculate

1. cell constant
2. specific conductivity, and
3. equivalent conductivity of 0.005 N K₂SO₄ solution.

The specific conductance of 0.02 M KCl is 0.002768 ohm⁻¹ cm⁻¹.

Answer 1

i. Calculation of cell constant:

Resistance of 0.02 M KCl solution = 82.4 ohm

Specific conductivity of KCl = 0.002768 ohm⁻¹ cm⁻¹

∴ Conductance of 0.02 M KCl solution = `1/82.4` ohm⁻¹

∵ Specific conductivity = Cell constant × Conductance

∴ `"Cell constant" = "Specific conductivity"/"Conductance"`

= `0.002768/(1//82.4)`

= 0.2281 cm⁻¹

ii. Calculation of specific conductivity:

Resistance of 0.005 N K₂SO₄ solution = 326 ohm

∴ Conductance of 0.005 N K₂SO₄ solution = `1/326` ohm⁻¹

∵ Specific conductivity = Cell constant × Conductance

= `0.2281 xx 1/326`

= 6.997 × 10⁻⁴ ohm⁻¹ cm⁻¹

iii. Calculation of equivalent conductivity:

`Lambda_"eq" = kappa xx 1000/"Normality of solution"`

= `(6.997 xx 10^-4 xx 1000)/0.005`

= 139.94 ohm⁻¹ cm² eq⁻¹