calculate the cell constant and degree of dissociation of ammonium hydroxide in the 6 x 10-5 M solution

Question

A conductivity cell has a resistance of 250 Ω when filled with 0.02 M KCl at 25°C and 105 Ω when filled with 6 x 10-5 M ammonium hydroxide solution. The conductivity of 0.02 M KCl solution at 25°C is 0.277 Sm-1 , and the limiting molar ionic conductivities of NH4 + and OH- are 7.34 x 10-3 and 0.0198 S m2mol-1 , respectively. Calculate the cell constant and degree of dissociation of ammonium hydroxide in the 6 x 10-5 M solution

Answer 1

(1)K=G.l/A

0.277=1/250× cell coconstant

Cell constant = 69.25 m^-

(2) degree of dissociation

Molar conductivity / limiting molar conductivity = degree of dissociation

As molar conductivity = conductivity /concentration

Conductivity of Nh4OH= 0.659

MMolar conductivity of Nh4Oh= 0.0271

0.659/0.0271= 24.31