# The conductivity of 0.1 M solution of AgNO3 is 9.47 X 10-3 S cm-1 at 291 K. The ionic conductivities os Ag+ and NO3 -at the same temperature are 55.7 and 50.8 Scm2equiv-1 respectively. Calculate the degree of dissociation of AgNO3 in 0.1 M solution. How to solve this and what's the meaning of ionic conductivities of the ions given here?

The ionic conductivities represent the fraction of conductivity due to the ions. Hence it is the limiting molar conductivity of the individual ions.

We are given that

κ = 9.47 X 10-3 S cm-1 for 0.1 M solution of AgCl

Therefore ΛAgNO3 = κ / c

where κ is the conductivity and c is the concentration of the AgNO3 solution. This gives us ΛAgNO3 as

ΛAgNO3 = [9.47 X 10-3 Scm-1 X 1000 cm3 L-1] / 0.1 (mol L-1)

= 94.7 Scm2mol-1

Now

Λ0 AgNO3 = λ0(Ag+) + λ0(NO- 3)

= 55.7 + 50.8

= 106.5 Scm2eq-1

Because AgNO3 dissociates as Ag+ and NO3- ions, hence its equivalent weight is the same as its molecular weight. Because of this the equivalent conductivity will be the same as the molar conductivity.

Therefore

Λ0 AgNO3 = 106.5 Scm2mol-1

Now  α = Λ /  Λ0

Therefore the degree of dissociation for AgNO3 will be

α =  ΛAgNO3 / Λ0 AgNO3

= 94.7 / 106.5 = 0.89

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