Question

The conductivity of 10-3 mol /L acetic acid at 250

C is 4.1 x 10 -5
S cm-1
.

Calculate its degree of dissociation, if
0 m
for acetic acid at 250

C is 390.5 S

cm2 mol-1

Answer 1

Answer:

Hello

the molar conductivity of the chloroacetic solution will have different value than what you list here, 362 S m2mol−1.

I assume that this value was supposed to be the molar conductivity at infinite dilution, Λ0. Moreover, I think that the value you provided is incorrect.

More specifically, this value should either be

Λ0=362⋅10−4S m2mol−1

or

Λ0=362 S cm2mol−1

With this being said, the equation that establishes a relationship between conductivity, k, and molar conductivity, Λ, looks like this

Λ=kc , where

c - the molarity of the solution.

This is where things usually get a little tricky. You need to make sure that you use the right units. For example, molar concentration must be used in moles per cubic meter, which means that you must convert the given moles per liter

0.0625molesliter⋅103liters1 m3=0.0625⋅103mol m−3

Likewise, notice that the conductivity uses cm−1, so convert it to

3.319⋅10−3Scm⋅102cm1 m=3.319⋅10−1S m−1

The molar conductivity of the solution will thus be

Λ=3.319⋅10−1S m−10.0625⋅103m−3=5.3104⋅10−3S m−2mol−1

Now, the equation that connects the degree of ionization of a weak electrolyte, α, and its molar conductivity looks like this

α=ΛΛ0