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Variation of Conductivity and Molar conductivity with Concentration

Last Updated : 01 May, 2022
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Electrochemistry is the study of chemical reactions that occur in a solution at the interface of an electron conductor (the electrode: a metal or a semiconductor) and an ionic conductor (the electrolyte). Electron transfer occurs between the electrode and the electrolyte or species in solution in these reactions.


A solution’s conductivity is defined as the conductance of a solution with a length of 1 cm and a cross-sectional area of 1 sq. cm. Conductivity, or particular conductance, is the inverse of resistivity. The letter k is used to signify it. If p stands for resistivity, we can write:

K= 1/p 

The conductivity of a solution at any given concentration is equal to the conductance (G) of one unit volume of solution held between two platinum electrodes with the same cross-sectional area and separated by the same distance.


G = K × a/l = K × l = K

(Since a = 1, l = 1)

For both weak and strong electrolytes, conductivity diminishes as concentration decreases. This is because as the concentration of a solution declines, the number of ions per unit volume that carry the current in the solution reduces.

Molar Conductivity

The conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length is the molar conductivity of a solution at a particular concentration.

Am = K × A/l

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).

Am = KV

With a decrease in concentration, molar conductivity rises. This is due to the fact that when a solution containing one mole of electrolyte is diluted, the total volume V of the solution increases.

The following graph depicts the fluctuation of Am with √c for strong and weak electrolytes:


Variation of Molar Conductivity with Concentration for Strong and Weak Electrolytes

  • Variation of Molar Conductivity with Concentration for Strong Electrolytes

Molar conductivity grows slowly with dilution in strong electrolytes, and it has a tendency to approach a limiting value as the concentration approaches 0, i.e. when the dilution is infinite. Molar conductivity at infinite dilution is the molar conductivity as the concentration approaches 0 (infinite dilution).  It is denoted by Am°

Am = Am°, when C ⇢ 0 (at infinite dilution)

The expression for the change of molar conductivity with concentration might be used.

Am = Am° − AC1/2


  • A is constant and A° stands for molar conductivity at infinite dilution. 

This equation, known as the Debye Huckel Onsager equation, holds true at low concentrations.

  • Variation of Molar Conductivity with Concentration for Weak Electrolytes

When opposed to strong electrolytes, weak electrolytes dissociate to a far smaller level. As a result, when compared to strong electrolytes, the molar conductivity is low.

However, the variation of Am with C1/2 is so enormous that the extrapolation of Am against C1/2 plots cannot yield molar conductance at infinite dilution ( Am°).

Variation of Molar Conductivity with Concentration

(A) Conductance behavior of weak electrolytes

The degree of dissociation with dilution determines the number of ions supplied by an electrolyte in a solution. The degree of dissociation rises as dilution increases, and molar conductance increases as a result. The limiting value of molar conductance (Am) corresponds to a degree of dissociation of 1, which means that the electrolyte completely dissociates.

At every concentration, the degree of dissociation may therefore be estimated.

α = Amc / Am°

where α represents the degree of dissociation, Amc represents the molar conductance at concentration C, and Am° represents the molar conductance at infinite dilution.

  • Conductance behavior of strong electrolytes

For strong electrolytes, there is no increase in the number of ions with dilution because strong electrolytes are completely ionized in solution at all concentrations.

Interionic forces are strong forces of attraction between ions of opposing charges in concentrated solutions of strong electrolytes. In concentrated solutions, the ions’ conducting capacity is reduced due to these interionic interactions. The ions grow farther apart as a result of dilution, and interionic forces diminish. As a result, molar conductivity rises as the solution is diluted. When the solution’s concentration is exceedingly low, interionic attractions are minimal, and molar conductance approaches the limiting value known as molar conductance at infinite dilution. This number is unique to each electrolyte.

Sample Questions 

Question 1: What effect does a solution’s concentration have on its specific conductivity?


The specific conductivity decreases as the concentration decreases. This is because the number of energized ions per unit volume  in a solution decreases with dilution. Therefore, concentration and conductivity are directly proportional to each other.

Question 2: Explain why the Cu+ ion is not stable in aqueous solutions?


Cu2+ is more stable in aqueous media than Cu+. This is because, while removing one electron from Cu+ to Cu2+ requires energy, the high hydration energy of Cu2+ compensates for it. As a result, the Cu+ ion is unstable in an aqueous solution. Cu2+ and Cu are disproportionately produced.

2Cu+(aq) ⇢ Cu2+(aq) + Cu(s)

Question 3:  The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. Calculate its degree of dissociation and dissociation constant. Given λ0(H+) = 349.6 S cm2 mol-1 and  λ0(HCOO) = 54.6 S cm2 mol.


Given that,

C = 0.025 mol L-1

Am = 46.1 S cm2 mol-1

 λ0(H+) = 349.6 S cm2 mol-1

 λ0(HCOO) = 54.6 S cm2 mol-1

Am°(HCOOH) =  λ0(H+) +  λ0(HCOO

= 349.6 + 54.6

= 404.2 S cm2 mol-1

Now, degree of dissociation:

α = Am(HCOOH)/Am°(HCOOH)

= 46.1/404.2

= 0.114(approx.)

Thus, dissociation constant:

K = c×α2/(1−α)

= (0.025 mol L-1)(0.114)2/(1 − 0.114)

= 3.67×10-4 mol L-1

Question 4:  The conductivity of 0.20M solution of KCl at 298K is 0.0248 S/cm. Calculate its molar conductivity 



 K= 0.0248 S/cm

C= 0.20M

Am = K×1000/C

= (0.02481000)/0.2

= 124 S cm2 mol-1

Question 5: Prove that “Molar conductivity increases with a decrease in concentration”.


Keeping length is equal to one.  

Multiply length ‘l’ in numerator and denominator

Am = K×A/l

Keeping length is equal to one.  

Multiply length ‘l’ in numerator and denominator

Am = K×A/l × l/l
As l=1 and A×l =V volume 

Am = KV

The concentration is low, the volume grows, and the cross-sectional area expands. As a result, both the volume and the molar conductivity rise. When the concentration is low, the volume is raised.

Because ‘K’ is related to conductivity, when concentration rises, ‘K’ rises while ‘V’ falls. When the concentration drops, ‘K’ drops as well, while ‘V’ rises. The change in the value of ‘V’ is significantly bigger than the change in the value of ‘K.’

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