Form 6 Chemistry – PHYSICAL CHEMISTRY 1.3- ELECTROCHEMISTRY

MOLAR CONDUCTIVITY AND DEGREE OF DISSOCIATION

The molar conductivity of weak electrolyte is proportional to the degree of dissociation.

i.e.   

 = K ……………. (i)

At infinity dilution, the weak electrolyte is completely ionized.

 = 1 or 100%

 = K (Constant)……….. (ii)

Taking ratio of equation (i) and (ii)

 Ostwald’s dilution law

It states that

“For a weak electrolyte the degree of dissociation is proportional to the square root of reciprocal of concentration”

Consider a weak binary electrolyte AB (i.e. ethanoic acid) in solution with concentration C (mol/).

AB     A⁺ _(aq)   +    B^– _(aq)

1mole     0               0       start

                          At equilibrium

But V =

AB  A⁺ _(aq) + B^– _(aq)

K =

But AB = CH₃ COOH

 =

For a weak electrolyte, degree of dissociation is very small thus the expression

1 –   1

 K_a =

K _a = C

Example

1.  At 25⁰c the solution of O.1M of ethanoic acid has a conductivity of 5.0791×10^-2m^-1mol^-1.
calculate the pH of the solution and dissociation constant Ka (Answer Ka=1.69 10^-2M)

2.  A 0.001 moldm^-3 solution of ethanoic acid was found to have molar conductivity of 14.35Sm²mol^-1. Use this value together with molar conductivity at infinity dilution of ethanoic acid  (C) is 390.7Ð…to calculate :-

i.        Calculate degree of dissociation of acid
ii.     Equilibrium constant

3.The resistivity of M KCl solution is 361Ω and conductivity cell containing such a solution was found to

have a resistance of 550Ω

a.Calculate the cell constant

b.The same cell filled with 0.1M ZnSO₄ solution had resistance of 72Ω. What is the conductivity of this solution?

KOHLRAUSCH’S LAW OF INDEPENDENT IONIC MOBILITY

Kohlrausch notes that the difference between molar conductivity at infinity dilution λ∞ values for the two salts which were strong electrolytes and of the same cation and anion was always constant.
Using the λ∞ values in D.. cm² mol^-1

      
            
This observation shows that each type of ion (caution or anion) contribute a definite amount of molar conductivity of an electrolyte of infinity dilution independently from other ions present in the solution i.e. a fraction of current that an ion carry is always constant and it doesn’t depend on the compound in which it is contained.

Hence  Kohlrausch’s law

States that;”The molar conductivity of an electrolyte at infinity dilution is equal to the sum of the molar conductivities of the caution and anion”.
OR State that the molar conductivity at infinity dilution of the solution equal to the sum of molar conductivity at infinity dilution of its components ions.

i.e.  =  +

E.g. =  + (C)

λ_∞(Al₂(S) = 2(AL³⁺) + 3(S)

Application of Kohlrausch’s law

In direct determination of molar conductivity at infinity dilution for weak electrolytes. Thus by using strong electrolytes we can easily. Calculate the molar conductivities at infinity. Dilution for weak electrolytes.

E.g.: The  (CCOOH) can be determined from   of potassium ethanoate (CCOOK) hydrochloric acid (HCl) and potassium chloride (KCl)

CH₃COOH + KCl  CH₃COOK + HCl

λ_∞(CH₃COOH) = (CH₃COOK) + (HCl) – KCl)

= (CH₃COO^–) + (K⁺) + (H⁺) + (Cl) – (K⁺) – (Cl)

λ_∞(CH₃ COOH) = (CH₃COOH^–) + (H⁺)

REVIEW QUESTIONS

1. Calculate (NH₄OH given that   of three strong electrolyte NaCl, NaOH and NH₄Cl in Ð…cm² mol^-1 are 126.4, 248.4 , 149.8 respectively.

2.The molar conductivity of 0.093 CH₃COOH solution at 298k is 536 x 10^-4 Sm²mol^{-1 .}The molar conductivity at infinity dilution of H⁺ and CH₃COO^– are 3.5 x 10^-2 and 0.41 x 10^-2Sm² mol^-1.what is are the dissociation constant of CH₃COOH

3. A 0.05M HF solution has a conductivity of 91.81mol^-1 m^-1 at 298k.At the same temperature (NaF),   (NaoH) and (H₂O) are 493360 and 162Sm²mol^-1 respectively. Calculate dissociation constant.
4. The λ_∞(NaI), λ_∞(CH₃COONa) and λ_∞(CH₃COO₂Mg) are 12.69,9.10 and 18.78Sm² mol^-1 respectively at 25⁰C. What is the molar conductivity of MgI₂ at infinity dilution.