PDF Conductance (Part -IV) - Narajole Raj College

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Prof. Manik Shit, SACT,

Department of Chemistry, Narajole Raj College,

Narajole.

PAPER: C5T (Physical Chemistry - II) TOPIC : Conductance (Part-IV)

C5T: Physical Chemistry-II

Transport processes

Conductance (Part -IV)

IONIC MOBILITIES

Q). Define ionic mobility and mention its unit ?

Ans: The conductance depends naturally on the velocity of an ion. The velocity should depend on the valency of ion, its concentration , mass and temperature and viscosity of the medium .For a particular ion at a particular temperature and solvent , all these are constant. The velocity will then depend on the applied voltage (∆φ) and the distance between the electrodes (l) , if u+ be the velocity of the cation , then

u+ ∝ (∆φ/l) u+ = V- (∆φ/l)

Where V+ is the proportionality constant and is constant under a particular set of conditions . It is equal to the velocity in cm/sec or m/sec .When

∆φ =1 volt and l=1cm or 1m, It is the velocity under unit potential gradient .It is termed as ionic mobility .

Similarly , for anion .

u+ = V- (∆φ/l)

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Where v -,u – are ionic mobility and ionic velocity of the anion

respectively. Therefore , the velocity with which an ion would move under a potential gradient of 1volt.cm^-1 or 1 volt.m^-1 . is termed as ionic mobility. The ionic mobilities at infinite dilution are sometimes referred to as absolute velocity of ions.

Unit : The unit of ionic mobility in C. G.S system is cm² volt^-1sec^-1 and in S.I.system is m² volt^-1sec^-1.

Q). Is there any differences between ionic mobility and absolute velocity of ions

?

Ans: The velocity of an ion at infinite dilution depends only on the applied voltage (∆φ) and the distance between two electrodes (l) , thus, if u⁰+ , u-0 are the ionic velocity of the cation and anion at infinite dilution , then

u⁰+ = v⁰+ (∆φ/l) u⁰- = v⁰- (∆φ/l)

Where v⁰+ , v⁰- are ionic mobilities of cation and anion at infinite dilution respectively. The ionic mobilities at infinite dilution are termed as absolute velocities of ions but at finite concentration the velocity of ions also depend on the interionic force of attraction . So , the above relation can not be applied and also the limiting value of ionic mobility can not be obtained .The limiting value of ionic mobility is obtained when the solution is at infinitely diluted , so that no interionic attraction would arise .Hence there is no difference between the limiting value of ionic mobility and absolute velocities of ions at infinite dilution.

But at finite concentration ionic mobility differs from absolute velocity of ions.

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RELATION BETWEEN IONIC MOBILITY AND

Q). Derive a relation between (i) ionic mobility and ionic conductance and (ii) ionic mobility and ionic transport no. of an electrolyte .

i) Relation between ionic mobility and ionic conductance :-

Ans: we consider a solution containing 1 mole of an electrolyte , AxBy is kept between two parallel electrodes at a distance of l apart the applied potential difference between these two electrode is (∆φ) . Suppose in one sec. the total current I is passed through these electrodes .The potential gradient is (∆φ/l) . The disassociation equilibrium of thin electrolyte is as follows:

AxBy ↔ xA^z+ + yB^z-

Where z+ , z- are the charges of cations and anions respectively . Let ‘α ‘ be the degree of dissociation of the electrolyte , then the amount if undissociation AxBy ( 1- α ), the dissociated A^z+ ions is xα moles and B^z- is yα moles . Suppose the ionic velocity of cation and anion are u+ and u- respectively.

The current carried by cations,

I+ = N+ z+ e u+ / l = N+ z+ e u+ / l ( Since NA is avogadro’s no.) And the current carried by anions

Thus, the total current carried by cations and anions together is I = I+ + I-

I = αNAe/ l [ xz+u- + y z- u-] Again, an univalent ion carry a charge.

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e = F / NA

Or , NA e = F (where F is faraday )

I = αF / l [ xz+u- + y z- u-] ... (1) By defination , molar conductance is given by

Ʌm = K . V = K (Al)

Or, K = Ʌm / Al ...(2)

Where A is the area of cross section of each electrodes , k is specific conductance.

Ʌm is the molar conductance of this solution.

Now , the conductance , G of this solution becomes – G= k.A/l

Which accroding to (2) equation becomes –

G= Ʌm / l²...(3)

According to ohm’s law the current flowing through the solution is

I = ∆φ/R = ∆φ . G [ Since, 1/R = G , where R is the resistance]

Which according to (3) equation becomes

I = ∆φ Ʌm / l² ...(4) Now , complaining equation (1) and (4) we get αF [ xz+u+ + y z- u-] = (∆φ / l) Ʌm

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If the solution is infinitely diluted, then α = 1 and also using Kohlrausch’s law (Ʌm0 = xλ+0 + y λ-0) where λ+0 , λ-0 are the molar ionic conductance of cation

anions respectively at infinite dilution equation above becomes – F [ xz+u+ + y z- u-] = (∆φ / l)( xλ+0 + y λ-0) ... (5) We can split the above equation into two parts , we have Fz+u⁰+ = (∆φ / l) λ+0

And, Fz-u⁰- = (∆φ / l) λ-0 } ...(6) Hence from equation (6) we get,

u⁰+ = (∆φ / l) λ+0/ Fz+

and u⁰- = (∆φ / l) λ-0/ Fz- } ...(7)

If V⁰+ , V⁰- are ionic mobiilities of cation and anion at infinite dilution , then V⁰+ = u⁰+ / (∆φ / l) = λ+0/ Fz+ } ...(8)

V⁰- = u⁰- / (∆φ / l) = λ-0/ Fz-

The equation is the required relation between ionic mobilities and ion conductance at infinite dilution.

Relation between ionic mobility and transport no.

It t⁰+ , t⁰- are the ionic transport no.of cation and anion at infinite dilution respectively . Then we have ,

t⁰+ = x λ+0/ Ʌm0 λ+0 = t⁰+ Ʌm0 /x and , t⁰- = yλ-0/ Ʌm0

λ-0 = t⁰- Ʌm0 /y

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with these equation (8) becomes..

V⁰+ = t⁰+ Ʌm0 /x Fz+ } ...(9) V⁰- = t⁰- Ʌm0 /y Fz-

The equation (9) is the relation between ionic mobility and ionic transport no. at infinite dilution .

Application of conductance measurements . DETERMIND OF DEGREE OF DISASSOCIATION :-

Q). Determind the degree of disassociation of a weak electrolytes by conductance measurement.

Ans: Suppose one equivalent of a weak electrolyte at infinite dilution produces N+ positive and N- negative ions where valencies are z+ and z- respectively.

Thus, at infinite dilution , the equivalent conductance would be

Ʌ⁰ = N+ z+ e u⁰+ / l + N- z- e u⁰- / l = N+ z+ e/l (u⁰+ + u⁰- )... (1) Where u⁰+ , u⁰- are the velocities of cation and anion at infinite dilution respectively ,

‘l’ is the distance between two parallel electrodes and ‘e’ is the quantity of electricity carried by unit charge .

It α be assumed to be degree of disassociation at any given concentration of the solution then the equivalent conductance would be

Ʌ = α N+ z+ e u⁰+ / l + α N- z- e u⁰- / l = α N+ z+ e/l (u+ + u- ) ... (2)

Where α N+ , α N- are the no. of ions of cation and anion respectively at any concentration . Now dividing equation (2) by (1) , we get

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Ʌ/ Ʌ⁰ = α (u+ + u- )/ (u⁰+ + u⁰- )

For strong electrolyte α=1 at any concentration and so the conductance ration would be { according to equation (3) }

Ʌ⁰/ Ʌ = (u+ + u- )/ (u⁰+ + u⁰- )

But for weak electrolyte u+ = u⁰+ , u- = u⁰- and thus equation (3) becomes –

Ʌ⁰/ Ʌ = α

This is the conductance ration for weak electrolytes . Therefore, degree of disassociation α can be known if Ʌ and Ʌ⁰ are known .

Q).Determine the disassociation constant for a weak acid by conductance method .

Ans: Let us consider a weak acid , HA which undergoes dissociation to produce and H⁺ and A^- .There will be an equilibrium between undissociated molecules and the produced ions i.e.

HA ↔ H⁺+ A^-

On applying the law of mass action the equilibrium constant i.e. dissociation constant ka of the weak acid given by –

Ka = CH+ . CA- / CHA

( for a dilute solution a nearly c)

If ‘c’ is the initial concentration of the acid in equivalent lit^-1 then the concentration of undissociated molecule (1- α)C , where α is the degree of dissociation at equilibrium, the concentration of dissociated H⁺and A^- are αc each respectively.

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Ka = α²C / (1- α)

For weak acid α << 1 and so, 1- α ≈ 1 Therefore, Ka = α²C

Again , α = Ʌ/ Ʌ ⁰ ( where Ʌ , Ʌ ⁰ are the equivalent

conductance of conductance at any conc. and at infinite dilution respectively ) Ka = Ʌ² C / Ʌ02

Therefore Ka can be determined , if Ʌ, Ʌ ⁰ and C are known.

Q). Determined the disassociation constant of a weak base by conductance method .

Ans: The ions produced on dissociation are in equilibrium with undissociated molecules of weak base in the solution .Thus a weak base will have the following equilibrium .

BOH ↔ B⁺+ OH^-

On applying the law of mass action the equilibrium constant i.e. dissociation constant Kb of the weak base is given by –

Ka = CB+ . COH- / CBOH [ for a dil. Solution a ≈ c]

It α be the degree of dissociation at equilibrium and ‘c’ is the initial conc. of the base in equiv/lit , then the concentration of B⁺and OH^- are αc each

respectively and the concentration of undissociated BOH is (1- α)C.

therefore ..

Kb = α²C / (1- α)

For weak acid α << 1 and so, 1- α ≈ 1

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Kb = α²C

Again , α = Ʌ/ Ʌ ⁰ ( where Ʌ , Ʌ ⁰ are the equivalent conductance of conductance at any conc. and at infinite dilution respectively )

Hence, Kb = Ʌ² C / Ʌ02

Therefore Kb can be known, and if Ʌ, Ʌ ⁰ and C are known.