Conductance Of Electrolytic Solutions

ELECTROLYTIC CONDUCTANCE

Property of electrolytic conductor which facilitates the flow of electricity through it. It is equal to the reciprocal of resistance i.e.

$Conduc\tan ce(c)=\frac{1}{R}$

Electrolysis conductance:

(i) Specific conductance

(ii) Equivalent conductance

(iii) Molar conductance

Specific conductance (k):

It is the reciprocal of specific resistance or it is the conductance produced by one centimetre cube of an electrolytic conductor.

If $R\propto \frac{\ell }{a}$

where $\ell =$ length of conductor

a = cross – sectional area of conductor

$\therefore R=\rho .\frac{\ell }{a}$

or,$\frac{1}{\rho }=\frac{1}{R}\times \frac{\ell }{a}$or,$k=C\times cellcons\tan t$

Representation of specific conductance:

$k=\frac{1}{\rho }=conduc\tan cec{m}^{-1}$

Unit at specific conductance:

$oh{m}^{-1}c{m}^{-1}or{\Omega }^{-1}c{m}^{-1}orSc{m}^{-1}(C.G.S)$

Equivalent conductance $({\wedge }_{eq}):$

Conductance of all ions which are produced by the ionisation of one equivalent of an electrolyte in a given solution.

i.e. Equivalent conductance $(\wedge )=specificconduc\tan ce\times V(ml)$

or,$\wedge =K\times V_{(ml)}$or,$\wedge =\frac{K\times 1000}{N}$

Unit of equivalent conductance is $oh{m}^{-1}c{m}^{2}equi{v}^{-1}.$

Molar conductance $({\lambda }_m):$

Conductance of all ions which are produced by the ionisation of one mole of an electrolyte in given solution.

i.e. Molar conductance $({\lambda }_m)=Specificconduc\tan ce\times V_{(ml)}$

${\lambda }_m=K\times V_{(ml)}$

or, ${\lambda }_m=\frac{K\times 1000}{M}$

Unit of molar conductance is $oh{m}^{-1}c{m}^{2}mo{l}^{-1}.$

Illustration 1. The equivalent conductance of monobasic acid at infinite dilution is 438 $oh{m}^{-1}\text{ c}{\text{m}}^{\text{2}}\text{ e}{\text{q}}^{\text{-1}}.$ If the resistivity of the solution containing 15g of acid (Molecular weight = 49) in 1L is 18.5 ohm cm. What is the degree of dissociation of acid?

(A) 45.9% (B) 40.2%

(C) 60.4% (D) 50.7%

Solution: Resistivity = $\frac{1}{\kappa }$

$\therefore \kappa =\frac{1}{15.8}oh{m}^{-1}\text{ c}{\text{m}}^{\text{-1}}$

${\lambda }_v=\frac{\kappa \times 1000}{C}=\frac{1}{18.5}\times \frac{1000}{\frac{15}{49}}$= $\frac{1\times 1000\times 49}{18.5\times 15}$

$\alpha =\frac{{\lambda }_v}{{\lambda }_{\infty }}=\frac{1\times 1000\times 49}{18.5\times 15\times 348}$= 0.507

= 50.7%

EFFECT OF DILUTION ON CONDUCTANCE

(i) When we increase the dilution of electrolyte, magnitude of specific conductance decreases, because number of ions in 1 cm³ volume of electrolyte decreases.

(ii) When dilution of electrolyte increases, magnitude of molar conductance and equivalent conductance increases, because number of ions increases after dilution.

(iii) The molar conductivity of strong electrolytes is found to vary with concentration according to the equation.

${\wedge }_m^c={\wedge }_m^o-b\sqrt{c}$

where ${\wedge }^c=$equivalent conductivity at fixed concentration

${\wedge }^o=$ equivalent conductivity of infinite dilution

(iv) Degree of dissociation $(\alpha )=\frac{{\wedge }_m^c}{{\wedge }_m^o}$

KOHLRAUSCH'S LAW OF INDEPENDENT MIGRATION OF IONS

At infinite dilution, conductance of any electrolyte is the sum of contribution of its constituents ions i.e. anions and cations.

For an electrolyte of the type of A_xB_y, we have

${\wedge }_m^0=x{\lambda }_{+}^0+y{\lambda }_{-}^0$

Application of Kohlrausch's Law

(i) Determination of ${\wedge }_m^0$ of a weak electrolyte:

In order to calculate ${\wedge }_m^0$ of a weak electrolyte say CH₃COOH, we determine experimentally ${\wedge }_m^0$ values of the following three strong electrolytes:

(a) A strong electrolyte containing same cation as in the test electrolyte, say HCl

(b) A strong electrolyte containing same anion as in the test electrolyte, say CH₃COONa

(c) A strong electrolyte containing same anion of (a) and cation of (b) i.e. NaCl.

${\wedge }_m^0$of CH₃COOH is then given as:

${\wedge }_m^0$ (CH₃COOH) = ${\wedge }_m^0$(HCl) + ${\wedge }_m^0$(CH₃COONa) – ${\wedge }_m^0$(NaCl)

Proof:

${\wedge }_m^0$ (HCl) = ${\lambda }_H^0+{\lambda }_{C{l}^{-}}$ …I

${\wedge }_m^0$ (CH₃COONa) =${\lambda }_{C{H}_{3}CO{O}^{-}}^0+{\lambda }_{N{a}^{+}}$ …II

${\wedge }_m^0$ (NaCl) =${\lambda }_{N{a}^{+}}^0+{\lambda }_{C{l}^{-}}^0$ …III

Adding equation (I) and equation (II) and subtracting (III) from them:

${\wedge }_{(HCl)}^0+{\wedge }_{(C{H}_{3}COONa)}^0-{\wedge }_{(NaCl)}^0={\lambda }_{(H^{+})}^0+{\lambda }_{(C{H}_{3}CO{O}^0)}^0={\wedge }_{0\left(C{H}_{3}COOH\right)}$

(ii) Determination of degree of dissociation ($\alpha$):

$\alpha =\frac{\mathrm{No}.ofmoleculesionised}{\mathrm{totalnumberofmoleculesdissolved}}=\frac{{\wedge }_m}{{\wedge }_m^0}$

(iii) Determination of ionic product of water:

From Kohlrausch's law, we determine ${\wedge }_m^0$ of H₂O where ${\wedge }_m^0$ is the molar conductance of water at infinite dilution when one mole of water is completely ionised to give one mole of H⁺ and one mole of OH^– ions i.e.

${\wedge }_m^0$ (H₂O) = ${\lambda }_{H^{+}}^0$ + ${\lambda }_{O{H}^{-}}^0$

Again using the following equation,

${\wedge }_m=\frac{\kappa \times 1000}{C}$, where C = molar concentration i.e. mol L^–1 or mol dm^–3

Þ${\wedge }_m=\frac{\kappa }{C}$, where C = concentration in mol m^–3

Assuming that ${\wedge }_m$ differs very little from ${\wedge }_m^0$

${\wedge }_m^0$= $\frac{\kappa }{C}$

$\Rightarrow$ C = $\frac{\kappa }{{\wedge }_m^0}$

Specific conductance (k) of pure water is determined experimentally. Thereafter, molar concentration of dissociated water is determined using the above equation.

K_w is then calculated as: K_w = C²

Illustration 2. The ${\wedge }_m^o$ values for NaCl and KCl are 126.5 and 149.9 ${\Omega }^{-1}c{m}^{2}mo{l}^{-1}$ respectively. The ionic conductance of $N{a}^{+}$ at infinite dilution is $50.1{\Omega }^{-1}c{m}^{2}mo{l}^{-1}.$ Calculate the ionic conductance at infinite dilution for $K^{+}ion.$

Solution: ${\lambda }_m^o(KCl)-{\lambda }_m^o(NaCl)={\lambda }_{K^{+}}^o-{\lambda }_{N{a}^{+}}^o$

= 149.9 – 126.5 = 23.4

or, ${\lambda }_{K^{+}}^o=23.4+{\lambda }_{N{a}^{+}}^o$

= 23.4 + 50.1 = 73.5 ${\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

Illustration 3. Calculate the limiting value of equivalent conductance of AgCl. Given that

${\wedge }_{AgN{O}_3}^o=133.4oh{m}^{-1}c{m}^{2}e{q}^{-1}$

${\wedge }_{KCl}^o=149.2oh{m}^{-1}c{m}^{2}e{q}^{-1}$

${\wedge }_{KN{O}_3}^o=145.9oh{m}^{-1}c{m}^{2}e{q}^{-1}$

Solution: According to Kohlrausch law

${\wedge }_{AgN{O}_3}^{\circ }={\wedge }_{A{g}^{+}}^{\circ }+{\wedge }_{N{O_3}^{-}}^{\circ }$

${\wedge }_{KCl}^{\circ }={\wedge }_{K^{+}}^{\circ }+{\wedge }_{C{l}^{-}}^{\circ }$

${\wedge }_{KN{O}_3}^{\circ }={\wedge }_{K^{++}}^{\circ }+{\wedge }_{N{O_3}^{-}}^{\circ }$

${\wedge }_{AgCl}^{\circ }={\wedge }_{A{g}^{+}}^{\circ }+{\wedge }_{N{O_3}^{-}}^{\circ }$

= ${\wedge }_{A{g}^{+}}^{\circ }+{\wedge }_{N{O_3}^{-}}^{\circ }-{\wedge }_{N{O_3}^{-}}^{\circ }$$+{\wedge }_{C{l}^{-}}^{\circ }+{\wedge }_{K^{+}}^{\circ }-{\wedge }_{K^{+}}^{\circ }$ = ${\wedge }_{AgN{O}_3}^{\circ }+{\wedge }_{KCl}^{\circ }-{\wedge }_{KN{O}_3}^{\circ }$

= 133.4 + 149.2 - 145.9 = 138.2 ohm^-1 cm² eq^-1