Electrochemistry : Kohlrausch's Law

Factors affecting the electrolytic conductance

In general, conductance of an electrolyte depends upon the following factors,

(1) Nature of electrolyte,

(2) Concentration of the solution,

(3) Temperature

(1) Nature of electrolyte : The conductance of an electrolyte depends upon the number of ions present in the solution. Therefore, the greater the number of ions in the solution the greater is the conductance. The number of ions produced by an electrolyte depends upon its nature. The strong electrolytes dissociate almost completely into ions in solutions and, therefore, their solutions have high conductance. On the other hand, weak electrolytes, dissociate to only small extents and give lesser number of ions. Therefore, the solutions of weak electrolytes have low conductance.

(2) Concentration of the solution : The molar conductance of electrolytic solution varies with the concentration of the electrolyte. In general, the molar conductance of an electrolyte increases with decrease in concentration or increase in dilution. The molar conductance of a few electrolytes in water at different concentrations are given in table

C	HCl	KCl	KNO₃	CH₃COOH	NH₄OH
0.1	391.3	129.0	120.4	5.2	3.6
0.05	399.1	133.4	126.3	–	–
0.01	412.0	141.3	132.8	16.3	11.3
0.005	415.8	143.5	131.5	–	–
0.001	421.4	146.9	141.8	49.2	34.0
0.0005	422.7	147.8	142.8	67.7	46.9
(Infinite dilution)	426.2	149.9	146.0	390.7	271.0

Inspection of table reveals that the molar conductance of strong electrolyte (HCl, KCl, KNO₃) as well as weak electrolytes (CH₃COOH, NH₄OH) increase with decrease in concentration or increase in dilution. The variation is however different for strong and weak electrolytes.

(i) Variation of conductivity with concentration for strong electrolytes : In case of strong electrolytes, there is a tendency for molar conductivity to approach a certain limiting value when the concentration approaches zero i.e., when the dilution is infinite. The molar conductivity when the concentration approaches zero (Infinite dilution) is called molar conductivity at infinite dilution. It is denoted by Λ⁰

Thus, Λ = Λ⁰ when C → 0 (At infinite dilution)

It has been observed that the variation of molar conductivity with concentration may be given by the expression

Λ = Λ⁰ – Ac^1/2

where, A is a constant and Λ⁰ is called molar conductivity at infinite dilution.

The variation of molar conductivity with concentration can be studied by plotting the values of Λ_m against square root of concentration $\[(\sqrt{C})\$ . The plots of variation of molar conductivity with $\[\sqrt{C}\$ for KCl and HCl are given in fig. It has been noticed that the variation of Λ_m with concentration, $\[\sqrt{C}\$ is small (Between 4 to 10% only) so that the plots can be extrapolated to zero concentration. This gives the limiting value of molar conductance when the concentration approaches zero, called molar conductivity at infinite dilution.

(ii) Variation of molar conductivity with concentration for weak electrolytes : The weak electrolytes dissociate to a much lesser extent as compared to strong electrolytes. Therefore, the molar conductivity is low as compared to that of strong electrolytes.

However, the variation of Λ_m with $\[\sqrt{C}\$ is very large and so much so that we cannot obtain molar conductance at infinite dilution (Λ⁰) by extrapolation of the Λ_m versus $\[\sqrt{C}\$ plots. The behavior of weak electrolytes such as CH₃COOH is shown in figure.

Note : The Λ⁰ value for weak electrolytes can be obtained by an indirect method based upon Kohlrausch law.

Explanation for the variation : The variation of molar conductance with concentration can be explained on the basis of conducting ability of ions for weak and strong electrolytes.

For weak electrolytes the variation of Λ with dilution can be explained on the bases of number of ions in solution. The number of ions furnished by an electrolyte in solution depends upon the degree of dissociation with dilution. With the increase in dilution, the degree of dissociation increases and as a result molar conductance increases. The limiting value of molar conductance (Λ⁰) corresponds to degree of dissociation equal to 1 i.e., the whole of the electrolyte dissociates.

Thus, the degree of dissociation can be calculated at any concentration as,

where α is the degree of dissociation, Λ^c is the molar conductance at concentration C and Λ⁰ is the molar conductance at infinite dilution.

For strong electrolytes, there is no increase in the number of ions with dilution because strong electrolytes are completely ionised in solution at all concentrations (By definition). However, in concentrated solutions of strong electrolytes there are strong forces of attraction between the ions of opposite charges called inter-ionic forces. Due to these inter-ionic forces the conducting ability of the ions is less in concentrated solutions. With dilution, the ions become far apart from one another and inter-ionic forces decrease. As a result, molar conductivity increases with dilution. When the concentration of the solution becomes very-very low, the inter-ionic attractions become negligible and the molar conductance approaches the limiting value called molar conductance at infinite dilution. This value is characteristic of each electrolyte.

(3) Temperature : The conductivity of an electrolyte depends upon the temperature. With increase in temperature, the conductivity of an electrolyte increases.

Migration of ions.

Electricity is carried out through the solution of an electrolyte by migration of ions. Therefore,

(1) Ions move toward oppositely charged electrodes at different speeds.

(2) During electrolysis, ions are discharged or liberated in equivalent amounts at the two electrodes, no matter what their relative speed is.

(3) Concentration of the electrolyte changes around the electrode due to difference in the speed of the ions.

(4) Loss of concentration around any electrode is proportional to the speed of the ion that moves away from the electrode, so

$\[\frac{\text{Loss around anode}}{\text{Loss around cathode}}=\frac{\text{Speed of cation}}{\text{Speed of anion}}\$

The relation is valid only when the discharged ions do not react with atoms of the electrodes. But when the ions combine with the material of the electrode, the concentration around the electrode shows an increase. For example, during electrolysis of AgNO₃ solution using Ag electrodes, the concentration of AgNO₃ around the anode increases, because every nitrate ion that reaches at the anode dissolve from it one Ag⁺ ion to form AgNO₃.

Transport number or Transference number.

(1) Definition : “The fraction of the total current carried by an ion is known as transport number, transference number or Hittorf number may be denoted by sets symbols like t₊ and t_– or t_cand t_a or n_c and n_a.”

From this definition,

$\[{{t}_{a}}=\frac{\text{Current carried by an anion}}{\text{Total current passed through the solution}}\$ ;

$\[{{t}_{c}}=\frac{\text{Current carried by a cation}}{\text{Total current passed through the solution}}\,=\,evidently,\text{ }{{t}_{a}}+\text{ }{{t}_{c}}=\text{ }1.\$

(2) Determination of transport number : Transport number can be determined by Hittorf’s method, moving boundary method, emf method and from ionic mobility.

(3) Factors affecting transport number

(i) Temperature : A rise in temperature tends to bring the transport number of cation and anion more closer to 0.5. It means that the transport number of an ion, if less than 0.5 at the room temperature increases and if greater than 0.5 at room temperature decreases with rise in temperature.

(ii) Concentration of the electrolyte : Transport number generally varies with the concentration of the electrolyte. In case of partially dissociated CdI₂, the value decreases from 0.49 at low concentration to almost zero at higher concentration and becomes negative at still higher concentration. The transport numbers of Cd²⁺ ions in 0.01 N, 0.05 N, 0.02 N and 0.50 N CdI₂ at 25°C are 0.449, 0.402, 0.131 and 0.005 respectively. This abnormal behavior may be explained by assuming,

(a) That in very dilute solution : CdI₂ ionises to Cd²⁺ ions and I⁻ions. Thus Cd²⁺ shows the usual transport number.

(b) That with increase in concentration : CdI₂takes on I⁻ions and form complex,CdI₂ + 2I⁻ ⇌ [CdI₄]²⁻. This explains the negative value for transport number of Cd²⁺ ions at higher concentration.

(iii) Nature of the other ions present in solution : The transport number of anion depends upon the speed of the anion and the cation and vice versa. For example, the transport number of Cl⁻ ion in NaCl is 0.0004 but in HCl it is 0.16. This is because H⁺moves faster than Na⁺.

(iv) Hydration of ion : In general, a decrease in the degree of hydration of anion will increase its transport number. For example, the transport number of Li⁺, Na⁺, K⁺ ions in LiCl, NaCl, KCl solutions are 0.328, 0.396, 0.496 respectively. Thus the ionic mobility of the cations is in the order Li⁺ ≤ Na⁺ ≤ K⁺ which is in the reverse order than that expected from the size of the ions. This anomaly can be explained by saying that Li⁺ is hydrated to greater extent than Na⁺ which in turn is more than K⁺. Thus the effective size of Li⁺ is more than that of Na⁺ which in turn is more than that of K⁺.

(4) Transport number and Ionic mobility : Ionic mobility or Ionic conductance is the conductivity of a solution containing 1 g ion, at infinite dilution, when two sufficiently large electrodes are placed 1 cm apart.

Ionic mobilities (λ_a or λ_c) ∞ Speeds of ions (u_a or u_c)

Unit of ionic mobility is Ohm^–1 cm² or V^–1S^-1cm² : Ionic mobility and transport number are related as,

λ_a or λ_c = t_a or t_c × λ_∞

Absolute ionic mobility is the mobility with which the ion moves under unit potential gradient. It’s unit is cm sec⁻¹.

$\[\text{Absolute}\,\,\text{ionic}\,\,\,\text{mobility}\,\,\,\text{=}\,\,\frac{\text{Ionic}\,\,\text{mobility}}{\text{96,500}}\$

Kohlrausch’s law.

(1) Kohlrausch law states that, “At time infinite dilution, the molar conductivity of an electrolyte can be expressed as the sum of the contributions from its individual ions” i.e., where, n₊ and n_{_} are the number of cations and anions per formula unit of electrolyte respectively and, $\[\lambda _{+}^{\infty }\$ and $\[\lambda _{-}^{\infty }\$ are the molar conductivities of the cation and anion at infinite dilution respectively. The use of above equation in expressing the molar conductivity of an electrolyte is illustrated below,

(i) Molar conductivity of HCl : The molar conductivity of HCl at infinite dilution can be expressed as,

$\[\Lambda _{HCl}^{\infty }={{\nu }_{{{H}^{+}}}}\lambda _{{{H}^{+}}}^{\infty }+{{\nu }_{C{{l}^{-}}}}\lambda _{C{{l}^{-}}}^{\infty }\,;~\,\,\,For\,\,\,HCl,~\,\,{{\nu }_{{{H}^{+}}}}=1\,\,\,and\,\,\,{{\nu }_{C{{l}^{-}}}}=\,1\,\,\,So,\$

$\[\Lambda _{HCl}^{\infty }=(1\times \lambda _{{{H}^{+}}}^{\infty })+(1\times \lambda _{C{{l}^{-}}}^{\infty })\$ ; Hence, $\[\Lambda _{HCl}^{\infty }=\lambda _{{{H}^{+}}}^{\infty }+\lambda _{C{{l}^{-}}}^{\infty }\$

(ii) Molar conductivity of MgCl₂ : For MgCl₂the molar conductivity at infinite dilution can be expressed as,

$\[\Lambda _{MgC{{l}_{2}}}^{\infty }={{\nu }_{M{{g}^{2+}}}}\,\,\lambda _{M{{g}^{2+}}}^{\infty }+{{\nu }_{C{{l}^{-}}}}\lambda _{C{{l}^{-}}}^{\infty }\$ ; For $\[MgC{{l}_{2}},\,\,\,{{\nu }_{M{{g}^{2+}}}}=1\,\,\,and\,\,\,{{\nu }_{C{{l}^{-}}}}=2\$

So, $\[\Lambda _{MgC{{l}_{2}}}^{\infty }=1\times \lambda _{M{{g}^{2+}}}^{\infty }+2\times \lambda _{C{{l}^{-}}}^{\infty }\$ ; Hence, $\[\Lambda _{MgC{{l}_{2}}}^{\infty }=\lambda _{M{{g}^{2+}}}^{\infty }+2\lambda _{C{{l}^{-}}}^{\infty }\$

(iii) Molar conductivity of CH₃COOH : CH₃COOH in solution ionises as, CH₃COOH ⇌ CH₃COO⁻ + H⁺; So, the molar conductivity of CH₃COOH (acetic acid) at infinite dilution can be expressed as,

$\[\Lambda _{CH{{ & }_{3}}COOH}^{\infty }={{v}_{{{H}^{+}}}}\,\lambda _{{{H}^{+}}}^{\infty }+{{v}_{C{{H}_{3}}CO{{O}^{-}}}};\,\,\,For,\,\,C{{H}_{3}}COOH\,\,{{v}_{{{H}^{+}}}}=1\,\,and\,\,{{v}_{C{{H}_{3}}CO{{O}^{-}}}}=1\$

So, $\[\Lambda _{C{{H}_{3}}COOH}^{\infty }=1\times \lambda _{{{H}^{+}}}^{\infty }+1\times \lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\infty }\,or\,\,\Lambda _{C{{H}_{3}}COOH}^{\infty }=\lambda _{{{H}^{+}}}^{\infty }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\infty }\$

The molar conductivities of some ions at infinite dilution and 298 K

Cation	$\[\lambda _{+}^{\infty }S\,\,c{{m}^{2}}\,\,mo{{l}^{-1}}\$	Cation	$\[\lambda _{+}^{\infty }S\,\,c{{m}^{2}}\,\,mo{{l}^{-1}}\$	Anion	$\[\lambda _{-}^{\infty }S\,\,c{{m}^{2}}\,\,mo{{l}^{-1}}\$	Anion	$\[\lambda _{-}^{\infty }\,S\,\,c{{m}^{2}}\,mo{{l}^{-1}}$
H⁺	349.8	Ba²⁺	127.2	OH⁻	198.0	$\[ClO_{4}^{-}$	68.0
TI⁺	74.7	Ca²⁺	119.0	Br⁻	78.4	$\[SO_{4}^{2-}$	159.2
K⁼	73.5	Sr²⁺	119.0	I⁻	76.8	CH₃COO⁻	40.9
Na⁺	50.1	Mg²⁺	106.2	Cl⁻	76.3
Li⁺	38.7			$\[NO_{3}^{-}$	71.4

(2) Applications of Kohlrausch’s law : Some typical applications of the

Kohlrausch’s law are described below,

(i) Determination of $$\Lambda _{m}^{\infty }$ for weak electrolytes : The molar conductivity of a weak electrolyte at infinite dilution $\[(\Lambda _{m}^{\infty })$ cannot be determined by extrapolation method. However, $$\Lambda _{m}^{\infty }$ values for weak electrolytes can be determined by using the Kohlrausch’s equation. For acetic acid this is illustrated below. According to the Kohlrausch’s law, the molar conductivity of acetic acid (CH₃COOH) is given by,

$\[\Lambda _{C{{H}_{3}}COOH}^{\infty }={{\nu }_{{{H}^{+}}}}\lambda _{{{H}^{+}}}^{\infty }+{{\nu }_{C{{H}_{3}}CO{{O}^{-}}}}\,\,\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\infty }=1\times \lambda _{{{H}^{+}}}^{\infty }+1\times \lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\infty }$

Taking the values of $\[\lambda _{{{H}^{+}}}^{\infty }$ and $\[\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\infty }$ (from table), we can write,

$\[\lambda _{C{{H}_{3}}COOH}^{\infty }=(349.8+40.9)\,\,S\,\,c{{m}^{2}}mo{{l}^{-1}}=390.7\,\,S\,\,c{{m}^{2}}mo{{l}^{-1}}$

Sometimes, the molar conductivity values for the ions may not be available. In such cases, the following procedure may be followed,

(a) Select a series of strong electrolytes, so that their sum or difference gives the weak electrolyte. Generally, three strong electrolytes are selected.

(b) Measure Λ_m values of these salts at various concentrations (C_m) and plot Λ_m against $\[\sqrt{{{C}_{m}}}\$ for each salt separately. Determine $$\Lambda _{m}^{\infty }$ for each salt (Strong electrolyte) by extrapolation method.

Suppose we have to determine the molar conductivity of a weak electrolyte MA at infinite dilution. For this purpose, we take three salts, viz., MCl, NaA and NaCl, and determine their $\[\Lambda _{m}^{\infty }\$ values by extrapolation method. Then, according to the Kohlrausch’s law,

$\[\Lambda _{MCl}^{\infty }=\lambda _{{{M}^{+}}}^{\infty }+\lambda _{C{{l}^{-}}}^{\infty }\,;\,\,\Lambda _{NaA}^{\infty }=\lambda _{N{{a}^{+}}}^{\infty }+\lambda _{{{A}^{-}}}^{\infty }\,;\,\,\Lambda _{NaCl}^{\infty }=\lambda _{N{{a}^{+}}}^{\infty }+\lambda _{C{{l}^{-}}}^{\infty }$

From these equations, we can write,

$\[\Lambda _{MCl}^{\infty }+\Lambda _{NaA}^{\infty }-\Lambda _{NaCl}^{\infty }=(\lambda _{{{M}^{+}}}^{\infty }+\lambda _{C{{l}^{-}}}^{\infty })+(\lambda _{N{{a}^{+}}}^{\infty }+\lambda _{{{A}^{-}}}^{\infty })-(\lambda _{N{{a}^{+}}}^{\infty }+\lambda _{C{{l}^{-}}}^{\infty })=\lambda _{{{M}^{+}}}^{\infty }+\lambda _{{{A}^{-}}}^{\infty }=\Lambda _{MA}^{\infty }$

So, $\[\Lambda _{MA}^{\infty }=\Lambda _{MCl}^{\infty }+\Lambda _{NaA}^{\infty }-\Lambda _{NaCl}^{\infty }$

Thus, we can obtain the molar conductivity of a weak electrolyte at infinite dilution from the $\[\Lambda _{m}^{\infty }\$ values of three suitable strong electrolytes.

(ii) Determination of the degree of ionisation of a weak electrolyte : The Kohlrausch’s law can be used for determining the degree of ionisation of a weak electrolyte at any concentration. If $\[\lambda _{m}^{c}\$ is the molar conductivity of a weak electrolyte at any concentration C and, $\[\lambda _{m}^{\infty }\$ is the molar conductivity of a electrolyte at infinite dilution. Then, the degree of ionisation is given by,

$\[{{\alpha }_{c}}=\frac{\Lambda _{m}^{c}}{\Lambda _{m}^{\infty }}=\frac{\Lambda _{m}^{c}}{({{\nu }_{+}}\lambda _{+}^{\infty }+{{\nu }_{-}}\lambda _{-}^{\infty })}$

Thus, knowing the value of $\[\Lambda _{m}^{c}\$ and $\[\Lambda _{m}^{\infty }\$ (From the Kohlrausch’s equation), the degree of ionisation at any concentration (α_c) can be determined.

(iii) Determination of the ionisation constant of a weak electrolyte : Weak electrolytes in aqueous solutions ionise to a very small extent. The extent of ionisation is described in terms of the degree of ionisation In solution, the ions are in dynamic equilibrium with the unionised molecules. Such an equilibrium can be described by a constant called ionisation constant. For example, for a weak electrolyte AB, the ionisation equilibrium is, AB ⇌ A⁺ + B⁻; If C is the initial concentration of the electrolyte AB in solution, then the equilibrium concentrations of various species in the solution are, [AB] = C(1−α) [A⁺] = Cα and [B⁻] = Cα

Then, the ionisation constant of AB is given by,

$\[K=\frac{[{{A}^{+}}][B]}{[AB]}=\frac{C\alpha .C\alpha }{C(1-\alpha )}=\frac{C{{\alpha }^{2}}}{(1-\alpha )}$

We know, that at any concentration C, the degree of ionisation (α) is given by, $\[\alpha =\Lambda _{m}^{c}/\Lambda _{m}^{\infty }\$

Then, $\[K=\frac{C{{(\Lambda _{m}^{c}/\Lambda _{m}^{\infty })}^{2}}}{[1-(\Lambda _{m}^{c}/\Lambda _{m}^{\infty })]}=\frac{C{{(\Lambda _{m}^{c})}^{2}}}{\Lambda _{m}^{\infty }(\Lambda _{m}^{\infty }-\Lambda _{m}^{c})}\$ ; Thus, knowing and at any concentration, the ionisation constant (K) of the electrolyte can be determined.

(iv) Determination of the solubility of a sparingly soluble salt : The solubility of a sparingly soluble salt in a solvent is quite low. Even a saturated solution of such a salt is so dilute that it can be assumed to be at infinite dilution. Then, the molar conductivity of a sparingly soluble salt at infinite dilution $\[(\Lambda _{m}^{\infty })\$ can be obtained from the relationship,

$\[\Lambda _{m}^{\infty }={{\nu }_{+}}\lambda _{+}^{\infty }+{{\nu }_{-}}\lambda _{-}^{\infty }$ …….(i)

The conductivity of the saturated solution of the sparingly soluble salt is measured. Form this, the conductivity of the salt (k_salt) can be obtained by using the relationship, $\[{{\kappa }_{\text{salt}}}={{\kappa }_{\text{sol}}}-{{\kappa }_{\text{wate}r}}\$ , where, k_water is the conductivity of the water used in the preparation of the saturated solution of the salt.

$\[\Lambda _{\text{salt}}^{\infty }=\frac{1000\,\,{{\kappa }_{\text{salt}}}}{{{C}_{m}}}$ ……..(ii)

From equation (i) and (ii) ;

$\[{{C}_{m}}=\frac{1000\,\,{{\kappa }_{\text{salt}}}}{({{\nu }_{+}}\,\,\lambda _{+}^{\infty }+{{\nu }_{-}}\,\,\lambda _{-}^{\infty })}\$ , C_m is the molar concentration of the sparingly soluble salt in its saturated solution. Thus, C_m is equal to the solubility of the sparingly soluble salt in the mole per litre units. The solubility of the salt in gram per litre units can be obtained by multiplying C_m with the molar mass of the salt.

Electrochemistry : Kohlrausch’s Law

Electrochemistry : Kohlrausch’s Law

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