ConductanceConductance is defined as the reciprocal of resistance. CGS unit of conductance is mho and SI unit is Siemens. They are equal.

Now, resistance, R ∞ l ( length of the conductor) and ∞ 1/ A (Cross-sectional area of conductor)

Therefore R ∞ l / A or R = ρ x l / A, where ρ is the proportionality constant called the specific resistance.Therefore conductance Λ = 1 / R = 1 / ρ x A / l = κ x A / l, where κ is the specific conductance.When l = 1 cm. & A = 1 sq. cm. then Λ = κ. So specific conductance, κ is defined as the conductance of a cube of the substance of side 1 cm. measured between its two opposite faces. It depends on the material and the temperature. Equivalent conductance, λ is defined as the conductance of a solution containing 1 gram equivalent of the electrolyte, placed between two coplanar and parallel electrodes 1cm. apart. Similarly molar conductance, λm is defined as the conductance of a solution containing 1 gram mole of the electrolyte, placed between two coplanar and parallel electrodes 1cm. apart.Relation between κ and λ

Let ‘c’ be the concentration of a solution in normality. So ‘c’ gram equivalent of electrolyte is dissolved in 1000 cc. or 1 litre of the solution. So 1 gram equivalent of electrolyte is dissolved in 1000/ c cc of the solution. According to the definition of equivalent conductance the entire solution containing 1 gram equivalent of electrolyte is to be placed between two electrodes 1 cm apart. Let A be the covered area of the electrodes by the solution under the condition. Therefore volume of solution containing 1 gm-equivalent of electrolyte is ( l x A ) cc, where l is the distance between electrodes. Therefore under the condition ( l x A ) = 1000/ c cc. Since distance between electrodes is 1 cm. so A = 1000/c sq. cm. Therefore equivalent conductance λ = κ x A / l = 1000 x κ / c. When c is expressed in molar unit we get molar conductance.

If however c is expressed in gram equivalent per m3 then the volume of solution containing 1 gm-equiv. of electrolyte is 1/ c m3. If κ is in Siemens.m-1 then λ is given as λ = κ / c Siemens.m2.gm-equiv.-1.Relation between λ and λm

If a single molecule of any electrolyte on ionization produces ν+ number of cations each of charge z+ ( or ν- number of anions each of charge z- ) then 1 mole of the electrolyte is equal to ( ν+. z+ ) x 1 equivalent of the electrolyte. If cm be the molar concentration then,

λm = 1000 x κ / cm and λeq = 1000 x κ / ( ν+. z+ ). cm

Hence, λeq = λm / ( ν+. z+ ).Unit of κ, λ and λm

We know 1/R = κ x A / lTherefore κ = 1 / R x l / A. So unit of κ is ohm-1.cm-1 or mho.cm.-1 or Siemens.m-1.Again equivalent conductance λ = 1000 x κ / c. So unit of λ is mho.cm-1 / gm-equiv.cm-3 or mho.cm2.gm-equiv.-1 or Siemens.m2.gm-equiv.-1. For molar conductance gm-equiv. is replaced by gm-mole.Conductance of solutions

For conductance measurement of solutions two electrodes are to be dipped into the solution and current is passed through the solution with the help of electrodes. In the case of conductance measurement of solution ‘l’ and ‘A’ represent the distance between the two electrodes and the area of cross-section of each electrode respectively. Usually electrodes of same area are used. Electrodes in the solution must be kept fixed with respect to one another. Otherwise the distance between the two electrodes and the effective area of the electrodes will change as a result of which conductance of the solution also changes.Cell and cell constant

To meet all these specifications a device of compact arrangement called the cell is used. It consists of two electrodes placed at a fixed distance from one another and fitted within a tube made of highly insoluble glass such as Pyrex. One end of the cell is open and the other end provides two contact points of the two electrodes which come in contact with the solution when the cell is dipped. Electrodes are stout Pt foils coated with finely divided platinum called Pt-black. These are called platinized Pt electrodes. Platinization is carried out by electrolysis of a solution containing about 3 % chloroplatinic acid and 0.02 to 0.03 % of lead acetate, the latter favoring deposition of Pt in a finely divided and adherent form.Advantages and disadvantages of platinization: Platinization increases the effective surface area of the electrodes. The large surface area of the finely divided Pt appears to catalyze the union of the hydrogen and oxygen which tend to be liberated by the successive pulses of the AC current. The polarization emf is thus eliminated. The finely divided platinum may however catalyze the oxidation of organic compounds present, or it may adsorb appreciable quantities of solute present in the electrolyte and so alter its concentration. This disadvantage of platinized electrodes may be partially removed by heating them to redness and so obtaining a grey surface. The resulting electrode is however not so effective in reducing polarization. By making measurements with smooth Pt electrodes at various frequencies and extrapolating the results to infinite frequency, conductance values have been obtained which are in agreement with those given by platinized electrodes.Polarization: At a reversible electrode in a state of equilibrium the discharge of ions and their reformation take place at the same rate and there is no net current flow. If the conditions are such that there is an actual passage of current, the electrode is disturbed from its equilibrium condition. This disturbance of equilibrium associated with the flow of current is called polarization. Polarization results from the slowness of one or more of the processes occurring at the electrode during the discharge or formation of ion. Polarization is mostly due to concentration changes in the vicinity of the electrode. At anode dissolution of metal results in increase in number of ions in contact with it in the solution and if rate of diffusion is relatively slow the concentration of cations in the immediate vicinity of the anode will be greater than

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in the bulk of solution. The result will be an increase of potential of the anode which will be greater the higher the concentration difference. At cathode opposite condition arises.The ratio l/A for any cell with fixed electrodes is fixed and is known as the cell constant. ‘l’ and ‘A’ for any cell cannot be measured very accurately. So to measure cell constant we use relation, Λ = κ x A/l, where Λ is measured conductance and κ is specific conductance. So, Cell constant, l/A = κ / Λ = specific conductance / measured conductance.Unit of cell constant is cm-1 or m-1.To determine cell constant we first prepare solutions of strong electrolytes (like KCl) of exact strength [say 0.1(N) and 0.01(N)]. Specific conductances for these solutions are known from the literature. Scientists have measured κ for these solutions using different cells with different l and A values accurately. The conductances of these solutions are then measured using the cell for which cell constant is to be measured. Using the relation, cell constant = specific conductance / measured conductance, cell constant is calculated.Modifications for measuring conductances of solutionsSome modifications are necessary for measuring conductances of solutions which are as follows-1. The conductivity cell is dipped in the solution care being taken that the electrodes are fully immersed under the solution. The resistance is then measured employing the Wheatstone’s metre-bridge principle. The null point is detected by using a cathode ray oscilloscope.2. The type of cell to be used depends on the type of solution whose conductance is to be measured. If the solution is a highly conducting one the distance between electrodes in the cell are far, whereas for a weakly conducting solution the electrodes are close.3. Use of direct current (D.C.) leads to electrolysis of the solution and polarization. To avoid this alternating current (A.C.) of frequency 500 to 2000 Hz/sec is used. This avoids electrolysis and polarization.Numerical Problems ( 1 → 5 )Variation of conductance with concentration

The conductance of a solution depends on a) number of ions, b) valency of ions, and c) speed of the ions.Specific conductance: Specific conductance by definition is the conductance of a cm. cube of the solution measured between opposite faces.Strong electrolyte: Strong electrolyte remains completely ionized at all concentrations. Dilution does not change total number of ions present in the entire solution but since volume increases so number of ions per cc. decreases. So κ for strong electrolyte decreases on dilution due to decrease in number of ions per unit volume. Truly κ is directly proportional to concentration.Weak electrolyte: Weak electrolytes are partially ionized. On dilution degree of ionization increases and hence total number of ions present in entire solution increases. But volume increase is more with respect to the corresponding change in the number of ions. So number of ions per cc. decreases and hence κ for weak electrolyte decreases on dilution due to decrease in number of ions per unit volume.Equivalent conductance: Equivalent conductance of any solution is the conductance of the solution containing 1 gm-equiv. of electrolyte when placed between two coplanar, parallel electrodes 1 cm. apart.Strong electrolyte: The electrolytes always remain completely ionized. In the solution each ion is always surrounded by an ion atmosphere of opposite charge. In an electric field the central ion and its atmosphere move in opposite direction. Due to coulombic interaction between opposite kind of ions, ionic movement is restricted. With dilution, ionic density decreases. So interactions are reduced and ions are freer to move. So λ increases with dilution.

Usually λ of a solution varies linearly with concentration, c approximately obeying the relation λ = λ0 - B , where λ0 is the equivalent conductance at infinite dilution. By plotting λ versus at different concentrations and then by extrapolating the graph to zero concentration we can get the value of λ0. However at fairly high concentration this variation is not linear. At high concentration the ions form non-conducting ion pairs and do not contribute to conduction thereby resulting in sharp decrease of conductance with increasing concentration. Weak electrolyte: Weak electrolytes are partially ionized and so the number of ions at a given concentration is small compared to the volume of solution. Hence the ions are always in a free to move condition. On dilution degree of ionization increases and so number of ions increases resulting in increase of λ.

However variation of λ for weak electrolytes with concentration is not linear and as concentration approaches zero instead of approaching a limiting value, the conductance tends to become infinite. So, graphical determination of λ0 is not possible for weak electrolytes.

The approach of λ0 of all electrolytes to a limiting value is due to the fact that under these high diluted conditions all the ions that can be derived from 1 gm.eq. take part in conducting the current. Since the total charge is constant for equivalent solutions at high dilution, the λ 0 of an electrolyte depends only on the ionic velocity. Kohlrausch’s law of independent migration of ions

Determination of λ0 for weak electrolytes is possible with the help of Kohlrausch’s law of independent migration of ions. Kohlrausch selected pairs of strong electrolytes having anion part common and differing in the cation part in the same manner irrespective of the nature of the anion and vice-versa. He found that difference of λ0 was constant. For example the λ0

values for some pairs are KF = 111.2, NaF = 90.1 (difference = 21.1); KCl = 130, NaCl = 108.9 (difference = 21.1); again KCl = 130, KF = 111.2 (difference = 18.8); NaCl = 108.9, NaF = 90.1 (difference = 18.8). To explain this Kohlrausch proposed:1. At infinite dilution when dissociation is complete each ion makes a definite contribution towards the equivalent conductance of the electrolyte, irrespective of the nature of the other ion with which it is associated, and

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2. The equivalent conductance at infinite dilution is equal to the sum of ionic conductances. So λ 0 = λ0+ + λ0

- , where λ0+ is the

equivalent conductance of the cation at infinite dilution and λ0- is the equivalent conductance of the anion at infinite dilution.

Let us consider the weak electrolyte CH3COOH. To determine its λ0 we have to choose three suitable strong electrolytes. Here we choose CH3COONa, HCl and NaCl. For these,

λ0 ( CH3COONa ) = λ0 ( CH3COO- ) + λ0 ( Na+) ----------------------------------------------- (1)λ0 ( HCl ) = λ0 ( H+ ) + λ0 ( Cl- ) ----------------------------------------------- (2)

λ 0 ( NaCl ) = λ 0 ( Na + ) + λ 0 ( Cl - ) ----------------------------------------------- (3) By (1) + (2) – (3), we get, λ0 ( CH3COONa ) + λ0 ( HCl ) - λ0 ( NaCl ) = λ0 ( CH3COO- ) + λ0 ( H+ ) = λ0 ( CH3COOH ).[ Note: for NaCl, 1 eq is 1 mole whereas for CaCl2, 1 eq is ½ mole; for 1 mole of CaCl2 gives 1 mole of Ca+2 ions, each ion having a charge of +2 and 2 moles of Cl- ions, each ion having a charge of -1. For an ion with multiple charges, we write ½ Mg +2, ½ SO4

-2 etc., to remind that it is the conductance of 1 eq and not of 1 mole. Thus for Na2SO4 we write: Λ0 ( ½ Na2SO4 ) = λ0 ( Na+ ) + λ0 ( ½ SO4

-2 )Numerical problems ( 6 → 10)Difference between ionic mobility and ionic velocity1. The distance covered by an ion in unit time is known as its velocity which is dependent on the applied potential difference across the two electrodes.Ionic mobility refers to the velocity of an ion under a fall of potential of 1 volt / cm or potential gradient of 1 volt.2. Mobility of an ion depends on all factors by which the velocity depends excepting the applied potential gradient.3. The unit of velocity is cm / sec whereas the unit of mobility is cm / sec / volt / cm, that is, cm2 / volt.sec. Transport number

An aqueous solution of non-electrolyte is non-conducting but the solution of an electrolyte is conducting. According to Arrhenius’ theory of electrolytic dissociation an electrolyte is always ionized, partially or completely in the aqueous solution to form ions. These ions are responsible for conducting electricity through the solution. Although cations and anions are moving in directions opposite to, one another but the flow of current is uni-directional. Depending on various factors such as charge on ion, extent of solvation, velocity of ions or solvated ions etc., different ions may carry different quantities of electricity. For example in a dilute solution of HNO 3 almost 84 % of the current is conveyed by H+ ions and only 16 % is carried by NO3

- ions. Transport number of any ion is defined as the fraction of current carried by that ion.If I be the current passed through the solution and I+ and I- be the current carried by cations and anions respectively, then

Transport number of cation, t+ = I+ / I = I+ / ( I+ + I- ).Similarily, transport number of anion, t- = I- / I = I- / ( I+ + I- ).Relation between velocity and transport numbers of an ion

We consider a solution of an electrolyte through which 1 ampere of current is passed for 1 second. Cations and anions are reaching cathode and anode respectively and in the process they are carrying current.

Let,n+ = number of cations / ccn_ = number of anions / ccZ+ = valency or charge of cationZ_ = valency or charge of anionu = velocity of cationv = velocity of anionЄ = magnitude of electronic chargeI+ = current carried by cationI- = current carried by anion

Now during the passage of electricity all the cations which are within a distance of u cm from the cathode must reach the electrode in 1 second. Similarly anions which are present within the distance of v cm from the anode must reach this electrode simultaneously. Therefore cations which are reaching cathode in 1second must be present in a volume A X u cc. where A is the area of each electrode. So number of cations reaching the cathode in 1 sec = A x u x n+

Now, each cation carries charge = Z+ x ЄTherefore, current carried by cations = A. u. n+. Z+. Є = I+

Similarly current carried by anions =A. v. n_. Z_. Є = I_

Therefore transport number of cation ( t+ ) = I+ / ( I+ + I_ ) = A. u. n+. Z+. Є / A. Є. (u. n+. Z++ v. n_. Z_ )

= u. n+. Z+ / ( u. n+. Z+ + v. n_. Z_ )Since the solution is electrically neutral, therefore, n+. Z+ = n_. Z_

Therefore, t+ = u. n+. Z+ / ( u. n+. Z+ + v. n_. Z_ )= u. n+. Z+ / ( u. n+. Z+ + v. n+. Z+ ) = u. n+. Z+ / n+. Z+. ( u + v )= u / ( u + v )

Similarly transport number of anion ( t_ ) = v / ( u + v ) Relation between ionic conductance and ionic mobility for a very dilute solution

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The conductance of a solution depends upon ionic conductances. Again the conductance of a particular type of ion depends on three factors :- 1) numbers of ions present

2) charge on each ion 3) velocity of the ion

If we consider a solution of 1 gm equivalent of electrolyte irrespective of weak and strong, at an infinite dilution then the electrolyte will be in a state of complete ionization. The product of ionic charge and the number of ions derived from 1 gm equivalent of electrolyte is always a constant quantity irrespective of the type of ion. Therefore ionic conductance in an infinitely dilute solution depends on velocity of the ion only. If λ0

+ and λ0_ be the ionic conductances of cation and anion respectively and u and v be their respective velocities for a solution

at infinite dilution then λ0+ = k.u and λ0

_ = k.v , where k is the proportionality constant to be determined.According to Kohlrausch’s law,λ0 = λ0

+ + λ0-

= k ( u + v ) .......................................................................................................................................................................( 1 )To determine the value of constant k we consider a very dilute solution of concentration ‘c’ gm equivalent / litre. This solution is now

placed in a cube of 1cm side and current is passed through the solution for 1 sec using two opposite faces as electrodes. The potential difference maintained between the two electrodes is 1 volt. Thus according to the definition the conductance of the solution would be its specific conductance (κ). Further the velocities of ions under these conditions will be their ionic mobilities.

This dilute solution may be treated as an infinitely dilute solution for which λ may be replaced by λ0.Therefore, λ0 = 1000. κ / c

or, κ = λ0 . c/ 1000 = k. ( u + v ). c/ 1000, [ from eqn. 1 ] ……………………………………………………( 2 )According to Ohm’s law E/ R = I. In this case, E = 1 volt.Therefore, I = 1/ R = κ.So from eqn. ( 2 ) I = k. ( u + v ). c/ 1000 …………………………………………………………………………………………………..( 3 )

Now during the passage of current through the solution all the cations and anions which are within a distance of u and v cm. from the cathode and anode respectively must reach the respective electrodes in 1 second. Since the area of each electrode is 1 sq. cm. Therefore all these cations and anions must be present within a volume of u and v cc. respectively.

Now in 1 cc of solution there are c/ 1000 gm-equiv. each of both cation and anion.So current carried by cations, I+ = u. c. F/ 1000 coulombs/ second or amperes.Similarily, current carried by anions, I- = v. c. F/ 1000 amperes

Now I = I+ + I- = ( u. c. F/ 1000 ) + ( v. c. F/ 1000 ) = c. F. ( u + v )/ 1000 ……………………………………………………( 4 )Comparing equations ( 3 ) and ( 4 ) we get, k = F

Therefore conductances of cation and anion are related to their ionic mobilities at infinite dilution by the relationsλ0

+ = F. u , and λ0- = F. v, where u and v are mobilities of cations and anions respectively.

Again transport number of cation in a solution is t+ = u / ( u + v ) = λ0+ / [λ0

+ + λ0- ] = λ0

+ / λ0.Therefore λ0

+ = t+. λ0. i. e. transport number x equivalent conductance.Similarily, λ0

- = t-. λ0.Variation of transport numberWith concentration:

The variation of transport number of an ion with concentration depends on:1. Influence of ionic hydration 2. Formation of complex ions 3. Changes in velocities of the ions as a result of interionic attraction

If the speeds of anions and cations vary to relatively different extents as the concentration is changed the transference or transport number will also vary. According to Zones and Dole the variation of transport number with concentration may be expressed in terms of the following empirical relation:

t = [ A / (1+B√C )] - 1 where A and B are two constants and A = 1+ t0 where, t 0 is the transport number at infinite dilution.For very dilute solution √c is very small.Therefore t = A . (1+ B√C )-1 - 1 or t = A . (1- B√C ) - 1, neglecting the higher powers of √C

or t = ( 1+ t0 ) x ( 1 - B√C ) – 1 or t = 1 + t0 - B√C - t0. B. √C – 1 or t = t0 - B√C ( 1 + t0 ) or t = t0 – A. B√C

So, transport number of an ion increases with the decrease in concentration of the solution in the very dilute region.With temperature:

It has been experimentally observed that the difference between ionic conductances and hence mobilities decrease as the temperature is raised. This means that at sufficiently high temperature the two ions of same electrolyte try to carry equal amounts of current. So transport number of each kind of ion tends to the value of 0.5 as temperature is increased. So it suggests that transport number of the ion whose magnitude is smaller than 0.5 tends to increase with rise in temperature and ultimately reaches the value of 0.5. Similarily the ion having transport number greater than 0.5 also has its transport number decreased to 0.5 with temperature increase.Numerical problems ( 11 → 15 )Abnormal transport number

In certain cases particularly with solutions of CdI2 the transport number varies markedly with concentration. At lower concentration CdI2 is ionized into Cd2+ ion and I- ions as, CdI2 = Cd2+ + 2I-. With the increase in the concentration of electrolyte it has been found that the

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transport number of Cd2+ ion gradually decreases below 0.4, passes through 0 and finally becomes apparently negative. At the same time the transport number of I- ion increases beyond 0.6, passes through 1 and finally apparently exceeds unity. This can be explained with the help of formation of complex Cd[CdI4] at higher concentration (above 0.02 eq./ litre ).

CdI2 + 2I- = CdI4-2

.

Due to the movement of Cd2+ ion and [CdI4]-2 in opposite direction to one another, and since appreciable amounts of cadmium are present in the anions and are transferred in the direction opposite to that of the flow of positive current so apparent transport number of Cd2+ ion may become negative. If equal quantities of electricity are carried in opposite direction by Cd 2+ ion and CdI4

-2 ions the transport number will appear to be 0. With increase in concentration the current is almost exclusively carried by Cd 2+ ion and CdI4

-2 ions. Since the speed of anion is greater than that of cation, so the apparent cation transport number becomes negative. Similar cases have also been observed with CdBr2 solutions.Effect of various factors on conductance1. Effect of temperature:

Increase of temperature results in increase of ionic conductance at infinite dilution. The variation with temperature may be expressed with fair accuracy by the empirical equation, λ0

t = λ025 [ 1 + α ( T - 25 ) + β ( T – 25 )2 ], where λ0

t and λ025 are the equivalent ionic

conductances at t0 and at 250 C respectively. α and β are constants for a given ion of a particular electrolyte in a particular solvent. For a narrow temperature range, say 100 on either side of 250 C, the constant β may be neglected. The constant α for salts is about 0.022 to 0.025 and for acids and bases it is usually 0.016 to 0.019. It means that the equivalent conductance increases approximately by 2% for every 10 rise in temperature. For strong electrolytes even at appreciable concentration, the above equation holds good, whereas in case of weak electrolytes, the variation of Λ with temperature is not so regular. The rise in conductance with temperature is due to the decrease in the viscosity of the solution, increase in the speed of the ions and an increase in the degree of ionization in case of weak electrolytes.

Since the conductance of a solution depends on the rate of movement of its ions, which increases with temperature, so we can write λ0 = A. e-E/RT where A is a constant, which may be taken as being independent of temperature over a small range, E is the activation energy of the process which determines the rate of movement of the ions, R is the universal gas constant and T is the absolute temperature. Differentiation of the above equation with respect to temperature, assuming A to be a constant gives,

d ln λ0 / dT = ( 1 / λ0 ). ( dλ0 / dT ) = E / RT2

Again differentiation of the equation, λ0t = λ0

25 [ 1 + α ( T - 25 ) + β ( T – 25 )2 ] with respect to temperature, neglecting β, shows that for a narrow temperature range, ( 1 / λ0 ). ( dλ0 / dT ) = α. Hence the activation energy is given by, E = αRT2. Since α is approximately 0.02 for most ions except hydrogen and hydroxyl ions at 250 C, so for conductance in water the activation energy is about 3.6 Kcal in every case.2. Effect of pressure:

The effect of pressure on electrolytic conductance points to a relationship between ionic mobility and viscosity. Plots of the ratio of the equivalent conductances at pressure p to that at unit pressure, i.e. λp / λ1 versus pressure for a number of aqueous solutions of electrolytes at low concentrations reveal that conductance first increases, passes through a maximum and then decreases with increasing concentration. The initial increase is mainly due to the changes in the viscosity of the medium which decreases by an increase in pressure.3. Effect of dielectric constant of solvent:

In solvents of low dielectric constant, due to small ionizing effects on the electrolytes, the electrostatic forces between oppositely charged ions is appreciable and so equivalent conductance has a small value. Solvents with high dielectric constants yield more conducting solutions. If the dielectric constant of the medium is greater than about 30, the conductance behaviour in that medium is usually similar to that of electrolytes in water, the differences not being fundamental. With solvents of low dielectric constant, however, the equivalent conductance often exhibit distinct abnormalities. It is found that with decreasing concentration the equivalent conductance decreases instead of increasing and at a certain concentration the value passes through a minimum, after which the subsequent variation is normal. In some cases like KI in liquid SO2 the equivalent conductance passes through a maximum and a minimum with decreasing concentration. According to Walden if cmin is the concentration for the minimum equivalent conductance and D is the dielectric constant of the medium, then cmin = kD3, where k is a constant for the given electrolyte. It is seen from the equation that in solvents of high dielectric constant the minimum should be observed only at extremely high concentrations. From measurements made by Fuoss and Kraus it is seen that as the dielectric constant becomes smaller, the falling off of equivalent conductance with decreasing concentrations is more marked. At sufficiently low dielectric constants the conductance minimum becomes evident. The concentration at which this occurs decreases with decreasing dielectric constant, in accordance with Walden’s equation.4. Effect of viscosity of medium:

When conductance of a solution is measured under an applied potential it may be assumed that Stoke ’s law ( F = 6π.r.η.u ) is strictly obeyed by the moving ions. If the measurements are carried out in different solvents and the size of the ions does not change with the change of solvent, then it may be concluded that the product of viscosity and velocity of the moving ion is a constant quantity, because in each case applied potential remains constant. Again we know that the velocity of an ion is proportional to ionic conductances at infinite dilution, since λ0

+ = F.u and λ0- = F.v, where u and v are the ionic mobilities of cation and anion respectively. Also at infinite dilution the

viscosity of the solution may be considered to be equal to that of the pure solvent. So the above conclusion may be extended in the form that the product of ionic conductances at infinite dilution and the viscosity of the pure solvent is a constant quantity irrespective of the solvent used. So, λ0. η0 = constant. This is the Walden’s rule.

Another suggestion, which has been made, to explain the constancy of the product λ0. η0 is that, the ion in solution is so completely

surrounded by solvent molecules which move with it, that is, it is so extensively solvated that its motion through the medium is virtually the same as the movement of solvent molecules past one another in viscous flow of the solvent.

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Experiments show that the constancy of λ0. η0 is an approximate one for a given ion in different solvents except water. So it confirms the validity of Walden’s rule.

Discrepancies or deviations from constancy may be explained in terms of two factors; radius of the ion and applicability of Stoke’s law. In most of the solvents ions are solvated. So sizes of solvated ions in different solvents are different. Thus we may expect different values for λ0. η0 in different solvents. Further the derivation of Stoke’s law assumes the movement of spherical particles in a continuous medium. This assumption is only valid when the size of the moving particle is large in comparison to the molecules of the medium. In aqueous medium the abnormal high value of the product is probably due to higher conductance for Grotthuss conduction mechanism.5. Effect of high frequencies:

A consequence of the existence of the ionic atmosphere, with a finite time of relaxation is the variation of conductance with frequency at high frequencies. If an alternating potential of high frequency is applied to an electrolyte, so that the time of oscillation is small in comparison with the relaxation time of the ionic atmosphere, the unsymmetrical charge distribution generally formed around an ion in motion will not have time to form completely. In fact, if the oscillation frequency is high enough, the ion will be virtually stationary and its ionic atmosphere will be symmetrical. Hence, the retarding force due to the relaxation or asymmetry effect will thus disappear partially or entirely as the frequency of the oscillations of the current is increased. So, at sufficiently high frequencies the conduction of a solution should be greater than that observed with low frequency alternating current. This is known as Debye-Falkenhagen effect. 6. Effect of high voltage:

When the applied potential is very high then the speed of the ion becomes very high. Hence it will travel several times the thickness of the ionic atmosphere in the time of relaxation. As a result the moving ion is virtually free from an oppositely charged ionic atmosphere, since there is never time for it to be built up to any extent. In these circumstances both asymmetry and electrophoretic effects will be greatly diminished and at sufficiently high voltages should disappear. Hence at high voltages the equivalent conductance at any appreciable concentration should be greater than the value at low voltages. The increase in conductance of an electrolyte at high voltage was observed by Wien as is known as Wien effect. Asymmetric or Relaxation effect

In absence of a driving force the coulombic field of the central ion has spherical symmetry. Thus the ion atmosphere formed around the central ion is always spherically symmetric in nature. But when an external field is applied, as in the case of measurement of conductance of a solution the spherical symmetry of the ion atmosphere is destroyed and asymmetry is developed. Under the circumstances the distribution of ion depends not only on the distance of the volume element from the central ion but also on the direction in which the volume element is situated in respect of the direction of ionic motion. The spherical symmetry of the ion atmosphere can be restored only if the ions and solvent molecules, present in it, immediately readjust to the new position of the central ion. This is possible only if movements of the ions in ion atmosphere are instantaneous, that is, no frictional force is present. But this is not the actual situation. A finite time is always required for its readjustment. Even if this time were available the spherical symmetry would not have been restored if the central ion keeps on moving. Since in the applied field ions are always moving, so the part of the ionic cloud behind the central ion is always dispersed by thermal motion and a new cloud is continuously built up in front of the moving central ion. Thus an egg shaped ionic cloud is always associated with the moving ion. The charge-centre of the ionic cloud do not coincide with that of the central ion and an electric force between this two charge centers is developed and acts in the opposite direction to the applied field. Thus the asymmetry of the ionic atmosphere, due to the time of relaxation, results in a retardation of the ion moving under the influence of an applied field. This influence on ionic speed is called the relaxation or asymmetry effect.Electrophoretic effect

Externally applied potential acts both on central ion and oppositely charged cloud which tend to move in opposite direction to one another. During the movement of ionic cloud it tries to carry its entire baggage containing the ions, solvent molecules and the central ion. Thus a retarding influence, equivalent to an increase in the viscous resistance of the solvent, is exerted on the moving central ion. Now the thickness of the ion atmosphere, which is a charged sphere moving under the influence of electric field, in a millimolar solution of 1:1 electrolyte is comparable with the dimensions of colloidal particles. Movement of charged colloidal particle in an electric field is known as electrophoresis. So the retarding influence of the ion atmosphere on the central ion is known as electrophoretic effect, since it is analogous to the resistance acting against the movement of a colloidal particle in an electrical field.Onsager equation

When an external electric field is applied to a solution containing ions, four different forces act on the ions. These are:1. An electric force ( Fe ), under whose influence ions move in the direction of field. The magnitude of the force is equal to V. Z. Є, where

V is the potential gradient, Z is the valency of the ion and Є is the magnitude of electronic charge.2. When the ion moves forward a frictional force ( Fv ) due to solvent, known as viscous force operates on the ion in the direction

opposite to the moving ion. Its magnitude according to Stoke’s law is -6πrvη, where v is the constant velocity of the moving ions, r is the effective radius of the ions and η is the viscosity of the medium. Simplifying it we can write, Fv = -k1v.

3. Relaxation force ( Fr ) due to asymmetry effect or relaxation effect also operates on the ion in the opposite direction of the moving ion. Its magnitude is –k2V√c, where k2 is a constant depending on temperature, r and η.

4. Finally, another force, electrophoretic force ( Fp) also acts in the same direction as the above two forces, due to the electrophoresis phenomenon of the ionic cloud. Its magnitude is –k3V√c.

At equilibrium, the magnitude of the first force must balance the magnitudes of the last three forces for steady movement of ionsSo electric force ( Fe ) = viscous force ( Fv ) + relaxation force ( Fr ) + electrophoretic force ( Fp). Therefore, V. Z. Є = k1v + k2V√c + k3V√c.

Or k1v = VZЄ – ( k2 + k3 )V√c

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Or v/ V = ZЄ/ k1 – ( k2 + k3 )√c/ k1

Or vF/ V = ZЄF/ k1 – ( k2 + k3 )F√c/ k1

Or λ = ZЄF/ k1 – ( k2 + k3 )F√c/ k , where λ is the ionic conductance and v/ V is the ionic mobility.Now when the solution is infinitely dilute, λ = λ0 and c is zero. Therefore λ0 is ZЄF/ k1.Therefore, λ = λ0 – ( k2 + k3 )F√c/ k. Considering both the cations and the anions of the electrolyte and by using mathematical expressions of different forces Onsager

finally derived an expression for equivalent conductance of the solution for uni-univalent electrolyte which is:λ = λ0 – [ A + B. λ0 ]√ c,

where λ0 is the equivalent conductance of the solution at infinite dilution and A and B are two constants. Two terms in the square bracket are due to electrophoretic and relaxation effects. The forms of A and B are:

A = 82.4 / ( D. T )1/2. η and B = 8.2 x 105 / ( D. T )3/2, where D is the dielectric constant of the medium and T is the temperature. Thus the two constants depend only on the nature of solvent and temperature. The above equation is known as Debye-Huckel-Onsager conductance equation. For electrolyte other than uni-univalent type the equation is more complicated. The value of the two constants can be determined from the plot of λ versus √c for different uni-univalent electrolytes in the same solvent at same temperatures. For different solutions of a given electrolyte we can determine a relation between A and B. Similarily taking other electrolytes we can also find out another relation connecting A and B. From the solution of these two relations the constants A and B can be evaluated. The value of A for solvent water ( D = 78 ) at 250 C is:

A = 82.4/ 0.9935 x ( 78 x 298 )1/2 = 50.5, and for B is 0.224 units.Validity of Onsager equation

1. Onsager equation is a limiting equation applicable only on very dilute solutions, since during its derivation the assumptions of point charge and dilute solutions have been made.

2. Validity of the equation does not necessarily mean the linearity of the plot, but the values of the constants A and B derived from the experimental slopes must agree with those obtained theoretically from the values of D and T.

Q: Use Onsager equation to show that Λ changes rapidly as C → 0.Q: What are the units of A and B in Onsager equation ?Q: Find the expression of λ0 from Onsager equation.Applications of conductance measurements1. Determination of solubility (S) and solubility product (Ks) of sparingly soluble salts:

When a sparingly soluble salt is added to water very little of it gets dissolved. So the solution may be considered to be a saturated solution with respect to the salt. Further the concentration of the salt is so low that it may be considered as an infinitely dilute solution. Hence, λ0 = 1000 x κ/ c, where λ0 is the equivalent conductance of the saturated solution of the sparingly soluble salt at infinite dilution, κ is the specific conductance of the electrolyte alone and c is the concentration of the solution in normality, that is in eq./ litre. If κ sol and κwater

represent the specific conductances of the solution and water respectively then,λ0 = 1000 x ( κsol - κwater )/ c

or, c = 1000 x ( κsol - κwater )/ ( λ0+ + λ0

- )From here we get the concentration of the salt in normality if λ represents equivalent conductance and in molarity if λ represents molar conductance. If the concentration is obtained in normality we transform it to molar concentration to get the solubility (S). If the concentration is in molarity the solubility is obtained directly. From here the solubility product (Ks) is obtained as

Ks = xx. yy. (S)( x+ y).Numerical problems ( 15 → 20)2. Determination of degree of ionization and ionization constant of weak electrolytes:

The ratio of the equivalent conductance ( λ ) at any concentration, to that at infinite dilution ( λ 0 ) is called the conductance ratio ( α ). According to Arrhenius the conductance ratio is equal to the degree of dissociation for weak electrolytes. This is because at any concentration the electrolyte is only partially ionized. But at an infinitely dilute state the electrolyte is completely ionized and hence its conductance then represents the equivalent conductance at infinite dilution. We know that

K = α2c/ ( 1- α ) and α = λ/ λ0 = λ/ ( λ0+ + λ0

- )Knowing the values of λ0

+ and λ0- from the literature and by measuring λ we can calculate α. Thereby we can calculate K since c is known.

Q: How can you find out the dissociation constant ( K ) and the λ 0 of a weak ( 1:1 ) electrolyte from a set of conductivity data of the weak electrolyte itself ?Numerical problems ( 20 → 25 )3. Determination of ionic product ( Kw ) of water:

Water is a weakly conducting substance and is feebly dissociated as:2 H2O ↔ H3O+ + OH-

c 0 0 where c is the concentration of water in molarity.c(1-α) cα cα

The ionic product of water Kw = [ H3O+ ].[ OH- ] = c2.α2.Now c = 0.997/ 18 (M) = 55.3 (M), since concentration of water is 0.997 gm/ cc at 250 C.But α = λ/ λ0, where λ is the equivalent conductance of water of concentration c ( = 1000.κ/ c ). Again λ 0 of water is the sum of ionic conductances. Therefore λ0 = λ0 ( H3O ) + λ0 ( OH- ) = ( 349.8 + 198 ) mho.cm2.mol-1. = 547.8 mho.cm2.mol-1.Again, κ for water is 5.8 x 10-6 mho/ cm.

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Therefore, Kw = c2.α2 = c2. ( λ/ λ0 )2 = ( 1000.κ/ c )2. c2/ ( λ0 )2.Putting the values of κ, λ0 and c we get Kw = 1.12 x 10-14 at 250 C.4. Determination of degree of hydrolysis and hydrolysis constant of a salt:

The salt must be of a strong acid and very weak base or vice-versa. Typical example of this type of salt is aniline-hydrochloride (C6H5NH2.HCl ).

C6H5NH2.HCl + H2O ↔ C6H5NH2 + HCl(H2O)c 0 0c( 1- x ) cx cx

The conductance of aniline hydrochloride is partly due to HCl formed as a result of hydrolysis and partly due to the unhydrolysed salt. The conductance of aniline formed may be neglected as it is a very weak base and exists completely in the unionized form. Let c be the concentration of salt and x is the degree of hydrolysis. Therefore hydrolysis constant of the salt, Kh is given as,

Kh = cx. cx/ c(1-x) = cx2/ (1-x) ≈ cx2, since x is very small compared to 1, so (1-x) ≈ 1.The conductance measurement is used for calculating x and Kh. Now total conductance of the solution λ is the sum of conductance of

c(1-x) gm-equiv. of hydrolyzed salt and cx gm-equiv. of free HCl, as aniline has no conductance.Therefore, cλ = c(1-X).λs + cx.λHCl, where λ is the equivalent conductance of the resulting solution which is directly measured, λs is the

equivalent conductance of hydrolyzed salt and λHCl is the equivalent conductance of hydrolyzed HCl. Now λHCl = 1000.κ/ c, c can be experimentally determined. Again since the concentration of HCl is very small in the resulting solution so it may be assumed to be an infinitely dilute solution of HCl. So we can write λHCl = λ0 ( H+ ) + λ0 ( Cl- ).

To determine λ, free base is added to HCl solution drop wise and conductance of the solution decreases due to backward reaction. When the conductance of the solution attains minimum value, it is assumed that the salt is present as totally unhydrolysed. At this stage equivalent conductance of the solution is determined by measuring specific conductance. This value gives the equivalent conductance of the unhydrolysed salt λs. The equivalent conductance of HCl can then be determined from the sum of ionic conductances. From the equation, cλ = c(1-X).λs + cx.λHCl , we have,

λ = (1-X).λs + x.λHCl = λs – x.λs + x.λHCl

or, λ - λs = x.( λHCl - λs) or, x = ( λ - λs)/ ( λHCl - λs)Therefore, Kx = [( λ - λs)/ ( λHCl - λs)]2.c. From the value of λ, λs and λHCl, x and Kx can be calculated.

5. Conductometric titrations:Different conductometric titration:i) Acid-base reactionsA. Titration of a strong acid with a strong base (eg. HCl with NaOH)

In this titration HCl is taken in the beaker and NaOH is gradually added from the burette. HCl being a strong electrolyte, completelyionizes in the solution. So the initial conductance of the solution is high. The reaction is

H + + Cl- + NaOH → Na+ + Cl- + H2OWhen strong base NaOH is gradually added to the acid, the highly conducting H+ ions of the acid are gradually removed from the

solution by the same number of less conducting Na+ ions. As a result the conductance of the solution decreases sharply. This trend continues as more and more NaOH is added. After complete neutralization of the acid the conductance of the solution goes on increasing due to the presence of unreacted NaOH which remains as Na+ and OH- in the solution. Thus at the neutralization point the conductance of the solution is minimum. The neutralization point is detected from the plot of conductance vs. numbers of drops of alkali added (this is practically volume of NaOH solution added). It is the point of intersection of two lines drawn before and after the equivalence point. Comparison of the two portion of the curve, before and after the equivalence point, shows that the former is steeper than the later. This is because the conductance of H+ ions is greater than that of OH- ions.

If the titration is carried out in a reverse manner i.e. HCl is added from the burette to the solution of NaOH taken in the beaker, the nature of the curve will remain same with the only difference is that the second portion of the curve after neutralization will be steeper than the first portion.B. Titration of a strong acid with a weak base (eg. HCl with NH4OH )

The strong acid is taken in the beaker and the weak base in the burette. The reaction is H+ + Cl- + NH4OH →NH4

+ + Cl - + H2OInitial conductance of the solution is high because HCl is a strong electrolyte. The conductance of

the solution decreases when NH4OH is added to the solution due to the replacement of more conducting species H+ by the less conducting NH4

+ ions. After neutralization the conductance of the solution practically remains constant, because the unreacted NH4OH is present in unionized form in presence of common ion NH4

+ coming from the ionization of NH4Cl which is produced in the reaction. From the point of intersection of the two portions the equivalence point can be determined.C. Titration of weak acid with a strong base (eg. HAc with NaOH)

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In this titration acetic acid is taken in the beaker and NaOH is added from the burette. The reaction is

CH3COOH + NaOH → Na+ +CH3COO- + H2OAcetic acid being a weak acid, ionizes very little in the solution and thus the initial conductance of

the solution is low. At the very early stage of titration conductance of the solution decrease because the acetate ion whish is formed in this reaction represses the ionization of weak acid due to common ion effect. With further addition of NaOH the conductance of the solution increases because almost non-conducting acid CH3COOH, in presence of common anion CH3COO- , I gradually converted into the highly ionized salt CH3COO-Na+. After the equivalent point there is a steep rise in the conductance value due to the presence of unused Na+ and OH- in the system.

Nearer to the equivalence point some what higher conductance value is recorded than anticipated due to hydrolysis o the salt.D. Titration of a weak acid with a weak base (eg. HAc with NH4OH)

When weak acid is titrated with a weak base the conductance titration curve is very similar to the curve with strong base. The rising part of this portion is due to the formation of salt NH4Cl which remains ionized in the solution. After the equivalence point there is practically no change in conductance because weak base is not practically ionized. Sharper equivalent point is observed in this titration than with strong base.E. Titration of a very weak acid with a strong base (eg. Phenol or H3BO3 with NaOH)If the acid is very weak such as phenol or basic acid or dilute solution of moderately weak acid then in its titration with a strong base conductance of the solution increases from the commencement of neutralization and the initial fall which is observed in the case of weak acid is missing here. The reason is that the acid is so weak that is hardly ionizes.The formation of salt, which is strong electrolyte is however responsible for the increase in conductance.

The titration of weak acid with a weak base is not possible owing to the extensive hydrolysis of salt even when the medium is acidic.

F. Titration of a mixture of strong and weak acid with a strong /weak base (eg. HCl and HAc with NaOH/ NH4OH )

NaOH solution is taken in a burette and the acid mixture is taken in the beaker .A mixture of acid can be titrated with a strong base only when the dissociation constant of the strong acid is at least 103 times higher than that of the weak acid. The principle behind the titration is that the ionization of weak acid can not take place in presence of large concentration of H+ ion coming from the ionization of strong acid. So when strong base is gradually added to the solution of mixed acids it reacts with HCl first. Thus the conductance of the solution decreases sharply due to the removal of high conducting H+ ions by the same number of low conducting cations. The reaction is

H+ + Cl- + CH3COOH + NaOH → Na+ + Cl- + CH3COOH + H2OAfter complete neutralization of strong acid, addition of strong base increases the conductance of the

solution by converting non-conducting weak acid molecules into their highly ionisable salt. The conductance of the solution rises sharply after the equivalence point because of the presence of unreacted ions coming from strong base. The first break point of the curve represents the equivalence point of titration of strong acid and difference of two break points represents the amount of base required for the neutralization of weak acid.

CH3COOH + NaOH → Na+ + CH3COO- +H2OIf weak base is used instead of strong base, the nature of the curve will remain same with the

exception that after complete neutralization of acid mixture the curve turns parallel to the volume of alkali-axis. G. Titration of a dibasic acid with a strong base (eg. Oxalic acid with NaOH)

Oxalic acid is a dibasic acid. It ionizes in two steps. According to the equationsCOOH----COOH ↔ COOH.COO- + H+

COOH.COO- ↔ COO-.COO- + H+

Thus it has two ionization constants corresponding to the two steps. The value of first ionization constant is very large in comparison to the second ionization constant. Thus it may be assumed that the second ionization cannot take place in presence of first ionization of the acid. Thus it may be regarded as an intramolecular mixture of strong and weak acid. So the nature and discussion of the curve will be the same as in the case of mixture of strong and weak acid. The only difference is that the first equivalence point lies always at the mid point of second equivalence point.ii) Displacement reactions H. Titration of salt of weak acid and strong base with strong acid (eg. CH3COONa with HCl)

In this case the salt solution is taken in a beaker and HCl is added from the burette. The reaction is CH3COO- + Na+ + HCl → Na+ + Cl- + CH3COOH

The solution of sodium acetate has considerable conductance value. According to the reaction when HCl is added to the salt solution acetate ions are replaced by equal number of chloride ions. The

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conductance of the solution may increase, decrease or remain same depending on the relative magnitudes of conductances of the two ions. If the conductance of Cl- is greater than that of CH3COO- ion, curve I is obtained. Curve II and III will be found when ionic conductances follow the relation λ (Cl- ) = λ (CH3COO- ) and λ (Cl- ) < λ (CH3COO- )- ion respectively. Whatever may be the cases, the conductance of the solution rises sharply when HCl is added after the equivalence point. This is due to the presence of unreacted H + ion and Cl- ions in the solution. Point of intersection of the curves gives the equivalence point.I. Titration of a mixture of weak base and the salt of weak acid by a strong acid (eg. CH3COONa and NH4OH with HCl)

The initial conductance of the solution is not too small as in the case of NH4OH alone due to presence of highly ionisable salt CH3COONa. When strong acid is added to the solution it first react with the base and only after complete neutralization of WB it starts to react with the salt as in the case of displacement reaction. The reaction is

1. NH4OH + HCl → NH4+ + Cl- + H2O

Addition of very small amount of HCl at the initial stage lowers the conductance of the solution. This is because of the removal of OH -

ions that are present initially with simultaneous decrease in the degree of ionization of NH4OH in presence of common ion from NH4Cl. But soon after, the conductance of the solution begins to increase because almost unionized NH4OH is gradually transformed into highly conducting ionisable salt NH4Cl. After incomplete neutralization of NH4OH displacement reaction begins according to the following equation

2. CH3COO- + Na+ + HCl → Na+ + Cl- + CH3COOHThe conductance of the solution may increase, decrease or remain same depending on the relative

ionic conductances of CH3COO- and Cl- ions. After complete reaction steep rise in the conductance value is observed due to the presence of unreacted H+

and Cl- ions. First equivalence point represents the complete neutralization of WB and the difference between first and second equivalence point indicates the amount of acid required for the reaction with salt.

iii) Precipitation reactionsJ. Titration of KCl with AgNO3 solution

In this titration KCl is taken in the beaker and AgNO3 in the burette. The reaction isK+ + Cl- + AgNO3 → AgCl ↓+ K+ + NO3

-.Since KCl is a strong electrolyte, therefore initial conductance of the solution has a high value. When

AgNO3 solution is added from the burette, AgCl is formed which is thrown out of the solution in the form of precipitate. So the reaction suggests that Cl- ions are replaced by same number of NO3

- ions. The conductance of Cl- ions and NO3

- ion is the same. Therefore addition of AgNO3 will not change the conductance of the solution. But after complete reaction conductance of the solution increases due to presence of unreacted Ag+ ion and NO3

- ions in the solution.K. Titration of MgSO4 with Ba(OH)2 or Ca(OH)2:

In this titration MgSO4 is taken in the beaker and Ba(OH)2 or Ca(OH)2 in the burette. The reaction isMg+2 + SO4

-2 + Ba(OH)2 → BaSO4 ↓ + Mg(OH)2 ↓.MgSO4 being a strong electrolyte its solution has considerable conductance value. When Ba(OH)2 is

gradually added from the burette it reacts with MgSO4 and produce BaSO4 and Mg(OH)2. Both the substances form precipitates and therefore the conducting species of the solution are rapidly removed from the solution phase. So conductance of the solution decreases sharply. After equivalence point conductance of the solution steeply rises due to the presence of OH- ions and Ba+2 ions.Q: Draw the derivative and double derivative plots of the above curves.Abnormal conductance of H + and OH - ions

Experimental observations show that the conductance of H+ ion in aqueous solution is abnormally high. Initially it was assumed that this high conductance of hydrogen ion may be due to its small size. In the aqueous medium hydrogen ion remains as hydrated form according to the equation - H+ + H2O↔ H3O+ . The effective size of hydrated proton and its conducting power should be approximately equal to that of the sodium ion because of almost same conducting power of the two ions in non-hydroxylic solvents where both of them are solvated. So the abnormally high conductance of hydrogen ion can not be explained on the basis of ionic velocity only. There must be some

mechanism which is responsible for this high conductance value. It has been explained on the basis of Grotthuss conduction. According to this and its subsequent development it is now believed that a proton jumps from H3O+ ion in to its adjacent water molecule. The resulting H3O+ in turn again transfer proton to the adjacent water molecule and this process is continued.

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After the transfer of proton to the neighboring molecule the orientation of the water molecule is completely different from that which was initially present. So to continue the process each water molecule must rotate through an angle of 180 0, so that it may receive another hydrogen ion coming from the same direction. Thus in the process hydrogen ion actually moves a very short distance, but as a result of its effective distance covered by the ion is very large. Therefore its conductance is abnormally high.

Similar to the hydrogen ion OH- ion also has high conductance value with respect to other anions in the aqueous medium. This can also be explained in terms of proton jump from adjacent water molecule to the OH- ion

followed by rotation of the molecule

through an angle of 1800 .

The abnormal

conductance of H+ ion in CH3OH and C2H5OH can also be explained in terms of proton transfer analogous to that suggested for water. A proton in alcohol medium may be represented by R-O-H2

+.The dependence of H+ ion conductance with the chain length of alkyls group shows that the conductance decreases with the chain

length because of higher energy barrier. This explains that the conductance of H+ ion in CH3OH is greater than that in C2H5OH.NH4

+ ion and NH2- ion have also abnormal conductance value in liquid NH3 medium.

Conductivity waterDistilled water is a poor conductor of electricity, but due to the presence of impurities such as ammonia, carbon-dioxide and traces of

dissolved substances derived from containing vessels, air and dust, it has a conductance sufficiently large to have an appreciable effect on the results in accurate work. This is very important in the case of dilute solutions or weak electrolytes, because the conductance of the water is then of the same order as that of the electrolyte itself. Moreover the impurities can influence the ionization of the electrolyte or chemical reaction may occur. So water as free as possible from impurities should be used. Such water is called conductivity water. The purest water hitherto obtained was prepared by Kohlrausch by distilling water forty two times under reduced pressure. It had a specific conductance of 0.043 X 10-6 mho.cm-1. But water of such a degree of purity is extremely hard to prepare. The so called ‘ultra-pure water’ with a specific conductance of 0.05 to 0.06 X 10-6 mho.cm-1 at 180 C can be prepared without serious difficulty, the main problem being the removal of carbon-dioxide. For general laboratory measurements water of specific conductance of about 1 X 10 -6 mho.cm-1 at 180 C is satisfactory and can be obtained by distilling good distilled water to which a little of permanganate or Nessler’s reagent has been added. The distilling flask and the condensation flask should be of resistance glass.Activity coefficient of electrolyte solution

It is found that the solution of nonelectrolytic substance behaves ideally and concentration terms can be used instead of activity coefficient even at moderately concentrated solution. But in electrolytic solution even in dilute state, when concentration terms are used in the expression of physical constants in ionic equilibrium and chemical kinetics, much error enters into the calculation. But when activity terms are used better results emerge. Therefore it can be said that this solution are assumed to be non-ideal due to interionic attraction and repulsion. Activity of solution (a) is given by, a = c.f, where c is the concentration of the solution and f is the activity coefficient of the solution. Now when f = 1, then the solution becomes ideal. But when f ≠ 1, then a ≠ c and so the solution is non-ideal. The degree of non-ideality can be measured from the departure of the value of f from unity.

To use the activity terms we must know the activity coefficient of the solution. Debye and Huckel developed a relation for the activity coefficient of a solution from their ion atom model.Relation between activity (a) and concentration (c) of an electrolyte solution

Let us consider an electrolyte BxAy which dissociates completely in solution as BxAy → xBz+ + yAz-. The (+) and (–) signs have been used to represent cations and anions respectively. Subscripts 1 and 2 are used to denote solvent and solute respectively. Now if c 2 moles of solute are dissolved in c1 moles of solvent then the total free energy of the solution according to Gibbs-Duhem equation is

G = c1μ1 + c2μ2 ……………………………………………………………………………………………………….. ( 1 )Since the solute is completely dissociated it can also be written as,

G = c1μ1 + xc2μ+ + yc2μ-

= c1μ1 + c2. ( xμ+ + yμ- ) ……………………………………..………………………………………..( 2 )Comparing equations ( 1 ) and ( 2 ) we have

μ2 = xμ+ + yμ- ………………………………………………………………………………………………. ( 3 )

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The mean chemical potential of the ions is given by, μ± = ( xμ+ + yμ- )/ ( x+ y )Or μ±( x+ y ) = xμ+ + yμ- Or ν μ± = xμ+ + yμ-, where x + y = ν …………………………………..……………………………………….. ( 4 )

From equation ( 3 ) and ( 4 ) we haveμ2 = μ = ν μ± = xμ+ + yμ- …………………………………………………………………………………… ( 5 )

If standard chemical potentials are used then the equation becomes,μ0 = μ±

0 = xμ+0 + yμ-

0 ………………………………………………………………………………………… ( 6 )Now using the expression of chemical potentials with activity of different species we have,

μ = μ0 + RT ln a, where a, is the activity coefficient.Or, μ± = μ±

0 + RT ln a±.Therefore, μ+ = μ0

+ + RT ln a+ and μ- = μ0- + RT ln a-

Putting these values in equation ( 5 ), we haveμ0 + RT ln a = ν.[ μ±

0 + RT ln a± ]= x.( μ0

+ + RT ln a+ ) + y. ( μ0- + RT ln a- )

or, μ0 + RT ln a = ν.μ±0 + RT ln ( a± )ν

= RT ln (a+)x. (a-)y + (x.μ0+ + y. μ0

- ) ……………….……………………………………( 7 )Utilizing equations ( 6 ) and ( 7 ), we have RT ln a = RT ln ( a± )ν = RT ln [(a+)x. (a-)y]

Or a = a±ν = (a+)x. (a-)y …………………………………………………………………………………………( 8 )

Again, a± = c±. f±, a+ = c+.f+ and a- = c-.f-, where f is the respective activity coefficients. Putting these values in equation ( 8 ), we havea = c±

ν. f±ν = ( c+.f+ )x.( c-.f- )y

or a = c±ν. f±

ν = c+x. c-

y. f+x. f-

y ( where, c±ν = c+

x. c-y and f±

ν = f+x. f-

y ) ………………………………………( 9 )So equation ( 9 ) becomes a = a±

ν = ( c+x. c-

y ). f±ν ……………………………………………………………………………….( 10 )

Now c+ = cx and c- = cy. Putting these values in equation ( 10 ) we have,a = a±

ν = [(x.c)x. (y.c)y]. f±ν

or, a = a±ν = xx. yy. c(x+y). f±

ν

or, a = a±ν = xx. yy. cν. f±

ν

This is the relation between activity (a) and concentration (c) of the electrolyte.Mean activity coefficient of an electrolyte solution

Mean ionic activity-coefficient (f±) of an electrolyte solution can be experimentally determined from the e.m.f. measurement of a cell and solubility of a sparingly soluble salt. It is found that mean activity coefficient (f ±) varies with concentration (m) of the electrolyte as follows: ln f ± = -A1√C. The relation is valid at very high dilution.

The following conclusion can be drawn from the above experimental data.1. The value of f ± decreases with concentration and attains minimum and then increases.2. For the same type of electrolytes the variation of f ± with m is found to be same up to certain range of concentration (when the solution is dilute). At higher concentration specific nature of the ions play a role in the variation. Lewis formulated the relation at low concentration, ln f ± = -A’√m 3. The f ± of an electrolyte solution depends on the valency of the ions and the ionic concentration of the electrolyte solution i.e. for uni- univalent electrolytes the value of A will be same but for different types of electrolyte the value of A will be different.4. Again for mixture of ions it is found that the influence of an ion on the activity coefficient f ± of the electrolyte depends on the square of its valency. For example, in a mixture of BaCl 2, KCl and NaOH solution, the influence of Ba2+ ion is 4 times than that of Na+ ions.

Thus to account for the valency of the ions in solution Lewis and Randall proposed a concentration term called ionic strength (μ). The ionic strength is defined as, μ = ½ ∑mi.zi

2, where m is the molality of the solution. = ½ ∑c2.z2

2, where c is the molarity of the solution.Therefore for KCl solution of molality m, μ= ½ [mK+.zK+

2 + mCl-.zCl-2] = ½.[m.12 + m.12] = m

So for uni-univalent electrolyte I = m.For bi-univalent electrolyte like CaCl2 of molality m, μ = ½.[m.22 + 2.m.12] = 3mFor bi-bivalent electrolyte like MgSO4 of molality m, μ = ½.[m.22 + m.212] = 4m.Ionic atmosphere

Aqueous solution of any electrolyte always contains ions. These ions are exposed to two opposing forces. Due to thermal motion ions of the solution distribute themselves in a random fashion. On the other hand coulombic force helps the ions to get an organized structure. This organized structure consists of an ion, known as central ion or reference ion, surrounded by counter ions in excess with relatively small number of same kind of ions. Thus as a result of time average distribution of ions every ion is surrounded by an ion atmosphere of opposite sign. The net charge of the atmosphere is equal in magnitude but opposite

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in sign to that of central ion. The distance up to which the ion atmosphere is spherically extended from the central ion is known as the thickness of the ion atmosphere. If the ion has definite size the distance is measured from the surface of the central ion.Debye-Huckel theory of ionic interactionsAssumptions of Debye-Huckel theory1. Solutions of true electrolytes or ionophores are considered. These electrolytes are completely ionized in all concentrations. Ion pair formation or ion association is ignored over this range of concentration.2. Ions are assumed to be point charges. Only the central ion has been given the discrete charge and the surrounding ions have been considered in the form of smothered out charge density and not as discrete charges. The charge density within the ion atmosphere is highest near the central ion and decreases exponentially with increasing distance from the central ion. After certain distance the charge density practically vanishes. This distance is known as the thickness of the ion atmosphere.3. Only long range coulombic forces are relevant to ion-ion interactions and short range non-coulombic forces like dispersion forces play a negligible role.4. Boltzmann distribution of molecular energies is applicable to ions existing in solutions with varying potentials.5. The average electrostatic interaction energy (Ziεψ) of an ion is is less than kT, which is proportional to the thermal energy. If it were large then the ionic interaction would overcome the thermal motion and the ions would aggregate into a solid.6. The non-ideality of the electrolyte solution is due to the presence of ionic atmosphere. At infinite dilution the ionic atmosphere will vanish and the solution will behave ideally.7. The only role of the solvent is to provide a dielectric medium for the operation of interionic forces.Determination of thickness of ion atmosphere.

Any one ion of the assembly is selected arbitrarily as central ion having discrete charge. Other ions of the atmosphere are considered as smothered out charge density. The charge density at distance r from the central ion is ρ and it decreases exponentially with the distance. The dielectric constant of the solvent is D. The central ion is a cation of charge Z+ε, here Z+ is the valency of cation and ε is the magnitude of electronic charge. Let dV be a small volume element within the ion atmosphere at a distance r from the central ion. Let n+ and n+

0 be the number of cations per unit volume in the volume element dV and in the solution bulk (where there is no ion atmosphere) respectively. Similarly n- and n-

0 represent the number of anions respectively and Z+ and Z- is the valency of cation and anion respectively. Further it is assumed that the time average electrostatic potential within the volume element dV, at a distance r from the central ion is ψ. According to definition of electrostatic potential it is the amount of work done to bring a unit positive charge from ∞ to that point. In this case infinity means a distance where the influence of central ion is absent. Now the charge density ρ in the volume element dV is

ρ = [n+ Z+ε.dV + n-( -Z-ε).dV]/dV = n+ Z+ε + n-( -Z-ε)According to Boltzmann distribution law ni = ni

0.e—E/kT. In this case ni and ni0 may be considered as i-th kind of ion in the volume element dV

and in the solution bulk and E is the change in the P.E of the i-th kind of ion when their concentration in the volume element dV is changed from the bulk value ni

0 to ni. In other words E represents the amount of work done to bring an i-th kind of ion from the solution bulk to the volume element. Hence in this case, n+ = n+

0.e-E/kT and n- = n-0.e-E/kT. Now E+ is the amount of work done to bring a cation of charge Z+ε from

solution bulk to volume element dV. This is equal to Z+εψ. Similarly E-, the amount of work done to bring an anion of charge –Z -ε form

solution bulk to volume element dv is equal to –Z-εψ. So and .

So ρ = n+Z+ε – n-Z-ε =

For formation of ion atmosphere it is necessary that the thermal energy must be greater than ionic interaction. So Z +εψ/ kT and Z-εψ/ kT must be « 1. So expanding the exponential terms and neglecting the higher powers for being very small, we get,

Therefore ρ = n+0Z+ε – (n+

0Z+2ε2ψ)/ kT – n-

0Z-ε – (n-0Z-

2ε2ψ)/ kT. Since the solution is electrically neutral, so amount of positive charge per unit volume in the solution bulk is equal to the amount of negative charge per unit volume. Hence n+

0Z+ε = n-0Z-ε.

Therefore ρ = - ε2ψ (n+0Z+

2 + n-0Z-

2)/ kT. If there are other cations and anions present in the solution then the net charge density will beρ = - ε2ψ/ kT Σ ni

0Zi2, where ni

0 is the number of i-th kind of ion in the solution bulk and Zi is its valency.Now the net charge density ρ and the time average electrostatic potential in the volume element dV are related by Poisson’s equation,

∇2ψ = -4πρ/ D, where ∇2 is the Laplacian operator and its explicit form is . Changing the above equation from

Cartesian to spherical polar co-ordinate system we can omit the θ and φ terms of the equation. This is because the ion atmosphere being symmetrical around the central ion the terms δψ/ δθ and δψ/ δφ equates to zero. Therefore the equation becomes

or , where K2 = 4πε2/ DkT. Σ ni0Zi

2 ………( 1 )

The solution of the above differential equation is ψ = (A.e-Kr)/ r+ (B.eKr)/ r, …………………………………………( 2 )where A and B are two constants to be determined from the boundary conditions of the problem.

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At distance far enough from the central ion situated at r = 0, the thermal forces completely dominate over the coulombic forces since the latter varies as 1/ r2. Thus ψ = 0 when r = ∞. Under this condition of the first term in the above equation [(A.e -Kr)/ r] becomes zero irrespective of the value of A. So for ψ to be zero the constant B must be zero. Hence ψ = (A.e-Kr)/ r. ……………………………..( 3 ) To determine the constant A we consider a hypothetical solution which is so dilute that interionic field may be neglected. Hence ionic atmosphere around a central is almost absent. Assuming the central ion to be a point charge the potential at a distance r from the central ion is due to the ion itself. This potential is given by, ψ = Ziε/ Dr, where Ziε is the charge of the central ion. …………………( 4 ) For this hypothetical solution concentration tends to zero. Hence ni

0 → 0. Therefore from the expression of K we may conclude that K → 0 under the above condition. Hence from equation ( 3 ) we get, ψ = A/ r …………………………………………( 5 )Comparing equations ( 4 ) and ( 5 ) we get, A/ r = Ziε/ Dr 0r A = Ziε/ DSubstituting the value of A in equation ( 3 ) we get, ψ = (Ziε.e-Kr) / Dr

Now if the ionic cloud does not exist, then the potential at a distance r is due to the central ion itself and is given by, ψ ion = Ziε/ Dr. If the charge on the cloud appears then the potential at the distance r is not only due to the central ion but is given by the law of superposition of potentials. Hence ψ = ψion + ψcloud.Therefore ψcloud = ψ - ψion

=

In sufficiently dilute solution K is very small. Therefore expanding the exponential term and neglecting the higher powers of Kr we get, ψcloud

=

This is the contribution due to the ionic atmosphere only. Now ψcloud is independent of r. Now the contribution of the cloud to the potential at the site of the point charge central ion can be given by the above equation. But if the entire charge of the ion atmosphere (which is -Ziε ) were placed at a distance of K-1 from the central ion then the potential produced at the central ion would be -Z iε/ DK-1. Hence the effect of the ionic cloud is equivalent to that of a single charge of -Z iε placed at a distance of K-1 from the central ion. So the quantity K -1

is termed as thickness of the ion atmosphere. So the thickness of the ion atmosphere or Debye length is given by r D = 1/ K, where K2 = 4πε2/ DkT. Σ ni

0Zi2. Here ni is the number of i-th kind of ions per unit volume in the solution bulk. If m i be the concentration of the i-th kind of

ions in the solution in molality then mi = 1000.ni/ d.N0, where d is the density of the solvent and N0 is the Avogadro’s number. Therefore n i = mi.N0.d / 1000.Therefore Σ ni

0Zi2 = N0.d/1000. Σ miZi

2 = 2.N0.d/1000. Σ miZi2/ 2 = 2.N0.d.μ/ 1000, where μ ( = ½.Σmi.Zi

2 ) is the ionic strength of solution.

Therefore, .

So thickness of ion atmosphere rD = 1/K = . So the thickness of ion atmosphere depends on dielectric constant D,

temperature T and ionic strength μ of the solution. The unit of the thickness of ion atmosphere can be derived from the above expression.

rD = 1/K = . Therefore its unit is [erg/ degree x degree x 1/(esu)2 x mole x litre/ mole]1/2 = [erg/(esu)2 x litre]1/2 = cm,

since erg = gm. cm2. sec-2 and esu = gm1/2.cm3/2.sec-1.Expression of free energy of electrical interaction and the expression of ionic activity coefficient

To determine ion-ion interaction we consider a hypothetical initial state in which ionic interactions are absent and a final state where ions are in solution with the interactions. If interactions are assumed to be electrostatic in origin then process of going from initial to final state involves electrostatic work of charging all ions from state zero to full charge. This work is due to change of free energy of system but the change is due to all possible ion ionic interaction of system. We want to calculate change of free energy due to ion-ion interaction arising from ionic species ‘ i ‘ only. This partial free energy change is by definition the chemical potential change ∆μ i arising from the interaction of ionic species ‘ i ‘ only with other ions of atmosphere. If ω represents electrostatic work of changing a single reference ion, then ∆μi = N ω

Now for a single central ion of positive charge without atmosphere, the potential ψ at a distance r from it is the amount of work to bring unit positive charge from ∞ to that point. Now work = force Х displacement against force. Force is due to electric field X r which varies with distance. If electric field at a distance r from central ion force is X r and dr is displacement of unit positive charge towards central ion against

this force then potential ψ ion is given by, ψ ion =

The electric force which operates on unit positive charge in medium of dielectric constant D is F = q 1q2/ Dr2, where q1 and q2 are magnitude of charges and r is distance between them. Here q1 = ZiЄ and q2 = 1. Hence Xr = q/Dr2

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Therefore, ψ ion =

If r is radius of ion then, ψ ion represent electrical potential on the surface of ion. To calculate work of charging reference ion we assume the ion to be in discharged state and then small amount of charge ‘dq’ is repeatedly brought to the ion from ∞ until it gets its full charge q. so work of charging ion when there is no atmosphere is

ω = .

If the reference ion is surrounded by ion atmosphere then the interaction between this ion and those in atmosphere gives the electrostatic potential ψcloud at the surface of reference ion which by analogy can be written as ω = Ziεψcloud/ 2.Therefore Δμi = N.ω = N.Ziε/ 2 x ψcloud = N x Zi ε/ 2 x –Zi ε/DK-1 = -N (Zi ε)2/ 2D x K. Now an ideal solution is one where there are no interactions between solute particles. A solution with discharged ionic species may be considered to be an ideal solution whose chemical potential is μideal = μi

0 + RT ln xi, where xi is the mole fraction of the i-th kind of ion. Therefore Δμ i = μreal – μideal = RT ln ai/ xi = RT ln fi, where fi is the activity coefficient. For real solution fi ≠ 1 and is a measure of chemical potential change arising from ionic interactions. Now Δμi = -N (Ziε)2/2D x K.

Therefore .

Therefore

.Changing number concentration to molar concentration, we get ci = 1000ni

0/ N we get,

Or , where A is a

constant and μ [ = Σ( ½ ciZi2 )] is the ionic strength of the solution. The value of A for solvent water at room temperature is almost 0.509

and it depend s on D and T. A ∞ 1/(DT)3/2.Formulation of relation between mean activity coefficient (f± ) of an electrolyte and ionic strength μ

Now individual ionic activity coefficient is an undetermined quantity and mean activity coefficient (f± ) is related to individual ionic activity coefficient as f±

ν = f+ν+.f-

ν-, where f+ and f- are the activity coefficients of cations and anions respectively and ν+ and ν- are numbers of cation and anion formed from a molecule of electrolyte, that is ν = ν+ + ν-.Therefore f±

ν = f+ν+.f-

ν- ⇒ log f±ν = log (f+

ν+.f-ν- )

Or ν log f± = ν+ log f+ + ν- log f- = ν+ ( -A Z+2√ μ) + ν-( -A Z-

2√ μ) = -A √ μ ( ν+Z+2 + ν-Z-

2) = -A √μ ( ν+Z+Z+ + ν-Z-Z-)Now since the solution is electrically neutral, so ν+Z+ = ν-Z-

Therefore ν log f± = -A √μ ( ν+Z+Z- + ν-Z+Z-) = -A √μ Z+ Z- ( ν+ + ν-)Or log f± = -A √μ Z+ Z- ( ν+ + ν- )/ ν = -A √μ Z+ Z- ( ν+ + ν- )/ ( ν+ + ν- ) = -A √μ Z+ Z-.

This is the Debye-Huckel limiting law. It is so called because it is valid only for very dilute solutions or its application is limited only to very dilute solutions.Limitations and demerits of Debye-Huckel limiting law

This law fails to agree with the experimental value at the higher concentration of the electrolyte solution. A plot of log f± versus √μ should be a straight line passing through the origin with a negative slope. The experimental plots match with the theoretical ones at the dilute range of concentration, but when the solution becomes concentrated the curves no longer remain straight. Moreover the slopes should depend only on the type of electrolyte that is the valence type of the electrolyte and not on the particular electrolyte. Though this is true in the dilute ranges but as concentration increases the curves of the particular electrolytes separate out. This is due to two faulty assumptions which are not valid in concentrated solutions. They are

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1. The ions are regarded as point charge. This assumption is not valid for concentrated solutions. When the size of the ions

are included in the derivation we have the potential due to ion-atmosphere at the central ion is ψcloud =

2. During the derivation of the law we have considered the dielectric constant (D) of the solvent as constant throughout the solution. But in concentrated solution D varies from point to point in the solution. The water molecules are dipoles. These solvent molecules will compete with the ions in forming the ion-atmosphere in solution. Moreover the use of Poisson’s equation is based on the screening out of charges into a continuously varying charge density. At high concentration the distance between ions being low they see each other as discrete charges and not as smothered out charges.

Considering these two facts Onsager modified Debye-Huckel equation as log f± = -A √μ Z+ Z- + cμ, where cμ is a constant. This equation is called Debye-Huckel-Onsager equation for concentrated solution. When the value of μ is small, that is μ < 1, the first term is predominant and when μ > 1 then the second term of above equation is predominant.Applications of Debye-Huckel limiting law

The most useful application of this law is that the mean activity co-efficient (f ±)of an electrolyte can be calculated from the ionic strength of he solution.Using this application various thermodynamic properties of electrolyte solution can be determined. One such property is activity solubility product sparingly soluble salt.Relation of activity product (Ka) and solubility product(Ks) of a sparingly soluble salt

Let us take sparingly soluble salt BxAy which dissociates as, BxAy ⇄ xBz+ + yAz-

The activity product Ka =a+x . a-

y

But a+ = c+f+ and a- = c-f- Therefore Ka = (c+f+)x . (c-f-)y =c+

x .c-y . f±

ν, where x + y = ν …………………………………..( 1 )Let c+

x .c-y =Ks = solubility product of the salt

Then c+ = xS and c- = ySTherefore Ks = (c+)x + (c-)y = (xs)x. (ys)y = xx. yy. Sx+y

or Ks = xx. yy. Sν ………………………………………………( 2 )Taking logarithm of both sides of equation ( 1 ), we get

log Ka = log Ks + ν log f± ………………………………………………( 3 )Using Debye-Huckel limiting law, log f± = -A √μ Z+ Z-, we have

log Ka = log Ks - ν A √μ Z+ Z- or log Ks = log Ka + ν A √μ Z+ Z- ………………………………………………( 4 )or log ( xx. yy. Sν ) = log Ka + ν A √μ Z+ Z- ………………………………………………( 5 )Equations ( 4 ) and ( 5 ) give the relation between solubility S, solubility product Ks and activity solubility product Ka.

A plot of log Ks versus √μ for any electrolyte should be a straight line with slope equal to ν A.Z + Z-. The extrapolated intercept is equal to log Ka, where from the activity product can be calculated.

Let us take the example of AgCl whose solubility is S in molar units. Hence [ Ag + ] = [ Cl- ] = S2. So from equation ( 4 ) we have for AgCl, log S2 = log Ka + 2 A √μ Z+ Z- or log S = ½ log Ka + A √μ Z+ Z- = ½ log Ka + 0.51√μ. From here we find that the solubility of a sparingly soluble salt increases with increase of ionic strength of the solution.Determination of f± of an electrolyte from solubility measurement

We know log Ka = log Ks + ν log f±. Therefore log f± = log Ka/ ν - log Ks/ ν. Now if the solubility is known, then Ks can be calculated. If Ka

of the electrolyte is known then f± can be calculated. Let us take the example of AgCl. If the solubility is S then for AgCl Ks is S2. ν for AgCl is 2. Therefore for AgCl the relation is log f± = log Ka/ 2 - log S2/ 2 or log f± = ½ log Ka – 2 log S/ 2 or log f± = ½ log Ka – log S.

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