Electrolytic Conduction

7/27/2019 Electrolytic Conduction

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14 ELECTROLYTIC CONDUCTION14.1 Introduction

Objectives14.2 Electrolysis14.3 Comparison of Galvanic and Electrolytic Cells14.4 Faraday's Laws of Electrolysis and Electrolytic Conduction14.5 Measurement of Conductance of Aqueous Solutions of Electrolytes14.6 Results of Conductance Measurements and their Interpretation14.7 Kohlrausch's Law and its Usefulness14.8 Conductometric Titrations14.9 Acids and Bases14.10 Acid-Base Equilibria14.11 Common Ion Effect14.12 Acidity and Alkalinity of Aqueous Solutions14.13 Buffer Solutions and Hydrolysis14.14 Hydrolytic Equilibria14.15 Precipitation Equilibria14.16 Applications of the Solubility Product

14.16.1 Solubilityand IGp14.16.2 The Effectof Added Electrolytes14.16.3 Applications in Qualitative Analysis

14.17 Summary14.18 Glossary14.19 Answers to SAQs

INTRODUCTIONtrical energy like the primary and secondary cellstransported through matter by the

n qf electric charge from one point to another in the form of an electricif there are charge carriers in the matter. These charge

ons, positive ions and /or negative ions. Substances likeh offer high resistance to the flow of electric charge aref electricity are called conductors. The conduction in metals is referredronic condition since the electrons of the metal are able to

the negative charges through the metal. A inetal can belattice of positive ions around which are smeared an equaler of mobile electrons. On applying an electrical potential, the electrons aree direction while the positive ions remain stationary. Theincrease of temperature as af an increase in the thermal vibrations of the lattice. In the case ofn, germanium etc., the electronic conduction increases with

f temperature. The electrons, rather tightly bound to local centres at ,ture is increased. Both theon characteristic, i.e., there is no chemical change

ing the charge transport.


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Equiiibria - d e m W ObjectivesAfter studying this unit, you should be able to :

* use Faraday's laws of electrolysis to calculate the amounts of materialdeposited or dissolved,

* predict the electrode reactions in electrolytic cells,* differentiate between galvanic and electrolytic cells.* explain the dependence of conductance of strong and weak electrolytes onconcentration,* appreciate theusefulness and applications of conductometric titrations,* define acids and bases,.* calculate the pH of solutions of acids or bases,* apply the Henderson equation to calculate the pH of buffer solutions,* explain how salts can be acidic or basic using hydrolytic equilibria, and* appreciate the usefulness of the solubility product principle in qualitativeanalysis.

14.2 ELECTROLYSISElectrolytes like sodium chloride, either in the molten state or in an aqueous solutionconduct electricity but are decomposed by the passage of an electric current. Acurrent of electricity can be passed into an aqueous solution or a fused salt via twoelectrodes dipping into the solution. The electrode connected to the negative pole ofthe battery is a source of electrons and is the cathode of the electrolytic cell, sincereduction can occur at this eleceode. The other electrode connected to the positivepole of the battery is deficient in electrons and so can accept electrons. It is the anodeof the electrolytic cell. Under the influence of an applied EMF the positive ions(cations) of the electrolyte move towards the cathode and get reduced. The negativeions(anions) move towards the anode and get oxidised. This movement of ions of anelectrolyte under the influence of an applied EMF is called electrolytic conductionand its chemical decomposition caused by the passage of electricity is calledelectrolysis. As a specific example. we consider the electrolysis of molten sodiumchloride using inert electrodes like Pt or graphite that do not react with either theelectrolyte or the products of electrolysis. Of the two ions present, Na+ and C1-, onlythe Na+ can be reduced. At the cathode of the electrolysis cell, the reduction reaction,Na+( 1)+e- =Na ( 1), occurs and the sodium metal collects around the electrode.The C1- anion on reaching the anode, loses the electron (oxidation) to the anode anda neutral chlorine atom is formed. Two such atoms combine to give a chlorinemolecule, which bubbles off as a gas. The net chemical change that takes place in theelectrolytic cell is called the cell reaction. It is obtained by adding together the anodeand cathode reactions in such a way that the same number of electrons are gained andlost. Thus, in the electrolysis of molten NaCl,cathode reaction : 2Na'+2e-=2Naanode reaction : 2Clm- 2 e=C12Cell reaction : 2 Na++ 2 Cl--> 2 Na +CI,

(electrolysis)In the simple case of molten sodium chloride, the current-carrying ions are found toget discharged at the electrodes. When many electro-active species (capable ofgetting reduced or oxidised at the electrode) are present and also theelectrodematerial is susceptible to be attacked, complications arise. Reactions of several typesare possible at each electrode. The reactions that may occur at the anode includei) dissolution of the anode by oxidation

e.g., Ag--> Ag +e- ;Cu -> cu2++ 2 e-, etc. .


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ii) discharge of anions, ande.g., 2 Cl- --+C12+ 2 e}

iii) oxidation of water in aqueous solutions

Among the conceivable oxidation reactions at the anode, the reaction with the morenegative (less positive) reduction potential is likely to take place. If the applied EMFis gradually increased, the oxidation process occur in the order of increasingreduction potentials (most negative first and least negative last). After all, this is aprediction based on thermodynamics and so one should actually know how rapidthese electrode processes are in order to predict the electrode reaction.At the cathode, reduction of a cation to give metal takes place in a few cases.

Ag++e ' j AgIn the case of cations of highly reactive metals like Na, K.etc., it is water that ispreferentially reduced to give hydrogen

Examples :(1) Electrolysis of an aqueous solution of Na + Cl- - using Pt electrodes

Cathode Reactions RemarksNa++e =Na Not possible because of negative EO value

possible. Products at the cathode are H2and OH-. Na+ ions move towards the.cathode to preserve elecmcal neutrality.Concentration of Wis too small for thisreaction to occur

Anode Reaction Remarks2C1--2e=C12(g) Actually observed2 4 0 - > 0 2 + 4 W + 4 e - Oxygen evolution is not observedTherefore, the net Cell Reaction :2 C1- +2H, 0 -> H2( g )+C12( g )+ 2 O K

'electrolysis

(2) Electrolysis of an aqueous solution of sodium sulphate using Ptelectrodes to give hydrogen at the cathode and oxygen at the anode :Cathode: 2H 20 +2 e m= H2 (g ) 20H-Anode : 2H2O=O2(g) +4H++4e-

Therefore, the cell reaction is

If theH+ ndO H ons are allowed to mix, the cell reaction will be

What is the function of sodium sulphate in this case? The anode reaction results in theformation of H+n the anode. The sulphate anions migrate to the anode to maintainelectrical neutrality. Simil'arly positive ions ( Na') move into the cathode region to

EleclrdytlcConduction


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compensate for the presence of O H ons. It will be noticed that the current-carryingions need not necessarily be discharged at the electrode. The ability to carry currentdepends on the concentration and speeds of the ionic species. But the ability to getreduced or oxidised depends on the electrode potential.

14.3 COMPAR ISON OF GALVANIC ANDELECTROLYTIC CELLSIn a galvanic cell, electrons released during oxidation at the anode push the electronsof Lhe anode material into the external circuit. The anode, being a source of negativecharge, is the negative electrode. At the cathode, electrons are used up for thereduction reaction and as a result of the electron deficiency it is called the positiveelectrode.In electrochemistry. electrodes are designated as anode or cathode depending onwhether oxidation or reduction takes place at an electrode. In an electrolytic cell, theanode is the positive electrode since it is connected to the positive pole of thegalvanic cell. Anions attracted by the anode, get oxidised and the electrons releasedto the anode are drawn by the cathode of the galvanic e l l to be used in the reductionreaction in the galvanic cell. The cathode of an electrolytic cell is a negative electrodesince it is comected to the negative pole (anode) of the galvanic el l . The electronsreleased during the oxidation at the anode of the galvanic cell are available for thereduction of catiws or other species at the cathode of an electrolytic cell. Thus thegalvanic cell may b& linked to an electron-circulating pump, taking electrons from theanode to the cathode of an electrolysis cell (Figure 14.1).

I Galvanic I

I

Electrolysircell

Figure 14.1 :Galvanic and E l e d y s i s cells

14.4 FARADAY'S LAWS OF ELECTRO LYSIS ANDELECTROLYTIC CONDUCTION- - -- - p- - --Michael Faraday was the first to describe quantitatively the ralationship between thequantity of electricity passed (current in ampere x time in seconds or amp. sec orcoulombs and the extent of chemical decomposition brought about by electrolysis.His investigations are sumrnarised in the form of two laws of electrolysis.First law : The amount of any substance deposited at or dissolved from anelectrode as a result of the passage of an electric current is proportional

to the quantity of electricity passed.If W grams of a substance are deposited or dissolved by passing a current of i-amperes for t seconds, according to the first law


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In Eqn. 14.1, e is called the electrochemical equivalent of the substance. From Eqn.14.1, it is seen that when i t or the quantity of electricity passed is one coulomb,W = e. Thus, the weight of the substance deposited or dissolved by the passage of onecoulomb of electricity is the electrochemical equivalent, It is no longer used.Second Law :The masses of different species deposited at or dissolved fromelectrodes by the same quantity of electricity are proportional to theirchemical equivalent weights.If W, and W2 are the amounts of two products of electrolysis obtained by passing qCoulombs of electricity, according to the second law

In Eq. 14.2, E, and E2 are the chemical equivalent weights of substances 1 and 2respectively. Equations 14.1 and 14.2 can be combined to give Eqn. 14.3

In Eqn. 14.3, F is a constant of proportionality. It will be seen that W = E when it =F. In other words F, called the Faraday, is the quantity of electricity required todeposit or dissolve 1 gram equivalant of a substance. The value of F has been foundto be 9.6487 x 184C/mol of electrons or 96500 C/mol of electrons. The equivalentweight is equal to the molecular or atomic mass divided by the number or electronstransferred per molecule, atom or ion.Significance of Faraday :It is found experimentally that 1 coulomb of electricity deposits 0.001 119 gm ofsilver. The amount of electricity needed to deposit 108gm or 1gm equivalent (or 1gm atomic wei&%%ofilver= log = 96500 coutombs. This is the value of 10.001 119Farad ay.Thus Faraday is the quantity of electricity needed to deposit or liberate onegram-equivalent weight or one gram atomic wt. or 6.023 x lo2, number of atoms(Avogado number). To deposit one gram-equivalent weight, 1 Faraday is needed.Example 14.1

A current of 0.5 amp., when passed through a solution of a chloride of formulaACl,, for 3520 seconds, gave at the cathode 1.28 g of a metal. (a) If the atomicmass of A is 197.0 g mol-', find n (b) If the cell is co ~e c t e dn series withanother electrolysis cell in which water is electrolysed. what will he volume atS.T.P. of oxygen collected at the anode?

Solution:(a) No. of faradays passed =4 = OS5 3520= 0.01824F 96500

AFor depositing- rams of A ,96500 C required ,n197or one Faraday =- ramn

Therefore , he formula of the chloride is ACl,


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E q u l l l b ~ ~ & ~ ~ t r ~ h e m l ~ r gb) The evolution of oxygen at the anode of another cell can be represented as

1 mol of 02= 22.4 lit (at S.T.P.) = 4F22 4 litvol. of 0, =-- x 0.01824 F = 0.102 litres.4F

Faraday's laws are exact and are applicable to aqueous solutions as well as fusedsalts. If there are several processes taking place at an electrode, Faraday's laws areapplicable to all the electrode processes taken together. With reference to eachelectrode reaction, it is usual to associate the term current efffcfency, to indicate howmuch of the total current is made use of in a process. It is defined as the ratio of theactual amount of the material deposited at or dissolved from an electrode and thatexpected theoretically on the basis of Faraday's laws.Example 14.2

By the passage of 0.06 faradays of electricity 1.057 g of nickel (atomic mass.58.71) was deposited at the cathode. If hydrogen is also evolvedsimultaneously what is the volume at S.T.P. of hydrogen evolved at thecathode?What are the current efficiencies for Ni deposition and hydrogen evolution?

Equivalent weight of Ni =-5871 ISolution : 58.7 gramsof Ni. The current.06F electricity should deposit 0.06 x-efficiency for Ni deposition= 1-057 x 100= 60.0 8 .The rest of the0.06 x 58.71current utilised for hydrogen evolution is equal to 0.06 - 0.036 =0.024F. Thevolume of hydrogen evolved at S.T.P. can be calculated from the fact that2Fof electricity is required to evolve 22.4 of H2 at S.T.P.

22.4 x 0.024 F=0.2688 1or 268.8 ml atherefore , olume of hydrogen liberated=-FS.T.P.SAQ 1What volume of oxygen would be liberated from an aqueous solution of NaOHby a curreht of 2 amperes flowing for 1; hrs at 27C and 1am pressure.

SAQ 21. An aqueous solution of sodium sulphate containing sulphuric acid iselecuolysed between Pt electrodes. What will be the products obtained

at each electrode?2. calculate the amount of products liberated or dissolvedat each electrodeif a current of 2.68 amp is passed for one hour through an aqueous

solution of CuSO, (atomic weight of Cu = 63.54)


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queous solutims of electrolytes, like metallic conductors, obey Ohm's law (Eqn.14.4) which enables us to calculate the resistance (R) from the EMF (E) and currentE = i R (14.4)

of direct current (d.c.) through the solution causes electrolysis andin the concentration. The resistances of aqueousons of electrolytes are always measured using an alternating current (a.c). Theance of any elecuical conductor is proportional to its length (I), and inverselyf cross section (A),IR a - orA

. 14.5, the constant of proportionality, p, is called the specific resistance orivity of the conductor. It is the resistance offered to the passage of electricrent by a conductor of length 1.0 cm having a cross sectional area of 1 cm2. From4.5, it is seen that the units of p are ohm cm (C G S) or ohm m (S I). Theistivity, being the resistance between opposite faces of a one cm3 of material, isf different conductors.electrolytic conduction, it is the conductance (G) which is reciprocally related tonce, that is more useful. For the comparision of conductances, the spec#ic(K ) which is the reciprocal of resistivity is useful. Theductivity is given by Eqn. 14.6

- .

of K are ohm- ' c m ' r oh m 'm- '.The reciprocal ohm is some timess the siemens (S). This quantity G is thus the conductancen the opposite faces of a one cm3 of the material. Conductivities of a fewances are given in Table 14.1

Table 14.1 :Conductivities of a few substances at 298KSubstance G/ohm cm- '

Silvercopper

Aluminium1.0 M KC1 (aq)

0.01 M KC1 (aq)0.1 M acetic acid (aq)

water

6.38 x 1 65.80 x 1 63.80 x 1 61 . 1 0 ~o-'1.41 x l u 35 . 2 0~ r 44.0 x 10-*

14.1, the conductivities are listed in the decreasing order. It will be seen thatcase of aqueous solutions of electrolytes, conductivity depends onentration. For the sake of comparison of conductivities we need to use the


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Equilibria& Electrochemistry solutions of electrolytes of the same concentration. One molar solutions of sodiumchloride and magnesium sulphate may be used. In ~ g ~ +G',each ion transportstwo units of charge whereas in Na+ Cl- each ion transports only one unit of charge.However, if the concentration is expressed in equivalents per litre (normality), oneequivalent of any electrolyte gives rise to positive ions with a total charge of IF andnegative ions with a total charge of 1F. The conductance of a volume of a solutioncontaining 1 equivalent of an electrolyte confined between two large parallelelectrodes 1 cm apart is called the equivalent conductance ALet 1 gram equivalent of an electrolyte be dissolved in V cm3 of the solution. Sincethe electrodes are 1 cm apart, the solution will cover the electrode surfaces upto anarea of V cm2. From Eqn. 14.6.

Under these conditions the conductance G is the equivalent conductance, A.Substituting for A = V c d = 1 cm, we have Eqn. 14.8. Here KV is numericallyequal to the equivalent conductance.A = K V (14.8)

In Eqn. 14.8, V is the volume of the solution in cm3 containing 1 gram equivalent ofthe electrolyte.If the concentration of the aqueous solution is c equiv. lit-': 1. gramequivalent of the electrolyte will be present in l/ c lit or 1000/c cm3. Thus V is 1000cm3/c. Substituting in Eqn 14.8, we get a more useful form of this equation.

The units of (A ) will thus be 1000cm3 it-' o h m " cm-' or ohm- ' m2 equiv- '.Theequiv. lit- 'last term is usually dropped since it is implied in expressing A as the equivalentconductance. The molar conductivity A,,, is defined by Eqn. 14.10Where M is the concentration in mol dm3 nd cm s the concentration in mol m-3. I twill be seen that for electrolytes like magnesium sulphate.

It is convenient to use equivalent conductance, since it is a measure of thecurrent-carrying ability of all the ions from 1 gram equivalent of any electrolyte.

14.5 MEASUREMENT OF CONDUCTANCE OFAOUEOUS SOLUTIONS OF ELECTRO LYTESThe specific and equivalent conductances of aqueous solutions of electrolytes arecalculated by determining the resistance of the electrolytic conductor contained in aconductance cell of known dimensions I and A. As indicated earlier, an alternatingcurrent (a.c) is to be used to prevent elec ~oly si s. sually an a.c of 1000 to 2000cycles per second is used so that the small amount of electro lysis taking place in onehalf of the cycle is reversed in the opposite half of the cycle. Platinum electrodes, onwhich platinum black is deposited. are used so that the electrode reactions occurrapidly and there is no accumulation of the products of electrolysis.A Wheatstone bridge arrangerncnt is normally used to measure the resistance of thesolution (Figurel4.2). The point of contact P is moved along the wire AB till nocurrent is detected in D. The resistance is so adjusted that the balance point is almostmidway between A and B. Since a.c. of this frequency is audible, an earphonedetector is used. A minimum nohe in the earphone indicates the balance point underthese conditions. In place of an earphone, analog or digital detectors can also be used.


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Resistance

Figure 142 :Muwemen td&tma ofm aqueousdut imofanele~lolylesing 8Whewume bridge unmganau.

Resistanceof the cell - AP- -Known resistance( R ) PB (14.1 1)The resistance of the cell can be calculated using Eqn. 14.1 1. Once the resistance ofthe cell is known, the value of K and A are calculated by using Equations 14.6 and14.9 respectively. Calculation of K requires a knowledge of the dimensions1and A ofthe cell. The direct measurement of 1 and A is rather difficult, once the cell isassembled. For a given cell, both 1and A are constants and the ratio 1A, called thecell constant (k), canbe calculated by measuring the resistance of a solution of 0.1NKCI, whose specific conductance is accuratelly known. The conductance cell is nowstandardised and can be used to calculate conductivity of any solution containing anelectrolyte. 14.6 can be' written in terms of the cell constant as Eqn. 14.12

Since ordinary distilled water has an appreciable conductance, specially preparedwater called conductivity water, with a resistance of nearly lo6ohms, is used toprepare all the solutions. Since the resistance varies with temperature, the cell mustbe kept in a constant temperature bath. Conductance or specific conductance is anadditive property. In an aqueous solution containing several electrolytes, the totalconductance is given by Eqn. 14.13.

n Eqn. 14.13,G is the conductance of water used in preparing the solutions andthe summation applies to all the electrolytes present. In the case of strongly

ng solutions,G is so small that it can be neglected in compais~ n iththe conductance of the electrolyte solution. For weakly conducting solutions,

imum sensitivity in measuring high conductances (low resistances) a cellith a high value of k is used. A cell with small electrodes separated by a largence is used (Figure 14.3a). Conversely, for the measurement of lownces (high resistances),1A should be as small as possible by having largerodes separated by a short distance (Figure 14.3b).14.3

A conductance cell, when filled with 0.1 molar solution of KCI,has aresistance of 33.2 ohms. The same cell after washing, rinsing etc., has aresistance of 300 ohms when filled with 0.1 molar solution of acetic acid(HOAc), and 55.2 ohms when filled with 0.03 molar solution of sodiumsulphate. The specific conductance of 0.1M KC1 at this temperature is 0.01 164ohm- ' m '.The conductivity of water used to prepare these blutions is


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Equilibria& Eleetrochemlstry 1

Fig. 14.3 : e ofconductance cellforthemcasumnent of(a) low resistana (high conductance)(b)high resistance (low onductance)7 . 6 ~r40hm-' cm-' t this temperature. Calculate (a) the cell constant, and(b) equivalent and molar conductances of the solutions of (i) acetic acid and(ii) sodium sulphate.

Solution:

Therefore,

(b) In order to calculate A, the specific conductances are required.K(i) A ,, 1000-C

This includes h e contribution from h e water used for preparingthe solution.Therefore,K due to HOAc alone= 1.287 x 10-~-7.6x 1 r 4 5 . 27 ~o4= 5.27 x 10-4 ohm" cm-'Therefore,

In case of acetic acid, equivalent weight =molecular weight, andhiOAcAHOAc 5.27 ohm 'cm2Therefore,

-6 .992~1r3-7 .6 x 1r4=6.233x1 r 3 hm 1cm-IK ~ s ,O, -1 moleofNa$304= 2molofNa+= 1 molof s @ - = ~ FEquivalent weight of Na2S04=molecular weight / 2 or 0.03 Msolution of Na2S04= 0.06 N Na2S04Therefore ,

=2 X 103.8 or 207.6 ohm- l c d


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6 RESULTS OF CONDUCTANCE MEASUREMENTS Electrolytic CondudlonAND THEIR INTERPRETATIONuivalent conductances of all electrolytes are found to vary withiluted. the specific conductance increases andin the case of many electrolytes. This limiting value is calledlent conductance at zero concentration (A0) or at infinite dilution (A?,does not inc rease further with dilution . The results are6 ,here c is the concentration in equivalents per.4). From Figure 14.4, it is seen that in the case of solutions of stronglinear and the values of A do noton. Tfiese are called stro ng electroly tes and in these

is possib le to obtain theva lues of A0 by extrapolation .In he case of weakike acetic acid, amm onium hydroxide etc., A increases rapidly a s thed, especially at low concentrations. In these cases, value of A0 cannot

14.4 :The variation of equivalent conductance of a few eleciroly les with ancm rat ion (Figure not to rcale)on in K and A with concentration using hisc dissociation theory. According to this theory, an electrolyte dissociatesree of dissociation (a ) increases with .The conductance of a solution of a given electrolyte depends on the num bertheir velocities. Arrhenius assumed that dilution does not affect thenumber of ions available from an electrolyte. At infinitea tends to be unity and at all other concentrations, a < 1.When the solutionis diluted, though the extent or degree of dissociation (a

he num ber of ions per cm3 of the solution decreases. The specificuctance between the opposite faces of a one cm3 of material1 cm3 of m aterial contains a lesser number of ions than before, K decreases

) with dilution was attiibuted to an increase in theions available from one equivalent of an electrolyte at high d ilutions. Ata= 1, the value of (A ) does not vary and becomes equal to A',a = 1 corresponds to A', Arrhenius proposedA- hould be a measure of a, at any concentration. Thus,A0

alues of a experimentally determined from colligative properties of aqueousof electrolytes agreed quite satisfactorily w ith those obtained from theents. Ostwald verified Arrhenius theory by applying the lawass action to the equilibrium existing between an electrolyte and its constituentConsider a solution of normal acetic acid (a weak electrolyte) in equilib riumH+ nd OAc' ions. If a is the degree of dissociation, the equilibrium


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&-kanis(~ concentrations of H+ .OAc- , nd HOAc are ac,ca and c ( 1- a ) respectively. IfVis the volume of the solution containing one equivalent of an electrolyte. V = l/c.The expression for the equilibrium constant (K), given by Eqn. 14.15

is known as the Ostwald's dilution law. The value of K was found to bereasonably constant only for weak electrolytes and not for strong electrolytes.Strong electrolytes have been shown to consist of ions even in the solid state andin aqueous solutions the electrolytes are completely ionised(a 1) at allconcentrations.According to the Debye-Htikel theory, which is applicable to both strong and weakelectrolytes, as a result of the interionic attraction the velocitiek of the ions areinfluenced. Each ion is surrounded by an ion cloud of oppositely charged ions (ionatmosphere). This ion cloud retards the movement of the central ion. In the case ofstrong. completely dissociated electrolytes, the variation in conductivity withconcentration can be traced to this retardation which increases with concentration. Atvery high concentrations, ions associate into ion pairs, and so, the effective number ofconducting species will be less than in the absence of ion association. As the solutionis diluted, the ions are father apart, and the density of the ion cloud decreases. Theinterionic forces of attraction will be less, the speed of the ion increases, and so theequivalent conductance increases. At infinite dilution, the ions are so far apart thatthere are no interionic forces of attraction. Thus, the equivalent conductance reaches amaximum value. For weak, incompletely dissociated electrolytes, changes inconductivity occur as a result of an increase in the degree of dissociation withincreasing dilution.

14.7 KOHLRAUSH'S LAW AND ITS USEFULNESSAt infinite dilution, in the absence of any interionic forces of attraction, each ion ofan electrolyte can be expected to behave independently. Kohlrausch found that thedifference. A, ( KX - A, (NaX) was the same irrespective of the nature of the anion,X. Similarly, a constant value for the difference A0(KX)A0 (KY), where X and Yare anions, was observed. These results could be explained if it is assumed that A,, foreach salt is considered to be the sum of the equivalent conductances of the constituentions (ion conductances) at infinite dilution.He put forward his law of independentconductance of ions as, "at infinite dilution, each ion makes a definite contribution tothe equivalent conductance of an electrolyte, whatever be the nature of the other ionof the electrolyte". This can be expressed as Eqn. 14.16

A0= A!+ k: (14.16)In this equation, k: and 5: are the ion conductances at infinite dilution of the cationand the anion respectively. TI4e ion conductances of a few ions are given in Table 14.2From this table, it will be seen that for polyvalent ions, the values are only for oneequivalent and not for a mole. The conductances reflect the amount of current thatthese ions can carry. Thus, one should compare the conductances)10 ( Na+) . 10 (f~2+) .10 (f Fe3+ ) etc.. since in each case. we are referring tothe amount of species associated with 1 mole of electrons.

1Also.A' (5 gSO, )= 53.1 + 80= 133.1 ohm- ' m2 equiv7'but A ( MgSO, )= 106.2+ 160= 266.2 ohm- ' m2 mol- '


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ElectrdytkCondudlonTable 14.2 : onic equivalent conductances at infinite dilution of a few ions at 298 K*Cations 1 Anions 71

If

H+ 349.8 OW 197.6Rb+ 77.8 ( 1/4 ) Fe( CN ) 111.0Cs+ 77.3 ( 1/3) ~ e ( N ) ~ - 99.1

NH: 73.4 ( 1/3 ) PO:- 92.8K+ 73.5 ( 1/2) 0:- 83.0

NO;

FormateAcetate

ure 14.3) that values of A, for strong electrolytes can bextrapolation. In order to obtain A0 values in the case of weak

acetic acid. we can use A0 values of suitable strong electrolytes as, _

= k O (H ' )+ ~ ' ( o A c - )gly soluble salts. the saturated solution is so dilute that furthertion does not change the value of A. Hence it is assumed that A= A

1 '1AO(f B ~ s ~ ~ ) = A ~ ( ~ B ~ c ~ ~ ) + A ~ ( ~ N ~ s ~ ~ ) - A ~ ( N ~ c ~ )

14.41 1At 298K. the A0 values of NH4CI;-BaC12 and-Ba ( OH ), in o h m ' m2 are2 2149.9, 139.9 and 262.2 respectively.

Calculate A0 of NH40H. I


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Equlllbrla& Electrochemistry Solution:

= 149.9+262.2 - 139.9= 272.2 ohm1cm2

Example 14.54 1The h0values of -BaCl -N$S04 and NaCl in o h m c d re 139.9, 130.12 2 ' 2 1and 126.5 respectively. Calculate the he0 of y BaSO,.

Solution:

= 143.5 ohm1c dExample 14.6

When water with a specific conductance of 1.1 x o h m cm- is saturatedwith BaSO,( s ), the saturated solution is found to have a specific conductanceof 4.6 x ohm cm- I. Calculate the solubility in n~ole/lit f BaSO, inwater at 298 K.

Solution:

ld x 3.5 x 1 r 6=2.44 Jconcn. of BaSO, in equiv/lit= 143.51 mol of BaSO, = 2 equiv.Therefore, concn. of BaSO, in mol/lit = 1.22 x lo-'

SAQ 31. A decinormal(0.1 normal) solution of sodium acetate, when placedbetween two electrodes each of area 1 5cm2, and placed at a distance of0.72 cm, has a resistance of 524 ohm. Calculate the specific andequivalent conductances of this solution.

2. The equivalent conductances (A0) on ohm' cm2 at infinite dilution ofNaCl, HCOONa and HCl are 126.4, 109.6 and 426.1 respectively.Calculate (Ao) (HCOOH). The equivalent conductance of 0.01 NHCOOH is 50.7. Calculate the degree of dissociation and[H'].


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CONDUCTOM ETRIC TITRATIONS Electrolytic Conductione conductan ce of a solution is the summ ation of the contributions from all the ions

It depend s on the number of ions per unit volume (concentration), and alsoring the course of a titration, the concen uations of thectance also changes. In this method, the variation of thethe solution during the cou rse of a titration is followed. It i s notine the actual specific conductance of the solution, and anytional to it can be used. For a given cell with a known cell constant, Ky proportional t o resistance R, and so one can use the reciprocal ofe to follow the titration. The conductance changes w ithto add a concentrated solution of the titrant from a burette.easurement of resistance, one can use this methodand even dilute solutions. Theance to locate h e end points in titrations is based on the c hanges in theuctance versus volume of titrant (titration curves). This

be due to the presence of ions of high conductance in ther the equivale nce point. Thus ,Titrations

tome tric titration curve fo r the neutralisation of a strong acid (H' C1-) byong base (Na' OH-) i s given in Figure 14.5(a). The reaction ca n be represented as:

H++C1- + Na++ OH- --+ Na++Cl- +H, 0+--+ +--+Species present Titrantin the conducta ncecell initially

e highly conducting hydrogen ions initially present in the solution are replaced bya much smaller ionic conductance. Hence, the conductance ofion de creases as the base is added (the descending position of the curve in. Beyond the equivalance point, further addition of a strong base inuoduc esions in solution. Since these are not used u p in the reaction,in direct proportion to the excess of base added ( the

after the end point are obtained. Tw o straight lines are drawns. T he point of intersection of the two straight lines gives these the given volume of the acid. If a weak(N H4 0H ) is used as a titrant in determ iningation of the strong acid, the descending portion of the curve is similar,ents the decrease in conduc tance as a result of replacing the hydrogenions. After the equivalence point, the conductance w ill remainof N H4 0H compared to NH,CItitrations of weak acid s or bases are rather difficult. A weak acidc) is present m ostly as unionised molecules and the solution has

ctance due to the small am ounts of H+ and OAc- ions. Th e neutralisationHOAc+-+ Na'0I-I-+ t-+ + N a + + O A C + K , OSpecie s present in theconduc tance cell initially Titrant

on proceeds, the comm on ion formed, i.e.. the acetate ion, supresseson of the ace tic acid (Sec 14.11). Hence, initial addition of sm all amountsith further addition of sodium


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Volume o f T i t r a n tF ~ U R4.5 :CarductmetricT~uaticmurves

(a) HCI against NaOH(b) HCI against N H 3

hydroxide. the cohductance of Na+ and OAc- ions exceeds that of acetic acid andconductance of the solution increases. After the equivalence point, thcre is a sharpincreases in conductance due to the Na+ and OH- which are no more required forneutralising HOAc [Figure 14.6(a)]. If NH,OH is used as a titrant, the initial portionof the curve is similar to that in Fig 14.6(a). After the equivalence point, there ispractically no change in conductance because of the very small conductance ofNH,OH [Fig.14.6(b)l.

Volune o f T i t r a n tFigure 14.6:Conductanetric itration curves

(a) Acetic acid against NaOII(b) Acetic add against NH3

Precipitation tiuations are based on the fact that the contribution of the ions of thesparingly soluble salt to the conductance of the solution is rather small. The reactionbetween K+ Cl- and Ag+ NO; can bc represented as

te -9+ -++ -> AgCl(s )+K++NOjSpecies present in theconductance cell initially Tiuant

The net result being the replacement of C1- by NO; of almost equal conductance,there is practically no change in conductance as the titration is performed. However,


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dlence point the addition of Ag' NO; increases the number of ions innd so the conductance increases [ Figure 14.7 ]ElectrolyticCoadudlon

Volume o f TitrantFigun 14.7 : ondudmetricurntion KCl againstA D @ (tilrant)

ACIDS AND BASES- - - - - - -fined by Arrhenius as substances capable of givingions and hydroxide ions, respectively, in solution. This definitionly to aqueous solutions. In order to account for the acid-base behaviour inke liquid ammonia, liquid sulphur dioxide, glacial acetic acid etc., Lowryronsted, independently put forward the proton transfer theory. According to this, an acid is defined as a substance with a tendency to loose a proton and a base

or an acid to donate its proton, some proton acceptor (a base) must beilary a potential base will accept a proton only in the presence of a protonan acid loses a proton, the remaining portion of the molecule will have ay to accept a proton and hence it is a base. In general one can write

and the base which differ by a proton are said to be conjugate to one. When a potential acid like hydrogen chloride gas is dissolved in water, theum ion, H30+.the hydrogen ion in aqueous solution is not the bare proton, H+,ut af the general formula H+( H20 ) The simplest formula, H30+ , sined when n is unity. This reaction can be represented as

HCl + H,O -$ H30t +C1- (1)Acid-1 Base-2 Acid-2 Base-1

be seen that C1- is the conjugate base of HCl and H30+ s the conjugate acid of20. The dissolution of a weak acid like acetic acid (HOAc) in water tove a i acid solution can be represented asHOAc+ $0 3 H30++OAc- (2)

ity of an aqueons solution of an acid is attributed to the presence of H,O+.r the concentration of this hydronium ion, stronger is the acid. Since HCl is aacid, the position of equilibrium for reaction (1)isi t is almost to the left of the equation.tively, one can consider that the tendency of C1- to accept a proton is far lesst hat of OAc'. Thus, acetate ion is a stronger conjugate base than CT. nl, the conjugate base of a strong acid is weak and vice versa. If a solute like

be represented as 1 %


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Equilibria& Electrochemistry

Solvents Eke water which can either take up protons or donate protons are said to beamphiprotic. Solvents like liquid HF, liquid HOAc which can give protons are calledprotogenic solvents. Solvents like liquid NH,,pyridine etc., which act as bases aresaid to be protophilic. Solvents like benzene, toluene etc. which cannot donate oraccept protons, are called aprotic solvents.Strong electrolytes are completely ionised in aqueons solutions and theconcentrations of the ionic species in solution can be calculated from theconcentration of electrolyte. For example, in a 0.05 M solution of barium chloride,[Ba2'] = 0.05 M and [Cl- ] =0.1 M. In the case of weak electrolytes, whatever isdissolved may not be in the form of ions. Only a small fraction of the solute givesriseto ionic species in solution. In such cases, the concentrations of the ionic speciescanbe calculated from the application of law of mass action to such equilibria. Equilibriainvolving electrolytes like acids, bases and sparingly soluble salts in aqueoussolution, are included in ionic equilibria.

14.10 ACID-BASE EOUILIBRIAThe ionisation equilibria in the case of weak acids (HA) and bases (B) canberepresented as

HA+H20 2 H30' +A- (3)and

B+H20 2 BH'+OH (4)respectively. The equilibrium constants for these equilibria are given by equations14.17 and 14.18 respectively.

The weight of one litre of water at 298 K s 997 g and this corresponds to 55.4 mol ofwater in a litre. In its reaction with solutes, the change in the concentration of water israther small and hence one can consider [ H20 ] to be a constant in these equations.Consequently, these can be written as Equations 14.19 and 14.20 respectively.

In Eqn 14.19, H30' is represented as I? for the sake of simplicity. The equilibriumconstant K, is called the dissociation or the ionisation constant of the acid and Kb isthat for the base. The molar concentrations of the various species can be calculatedfrom the degree of dissociation and concentration (Sec. 14.6).If a is the degree ofdissociation of the acid HA of concentration c,, the concentrations of the variousspecies in equilibrium are

[H']=[A-]=c,a and [H A] =c ,( l- a) .Similarly, if P is the degree of dissociation of the base B of concentration cb, we canwrite

It is assumed that [ H+ or [ BH ] are available only from the dissolved solutes.Ifthere are other sources for these ions, the concentrations from these sources must alsobe included (Sec.14.11).


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n that K, and K,, in terms of the degree of dissociation, are given by14.20 and 14.21 respectively.

a and p can be measured conductometrically~Sec. 4.6), it is possible toK, and K,. If the ionisation constant is known, it is possible to calculate hef dissociation and also the various concentrations (Example 14.8).14.7

A 0.1 M solution of NH3 has an equivalent conductance of 3.60 o h m ' m2 at298 K. At the same temperature. the ion conductances in ohm'. ' d fNeand O H ons are 73.4 and 197.6 respectively. Calculate (1 ) the degree ofdissociation, (2) the concentration of O H ons, and (3 ) the dissociationconstant,Kb.

= 1.788x lo-'( l - $ ) ~ l , ~ , ~ P ~ ci . e . ) 1 . 7 6 8 ~ 1 0 - '

within 1.2%error of that in (3).14.8

The ionisation constant of formic acid is 2.0 x at 298 K. Calculatethe degree of dissociation and [ H+]f a 0.01M solution of fonnic acid at298 K.From Eqn. 14.20,

fication we get a quadratic equation in a, .e., a2+0.02 a - 0.02=0

ElechPlgtkCondudlam

a measure of the strength of an acid or a base. Thus,(K, = 1.4 x 1 0 - 3, is stronger than formic acid

2 .0 x 1 r 4 ) ,which is itself stronger than benzoic acid (K,= 6.3 x lo-').y, among the amines, the base-strength decreases as

diethyamine > ethylarnine > ammonia > aniline= 1 . 0 0 ~ ( K b = 4. 6 6x (K b= 1.77 x lo-') ( K b=3 . 8 x 1F14 '


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Equllibtia & k%chdm i s t r y SAQ 41. Fill in the blanks :

Conjugate Acid Conjugate Base

~ ~ o ; y ~HCOOH

2. A decimolar solution of acetic acid is 1.33% ionised. Calculate thedegree of dissociation (a).H'] and the ionisation constant(K,).At thesame temperature, enough solid sodium acetate is dissolved in one litreof 0.1 M acetic acids so as to make the concentration of the salt equal toI O M. What will be the [H+]of the resulting solution ? (Hinta


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~olution:(a) H A 2 H+ + A-

~ ( 1 - a ) a c a cSince,a c = [ H+ 1= 1.732 x 1 r 3 ,c being 0.01 M ,a =0.1732Therefore.

(b) The addition of the comm on ion A- is expected to decrease the degree ofdissociation. If a' is the degree of dissociation, the dissociationequilibrium ca n be represented as, H A 2 H++A-

c ( 1 -a') ( a ' c ) ( a ' c + x )Where x is the concentration of the added common ion

( a' ) c


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14.12 ACIDITY AND ALKALINITY OF AQUOIUSSOLUTIONSSince the solven t water is capable of losing as w ell as gaining a porton, there must bea proton transfer equilibrium as show n below:

The very sma ll electrical conductivity of pure water indicates the presence of ions inlow concentrations. This indicates that the dissociation of water to give H+ nd O Hions must be taking place to a small extent. Applying the law of mass action to theself-ionisation of water, we have

Since most of the water remains undissociated, [ H 20 1may be considered tobeconstant, and so, Eqn.(14.22) simplifies to Eqn.14.23

The equilibrium constant K, being the product of ion ic concetrations, is called theionic product of water and has a value of 1.0 x 1W l4 at 298 K. In any aqueoussolution, equation 14.23 should hold good. Since pure w ater is neither acidic norbasic but neutral, [ H+ ]= [ O H = 1.0 x 1 r 7 .This condition of,equality is also acondition for neutrality in aqueous solutions. If [H+ is greater than 1.0 x M,the solution is acidic and if it is less than 1 r 7 , he solution is alkaline.A conven ient scale of acidity and alkalinity. which avoids the use of negativeexponents in expressing [ H+ 1 was introduced by Sorenson . He defined the pH of asolution as the negative logrithrn of the hydrogen ion concentration (Eqn. 14.24)

In this scale, a pH equal to 7 represents the condition of neu trality. For an acidicsolution of [ H+ ]= 1.0 x 1 r 4 H = 4 and for an alkaline solution in which[ O H ] = l . O x 1 0 - ~ o r H+]=-- Kw - 1.0 x l r 9 , he pH will be 9. This methodI O Hhas the advantage that all states of acid ity and alkalinity upto one mo lar solution canbe expressed by a series of numbers from 0 to 14. Thus it will be seen that, for acidicsolutions, pH 7 .This kind of p(X)notation can be applied to express the concentrations of o ther ions, ionisationconstants and other equilibrium constants. Thus from Eqn 14.23

- l o g [ H + ] - l o g [ O H ] = 1 4 o rpH+pOH=14 (14.25)

Equa tion 14.25 enables one calculate pOH from pH and vice versa. The samereciprocal relationship that one finds between [ H+ ] and pH ho lds in all these cases.Thus, formic acid ( K, = 2.0 x ; pK, = 3.70) is a stronger acid than benzoicacid ( K, = 6.3 x ; pK, = 4.21 ) .Similarly, among bases, ammonia( K, = 1.77 x 1 r 5 ; K, =4.75 ) is stronger than an line ( K, = 3.8 x 10- ;pK, =9.42 ). The pH of solution can be measu red using a pH meter (Sec.13.10).Though the pH scale covers a range of 6 to 14, in very strongly acidic solu tions thepH may be negative and in very strong ly alkaline solutions it may be greater than 14.Example 14.10

Calculate the pH of the following solutions (1 ) 1.0 M HClO, (2) a solution inwhich [ H + ]=3.564 x 1 r 4 nd (3) a solution in which [O K = 2.0 x 1W3M.


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Electr~blleondudion

(3) pOH=- og[ O H = ( -5.301 )=2.691pH= 14- 2.69= 11.31

14.11Calculate the [ H+ in the following(1) a solution of pH=4.75, and (2) pH = 12.31

-.-(1) log[H+]=-pH=-4.75= 5.25Therefore,

(2) log [ H+ I = - 12.31= i3.69Therefore,[ H+]= 4 . 8 9 ~ M ; OH-] = 1 . 0 ~o-" = O . O ~ M4.89 x 10- l3

Q 51. a) An analysis of water in a lake indicated that [W] = 3.2 x 1 r 5mol/lit.

What is the pH of the water in the lake ? Is it acidic or basic ?b) Calculate the[H'] that corresponds to each of the following values of

pH.i) lemon juice pH = 2.25ii ) pH of milk of magnesia = 10.50

The ionic product of water is 1.0x 1W14

2. a) Given that K, of acetic acid is 1.8x 1w5,calculate the pH of 0.15 Msolution of sodium acetate.

b) If the dissociation constant of ammonium hydroxide is 1.8x 1v5.Whatwill be the pH of 0.15M solution of NH4N03?


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Equllibrln & ElortroehmWr~ 14.13 BUFFER SOLUT IONSA careful control of pH is important in life processes, in industry and in certainanalysis. For this purpose, suitable buffer solutions are used. When reasonableamounts of an acid or an alkali are added to an aqueous solution of ammoniumacetate or a mixture of ammonium chloride and ammonium hydroxide, there ispractically no change in the pH of the solution. The resistance that a solution exhibitsto changes in pH upon the addition of acid or alkali is called buffer action andsolutions exhibiting this property are called buffer solutions. Buffer soultions usuallyconsist of (1) a weak acid (acetic acid) and its conjugate base (acetate ion), or (ii) aweak base (ammonia) and its conjugate acid (ammonium ion), or (iii) a weak base(ammonia) and a weak acid (acetic acid). Buffer solutions are used mainly forresisting changes in pH or for providing solutions of known pH.The buffer action will be explained using two examples. Consider the buffer solutionconsisting of HOAc and NaOAc. If H+ ions are added, these are removed by OAc- toform the unionised acid.

If O H ons are added, these are removed by the reaction with the unionised HOAc togive OAc- ions.

HOAC+OH- -+ H20+OAC-The pH of the solution thus remains practically constant. In the case of a buffersolution consisting of NH3( NH40H) and NH4C1 he following reactions

account for the buffer action. In a similar fashion, if a buffer solution of ammoniumacetate is used, the buffer action can be explained as followsH++OAC- (from buffer) -+ HOAc

O H + NH: (from buffer) -+ NH3+H20The pH of a buffer solution can be calculated by considering the effect of adding acommon ion (Sec.14.11) to the weak acid or the weak base equilibria. Consider abuffer solution consisting of a weak acid (HA) and its salt like NaA. This is a specialcase of the ionisation of the acid in the presence of a common anion, A-. Thedissociation constant K, of the acid is given by Eqn 14.19

If the initial concentration of the acid is c, and that of the salt NaA is c, and a is thedegree of dissociation in the presence of NaA, the equilibrium concentrations of thevarious species can be expressed as follows.[H+]=c,a , [A-]=(c,a+c ,)and [HA]=c , ( l - a )Substituting these values in Eqn 14.19, we have

As explained in Example 14.9, a < < 1 and so c, ( 1- a )G c,. Considering the [ A- 1,it will be realised that it is only the added salt NaA, that is the major source of A- insolution, since [ A- ] available from HA will be very small as a result of the commonion effect. So c, a


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[ Salt ]K. = [*I [Acid ] ElectrolyUcConductionn can be rearranged to give Eq. 14.27

[ ~ t . 1=K,W[ Salt ]ms and changing the sign throughout, we have

Acid- l o g [ ~ ] = - l o g ~ . - l o g ~ o r[ Salt ]pH = pK, + log [ Salt][ Acid ]

14.28, known as the Henderson equation, is useful in calculating the pH ofilar equation for the system B - BIF) ( NH,,-N@ ) is

the expression for the ionisation constant of the base .(Eq. 4.20)

A - A- system, invoking the approximations, [ BH' ] 5 [ Salt 1 andB I e [ Base 1. we haveK ~ = [ O H I S l l r l o r =K,W[ Base ] [ Salt 1

hms and changing the sign throughout equation 14.29, we get

quantities of H+or O H ons are. The buffer capacity of a mixture may be regarded as the amount of acid ort a buffer solution can use up without suffering a change in pH. It is foundthe ratio [Salt]or [Salt] is[acid] [base]

to unity and that the concentrations used are about 1.OM. For sat isfactory bufferf the pH is to remain constant withinf 1 , hese ratios must lie between 10.1: Thus, the useful ranges of buffer solu tions lie between pH =pK, f or=p KbI . For example, p K, of acetic acid is 4.75 and the useful range of theto 5.75. Similarly, for NH, - NH4 C1 buffer

10.25.14.12

A solution is prepared by dissolving 0.20 mole of sodium formate and 0.25mole of formic acid in water and making the total volume upto 250 ml exactly.The ionisation constant of form ic acid is 2.0 x Calcu late (i) the pH of thissolution, (ii) the change in pH by adding 1.0 ml of concentrated HCl(normality = 10.0) to 250 ml of this solution, and (iii) the change in the pH byadding 1.0 ml of 10N NaOH solution to another 250 ml portion of the buffersolution.

i, pH =pK, + og [ Salt ][ Acid 1The pK, of formic acid = -log ( 2 x 1V )= 3.70

Om2 'O0O = 0.8 m ol l litSalt1 = 250


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Equilibria& J3kIroehemLFtry Similarly, [Acid]= Oe2' lMX)1.0molPit250Therefore, pH = 3.70+ log@ =3.601o

(ii) no. of moles of HCl added = 1.0ml x li t x 10mol lit- '1000ml= 0.01mol

[HCl] added =0.04momit ( Since, .O1mol is contained in 250ml)The added HCI reacts with the conjugate base HCOO- as

HCOO-+H+ ---4 H CO O HHence the concetration of HCOO H resulting from this reaction is equalto the concentration of HCl added.Thus [HCOOH] = 1.0+0.04+ 1.04mol/lit.This reaction between H COO- and H+ also decreases the concentrationof the salt. sodium formate by 0.04 momit.[Salt] =0.8-0.04 =0.76 molllit

076,Therefore, pH after the addition of HCl =3.70+ og- - 3.561.04Therefore, Change in pH is only by 3.56-3.60 or -0.04units. The pHdecreases by 0.04 units.

(iii) no.of moles of NaOH = 0.01[ O H ] added =0.04momitThe added O H eacts with HCOOH as shown below.O K +HCOOH- O +HCOO-. Thus the [ H C O W increases by0.04momitTherefore, [ HCOO- ] = [Salt]=0.8+0.04 = 0.84 molPitSimultaneously [HCOO H] decreases by 0.04 molpit[HCOOH]=1.0- 0.04 = 0.96 momit

0.84-pH after adding the alkali=3.70+ og- 3.640.96Therefore. change in pH = 3.64 - 3.60 =0.04The pH of the solution increases by 0.04 units

Now, let us define hydrolysis and consider different types of salts formed bycombinations of acids and bases. The deta ils of hydrolytic equilibria will beconsidered in the next section.Hydrolysis of SaltsDefinition :The hydrolysis of a salt is defined as the interaction of the ions of a salt with eitherthe H +o r the O W ions of water to form a basic or an acidic solution.

If the ions of the salt interact with O W of w ater, the salt is said to be hydrolysed. Theresulting solution is acidic. If the ions of the salt, dissolved in water react with H+ion of water, the salt is said to be hydrolysed. Th e resulting solution is basic. Here,we sun m ar ise the formulae used and derive them in the next section.


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of saltsition of hyd rolys is, thc salts arc classified as follows into four

:(i ) Salts of strong ac ids and strong bases

eg. NaClNaCl is the sa lt of a strong acid HCI and the strong base NaOH.

NaCl 2 Na+ + Cl-) Sa lts of weak acids and s tron g bases :

eg. CH3COONaetate is form ed by the combinalion of CH3COOH, a weak acid and NaOH,

base.H 20 -+ H++OH-

e acetate ions intera ct with the H+ ions to Torn1 uniot~iscd r wcakly ionised aceticThese salts are hyd rolysed and ihc rcsul ling sol111on is basic.ii) Salts of stron g acids and weak bases :

eg. NH4Clm chloride is formed by thc inrcrnc~ ion f th c strong acid HCI and weak, NH,OH.

H 20 --+ H++ OH-

NH; +OH- ---+ NH,OHto form ammonium hydroxide, thehydrolysed andathe resulting solution is acidic.

Salts of weak acids an d weak basiseg. CH,COONH,

m acetate is formed by rhc in~ crac lion l' amm onium hydroxide, a weak

NH: +OH- ---+ NH40Hions interac t with. H+ ions lo for111 ac i~ ic cid and hydroxyl ions andum ions interact to form a m m on ii~ n~lydroxidc. TIlc salt is hyd rolysed ande resulting solution is neutral.

HYDROLYTIC EOUILIBRIAare formed by the neutralization of a n acid by an cquivalcnt amount of a base or. However, aqueous solutions of salts likc sodium acetate, sodium cyan ide,~ h c l l is nor cqual to 7.0 in these cases.


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Electrochemistry Only in the case of salts formcd from strong acids and bases, the aqueous solutionsare neutral. An aqueous solution of sodium acetate has a pH greater than 7.0(alkaline). whereas that of ammonium chloride has a pH less than 7.0 (acidic). In thecase of sodium acetate, thc acctate ion present in solution is a strong base andabstracts protons from water as shown belowOAc-+ H20 2 HOAc+O H

The hydroxide ions formed in the reaction are responsible for the solution having apH>7.0. In the case of ammonium chloride, the N q eing the conjugate acid of aweak base, is capable of donating a proton to water to give rise to H30+ ons.

The formation of H,O+ explains to acidic nature of the solution. This interaction ofions of the electrolyte leading to Lhe abstraction of a proton from the solvent, or thedonation of a proton to Lhc solvcnt is called hydrolysis. It will be noticed that onlystrong conjugate acids and bases get hydrolysed in aqueous solution. Weak conjugatebases like Cl- ,NO; do not undcrgo hydrolysis. This interaction leads to a decrease inthe concentration of a conjugatc bases like cG- ,s2-C2@- etc., which arenormally uscd to prccipitatc cations. Hcnce ways and means of preventing hydrolysisare important.The extcnt to which thc hydrolysis rcaclion proceeds is determined by the equilibriumconstant, called the hydrolysis constant (K, of the salt undergoing hydrolysis. FromKh. t is possible to calculate the pH of such aqueous solutions. Consider the salt of awcak acid (HA) and a strong basc (MOH), for example NaOAc. The anion, A-undergocs hydrolysis as

The reaction can be considcred as hydrolysis of the salt MA or dissociation of theconjugate base, A-. For this equilibrium, the hydrolysis constant (Kh) can be writtenas Eqn. 14.29 in which [ H20 ] is considered as a constant.K, = [ HA ] [ O H ][A-IThe above'cquation can bc written as

K h - [ H A ] H I ] o K ~ - K w-- (14.30)[H +l[ A- l KaFrom the hydrolysis equilibrium, it is sccn that [ HA ] = [OH- and if Kh is rathersmall, the extcnt of hydrolysis will also be small. Thus the equilibrium concentrationof A- can bc taken to bc equal to 111c original concentration of A i.e., Equation 14.29can bc writtcn as Eqn. 14.31

Since, [ H+] [ OH- ] =K,, Eqn. 14.31 becomes

From equations 14.30 and 14.32, we get

Taking logarilhms and multiplying by -1 throughout, we get1 1 1pH=-pK,+-pK,+-1og.c2 2 2


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weak base (B) and a strong acid (HX ) like NH,Cl. ElectrolyticCondudlon

K,, an be writlen as follows.

[ B ] = [ H30+ and [ BH+ ] s c ,one can wrilc

14.35 an d 14.36 we get

r the salt (BHA) of a weak acid (H A) and a w eak basc(B ), thc hydrolysis reaction is

on equilibirum of HA

[A-] = c, Eqn (14.40) can be written as

ng fo r [HA ] in Eq. 14.39, we get Eq. 14.40

a is the degree of hydrolysis, Eq. 14.29 and 14.35 can be written interms of a as14.42.

K represents K, or K,, as the case may be. If a


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From Eq. 14.43 , t will bc seen that extent of hydrolysis increases as (i)K, increases,i.e., as the temprature increases, (ii) as K decreases, i-e.. the weaker the acid or thebase, and (iii) as c decreases i.e., as the solution gets diluted. Thus it is seen that at agiven temperature, the degree of hydrolysis (a) an be decreasedif a highconcentration of the salt is used. For a salt of the type BHA, if is the degree ofhydrolysis, we can write the equilibrium as

~ ( 1 - P I ~ ( 1 - P I c P c PFrom Eqn. 14.38

it will be seen that K, will be large since K, .Kb product of two small quantities, israther small. This means that hydrolysis will take place extensively and isindependent of concentration. At higher tempratures, the extent of hydrosis will bequite considerable.Practically, all the cations except those of alkali and alkaline earth metals undergohydrolysis in aqueous solution to give acidic solutions as shown in the case of salts of~ 1 ,

and so on. As in the case of other equilibria, it will be seen that the hydrolysisreaction can be supprcsscd by adding one of the products of the equilibrium. In thiscase, the hydrolysis reaction can be suppressed by adding acids. Thus, whilepreparing aqueous solutions of salts of polyvalent cations, it is advisable to add acidsto prevent hydrolysis.Example 14.13

Calculate (i) the hydrolysis constant(K,) (ii) the degree of hydrolysis(a)nd(iii) the pH of a 0.1M solution NH,CI. The ionisation constant of NH3 ,K,, atthis temperature is 1.80x 10-5 . What will be the degree of hydrolysis if 1.OMsolution of NH,CI is used? The ionic product of water is 1.0x 10- l4 at thistemperature.

Solution:The hydrolysis reaction isNH: +H,O 2 NH, +H30+

From Eqn. 14.35,

K, = a2 ,2 ( Because, a h ather small)( 1 - a )


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iii) From Eqn. 14.37,

PRECIPITATION EOUILIBRIAchemical analysis, precipitation is an important process. In qualitative analysis, theon and identification of the cations and anions is bascd on precipitation

antitative analysis, one should make sure whcthcr the desiredent is quantitively precipitated or not. An understanding of thc solubilitythc law of mass action, is th eforen of an electrolytc is in contact with thc solid solute, there

brium between the ions in the solution and the solid. Consider such anilibrium between MA and its ions M+ and A-.

K.

rated solution, [MA(s)] can be considerd to be a constant, since it is note as a result of the dissolution proccss. Hcncc Eqn.14-44 can be writte'n

Ksp =[M+ 1 A-] (14.45)has dissolved is assumed to bc cxisting as ions. The ionic product

by Eqn. 14.45 is called the solubility product and applies to anyd solution. However, it is of practical inlcrcst only in lhc case of sparinglyility product expression (Eqn. 14.45) being an equilibrium constant,coefficients of thc ionic spccics, is., the formula of thegly soluble electrolyte. For a 1.1 (AgC1) or 2.2 (BaSO, ) or $3 ( FePO, )

ytes of the formula MA, the expression for K,, can bc writtcn as follows. Letbe the solubility in water in mol/dm3 or molllit of thc salt, at a given temperature.

[ M n + ] = [ A n - ] = S

K,=[M "+] [A"-]=s ' (14.46).

ElecbolyticConduction


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For a 1.2 (Ag2Cr04 or 2.1 (CaF, ) electrolyte, the equilibria can be expressed asM2A 2 2 W + A%

andMA, 3 ~ ' + + 2-

respectively. The solubility product constants for h$ A and M4 re given byequations 14.47 and 14.48 respectivelyK,= [M +]~ A2- ] (14.47)

K,= [M"] [ A-1' (14.48)If S is the solubility in moVdm of the salt M2A, then [ M+]=2s and [ A'- ]=S.Substituting these values in equation 14.47 we get Eqn. 14.49 as the expression forthe solubility product of M, A

K ~ ~ = [ M ' ] ' [ A ~ - ] = ( ~ S ) ~ ( S ) = ~ S ~ (14.49)Similarly for M A,, if X is the solubility in movdm3,[ M2+ ] =X and [ A- ]= 2 X.The expression for the solubility product of MA, is givenby Eqn.14.50

[M 2+] A - ] ~ = ( x ) ( ~ x ) ~ = ~ x ~ (14.50)Example 14.14

Write down the solubility product in terms of the solubility (S movdm3) ofthe elecirolyle Sb, S3Solution:

Sb,S3(s)2 sb3++3s2 -2s 3s

3 + 2 s2-13K,,=[Sb I [ o r ( 2 ~ ) ' ( 3 S ) ~= 108s5

14.16 APPLICATION OF THE SO LUBILITY PRODUCT14.16.1 Solubility and KspThe solubility product is a measure of the solubility of the salt in water at a giventemperature. From equations 14.46, 14.49, 14.50 and Examplel4.14, it will berealised that for the sake of comparision one should consider b e IZ,values of salts ofsimilar formulae. Thus Agl (K,, = 1.0x 10- 16) is less soluble than AgCl (KT= 1.1 x 10-lo). Also, Ca F2( K,, =3.2 x 10- l 1 ) is less soluble thanMg F2( K =7.0 x 10- ). Howcvcr, instead of K,,, one can also compare the molarsolubilities (S) of salts. Thus Ag2CrQ ( K, = 1.7 x 10' l2 and S=7.5 x 10'' M ) ismore soluble than Ag CI ( KT= 1.1 x 10-''and S= 1.05 x 10"M ).14.16.2 The Effect of added electrolytes on the equilibriumbetween the sparingly soluble electrolyte and its constituent ions.The added electrolyte, irrespective of its nature, influences this equilibrium. Whenthe added eleclrolyte has an ion in common with the salt MA, we have b e commonion effect (sec 14.1 1). which is relevant to the precipitation equilibria. The constantK,, defines the maximum concentration of the ions in solution bat can remain inequilibrium wiih ihe sparingly soluble salt. Consider the effect of addingH+C1-to Ag C1 (s ) .


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AgCl ( s ) 2 Ag++Cl-s indicated earlier, the [ Cl- ] in equilibrium with AgCl(s), in the absence of any, s approximately 10-5 M. If [ Cl- ] exceeds this value, [ Ag' 1decrease. (LeChatalier principle). In other words, the solubility of the

in the presence of a common ion, or the addedCte product of the concentrations of the ions (ionic product) exceeds the solubilitybe seen'that if the

f one or both the constituents are decreased either by formation of aweakly ionised substance, th solubility product equilibrium isore of the salt dissolveshQis is the basis for the dissolution ofNH,. Here Ag' forms a complex [ Ag ( NH, )2 ]+ and if the [ NH, 1 is quite

practionally all the Ag is converted into to the complex. Hence, more of AgClbelved in NH,. This can be represented asAgCl (s ) 2 Ag++Cl-

Applications in Qualitative Analysis. those of Hg, Cu, Pb, Cd, Bi. As, Sb and Sn are less soluble. Hydrogen sulphide is used as a source of sulphide

ak electrolyte, its overall dissociationbe represented ask, ) H2S 2 2H++s2-

constantcentration of dissolved H2 S. we can write [ s2- = [ H+ I .Thus,rease [s2-] by increasing [ H+ ] . o that only the less soluble sulphides of

II get precipitated. Once these are removed, the other cations can be[ s 2 - ] available from H,S by making the medium very

kly acidic or alkaline. Thus h e group IV sulpliides are precipitated by passingS into an arnmoniacal medium containing the cations of h is group.

[ O K available from a weak electrolyte like aqueousH3 ( NH40H) by adding NH4CI, it is possible to precipitate the less soluble

oxides of Fe. Al and Cr in group 111. Under these conditions. the hydroxides of. Ni, Mn, Co and Zn are not precipitated.14.15

The specific conductance of conductivity water, at 298 K, is1.61 x 10dohm1cm-'. When saturated with pure AgCl(s). its conductivityincreases to 3.41 x l ~ ~ o h m - 'm-'.The ionic equivalent conductances atinfinite dilution of Ag+ and Cl- in ohm- ' cm2 are 61.9 and 76.3 respectively.Calculate (i) the solubility of AgCl(s) in moll lit. (ii) the solubility product ofAgCl(s) and (iii) the solubility of AgCl in 0.001 N NaCl solution.

As indicated earlier (Sec 14.7), A GA' in this case.h 0 ( ~ g ~ 1 ) = 6 1 . 9 + 7 6 . 3 =38.2ohm-' cm2

Where S is the solubility of AgCl(s) in cquiv / lit or moll lit.


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Thercforc, S= lWOx lr6=. 302~(r%ovlit138.2ii) Ksp [ Ag ] [ Cl- ]=s2 Eqn. 14.46)

=( 1.302~o-')'

iii) In the prcsencc of 0.001 NaCl, the [ Cl- ] available from NaCl is 0.001and this is >> [ Cl- ] available from AgCl, in the absence of a commonion. One would expect the [ Cl-] available from AgCl(s) tobedecreased by h e c o m n on effect-(Sec 14.4.31). Lets ' be themhbiof AgCl(s) in the prescnce of added C t . One would expect S


35/38

(ii) [~~~+][0~]Z=0.01(1.8~10~~)'=3.24~1~'~The ionic product in the case of Fe (OH ), iz.,5.832 x 10- I is>>K, 3.8 x 1r3*and so it is Fe (OH )3 that gets precipitated.In the case of Mg ( OH ),. the ionic product ( 3.24 x 10- l2 ) is less than itsK, ( 3.4 x 10-I' . and so. i t is not precipitated.

SAQ 61. The solubility of Mg F2 at room tempcraturc is 0.0012 moll litCalculate the solubility product of Mg F,. What will be the solubility ofMg F2in 0.1M MgSO, ?

2. The solubility product of lcad chloridc ( PbC12) is 1.6 x 1r5.f 500ml of 0.03M NaCl is mixed with 500 ml of 0.3 M lcad nitrate. will therebe a precipitation of PWI, ?

14.17 SUMMARYEleclrolytes, either in the molten state or in aqueous solution. conduct electricity andundergo chemical reactions at the elcctrodcs. This process is callcd electrolysis.Faraday's laws of electrolysis summarise thc quantitative rclationship between thequantity of electricity (ampere-seconds or coulombs) passcd and the extent ofchemical reaction that takes place. One Faraday, equal to 96500 C, is Lhe quantity ofelectricity required to deposit or dissolve onc cquivalcnt of a substance.If instead of a direct current. an altcmating currcnl is uscd. i t is possible to measurethe resistance (conductance) of electroryrrz conductors. In the casc of aqueoussolutions of electrolytes, the conduclivily dcpcnds on conccnlration. For acomparison of conductivities. equivalcnt conductanccs arc q u itc useful. The variationof specific and equivalent conductances with diluiio~l f aqucous solution ofelectrolytes could be accounted for by lllc Arrhenius thcory of electrolyticdissociation. The Debye-Huckel theory of intcriot~ic rtraclion provides a morereasonable explanation of this bchaviour. Kohlrausch's law of independention-conductances at infinite dilution cnablcs oric lo calculate thc cquivalentconductance at iflinite dilution of wcak clcctrolytcs and sparingly solubleelectrolytes. The change in the conduclancc wilh concentralion forms the basis forcarrying out acid-base and precipitation titrations conductometrically.An importantapplication of conductomclry is thc calculalion of the degreeofdissociation of weak acids and bascs. By applying Ihc law of mass action to equilibriainvolving weak acids and weak bascs, onc can calculalc rhcir ionisation constants.Similarly,one can calculate the solubilities and solubiliry products of sparingly


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Equlllbrlr & Ekdroehaaldq soluble salts. The term "common-ion effect" refers to the influence of electrolytes,having the same cation or anion as those derived from the weak acids, weak bases orsparingly soluble electrolytes, on equilibria involving such species.'Systems involving weak acids (bases) in the presence of common ions show bufferaction. They are useful in resisting the changes idpH of such a buffer system. The pHcanbe calculated using the Henderson equation. The conjugate b$es of weak acidsare strong enough to abstract a proton from water which results intan alkalinesolution. Similarly, the conjugate acid of a weak base is a strong acid, capable ofdonating protons to water. The interaction of the ions of a salt with either the Ht orthe OW ions of water is called hydrolysis. The solubility product principle, commonion effect and hydrolysis find extensive applicatio& in chemical analysis.

14.18 GLOSSARYElectrolysis : Chemical decomposition caused by the passage ofelcctricity.Faraday : The charge contained in one mole of electrons.Electrolytic conduction : The movement of ions of an electrolyte under the

influence of an applied EMF.Conductivity water : Watcr obtained by repeated distillations, whereconductivity is only due to the conductivities of H+ ndOH- ions.

a2cOswalds dilution law : K =- here K =equilibrium constant,a =degree(1 -a)'of dissociation and c, the concentration of ttleelectrolyte.Arrhenius theory : The theory wherein the molecules of acids, bases andsalts arc dissociated into constituent ions.Kohlrausch's law : A, =A", + Loa. where. the terms refer to the equivalent

conductances of the solution. cations and anionsrespectively at infinite dilution.Conductometric ~itratibnswherein the conductance of the electrolytetitrations : solution is measured as a function of the volume of thetitrant added.Acids & bases : An acid is a substance capable of accepting electrons,while a base is a molecule capable of donating electrons.

: - og [H']. The Henderson equation pH =pKa+'log is useful for calculating the pH.[Acid]

Buffer solutions : Solutions of a weak acid and its salt or a weak base andits salt whish resist changes in pH when acids or basesare added to them.Hydrolysis : Thc interaction of the ions formed from a salt with or

OH- ons of water to yield basic or acidic solutions.Solubility product : In MA$ Mt + A-, KSp= [m]A714.19 ANSWERS TO SAQsSAQ 1

40H- --+ 0 2+2H2011- hrs. = 5400 cc.2


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SAQ 21)2)

SAQ 31)2)

SAQ 41)

2)SAQ 5

1.

2.SAQ 6

1)-2)

4 x 96500 Coulombs liberate= 32 g of0,32x 5400x 22 x 5400 Coulombs liberate = 96500, g

nRTVolume of O,, V =-H, at cathode and0, t anode3.177 Cu is deposited at cathode and 3.177 g of Cu is dissolved fromanode

Conjugate base :s@- andHCOO-Conjugate base :NH; andHz PO,a=0.0133; [ ~ + ] = 1 . 3 3 x 1 0 - ~ ; ~ = 1 . 7 7 x 1 O - ~ ;~ + ] = 1 . ? 7 x l O - ~

a) pH =4.49. Water in the lakc is acidicb)i)5.6x1w3 ii)3.2~10"'a) 8.97 b) 5.03

Yes ,Since ion product [ pb2+ 1 [C1- ] = ( 0.15 ) ( 0.015 ) ' s' greater than Kip.


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FURTHER READINGS1. James E Brady and John R H olum, Fundamentals of Chem istry, JohnWiley and Sons. (1981 and later editions)2. Leonard W F ine and Herbert Beall, Chemistry for Engineers andScientists, Saunders College Publishing (1990)3. J.C K uriacose and J. Rajaram. Chemistry in Engineering andTechnology Volumes 1 and 2,Tata M cGraw-Hill Publishing Com pany

Ltd., (New Dclhi (1988)4. Atkins, P.W., Physical Chem istry, ELBS and the Oxford UniversityPress. 5th Edition (1993)5. Caslcllan Gilbert, W.; Physical Chemistry , Addison-W esley, Reading(Mass). 5th Edition.6. Cottre ll A. H., Introduction to Metallurgy, Edward Arnold, London.7. Daincls Farrington and R obert A. A lberty. Physical Chem istry, JohnWilcy, Ncw York.8. Ferguson F.D. and Jones T.K. Phase Rule. Butterworths, London.9. Findlay Alcxandc r; Phase Rule and its Applications, DoverPublications, New York.10. Glastonc Sam ual and David Lewis; Elements of Physical Chemistry;Macmillan, London.11. Kuriacose J.C. and Rajararn J.; C hemistry in Engineering andTcchnology, Vol. 1, General and Physical Chem istry; Tata McGraw HillPublishing Co. Ltd., New Delhi.12. Mahan Bruce. H., Univcrsity Chemistry. Narosa Publishing House, NewDclhi (1985).13. Maron Samucl H. and Jerome B. Lando; Fun damentals of Physical

Clicmistry, M acn~ ilan,Ncw York.14. Moore Wallcr, J., Physical C hemistry, Orient Longmans, London.

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Electrolytic Conduction

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