Experiment 4: Conductivity of electrolyte solutions (Dated ...

1 Experiment 4: Conductivity of electrolyte solutions (Dated : October 29, 2009)I. INTRODUCTIONPure water does not conduct electricity, but any solvated ionic species would contributeto conduction of ionically conducting solution is called an electrolyte solution and the compound, which produces the ions as itdissolves, is called an electrolyte . A strong electrolyte is a compound that will completely dissociate into ions in , a weak electrolyte dissolves only partially. The Conductivity of an electrolyte solution depends onconcentration of the ionic species and behaves differently for strong and weak electrolytes. In this work the electricconductivity of water containing various electrolytes will be studied. The data will be extrapolated to infinitelydilute solutions and the acidity constant for a given weak electrolyte will also be determined.

2 Additional theoreticalbackground for electrolyte solutions can be found from Refs. [1 3].II. THEORYM ovement of ions in water can be studied by installing a pair of electrodes into the liquid and by introducing apotential difference between the electrodes. Like metallic conducting materials, electrolyte solutions follow Ohm slaw:R=UI(1)whereRis the resistance ( , ohms ),Uis the potential difference (V, Volts ), andIis the current (A, Amperes ).ConductanceG(S, Siemens or 1) is then defined as reciprocal of the resistance:G=1R(2)Conductance of a given liquid sample decreases when the distance between the electrodes increasesand increaseswhen the effective area of the electrodes increases. This is shown in the following relation:G= Al(3)where is the Conductivity (S m 1),Ais the cross-sectional area of the electrodes (m2; the effective area availablefor conducting electrons through the liquid), andlis the distance between the electrodes (m).

3 Molar Conductivity m(S m2mol 1) is defined as: m= c(4)wherecis the molar concentration of the added electrolyte . A typical value for molar Conductivity is 10 mS m2mol molar Conductivity of an electrolyte would be independent of concentration if were proportional to theconcentration of the electrolyte . In practice, however, the molar Conductivity is found to vary with the concentration(see Fig. 1). One reason for this variation is that the number of ions in thesolution might not be proportional tothe concentration of the electrolyte . For example, the concentration of ions in a solution of a weak acid depends onthe concentration of the acid in a complicated way, and doubling the concentration ofthe acid does not double thenumber of ions.

4 Another issue is that ions interact with each other and tend to slow down each other leading reducedconductivity. In this limit, the molar Conductivity depends on square root of electrolyte molar the 19th century Friedrich Kohlrausch discovered the following empirical relation between the molar concentrationof a strong electrolyte and the molar Conductivity (Kohlrausch s law) at low concentrations: m= 0m K c(5)Typeset by REVTEXFIG. 1: Variation of molar Conductivity as a function of molar concentration. a)Strong electrolute and b) weak a non-negative constant depending on the electrolyte and 0mis the limiting molar Conductivity ( themolar Conductivity in the limit of zero concentration of the electrolyte ). Furthermore, Kohlrausch was able to showthat 0mcan be expressed as a sum of contributions from its individual ions.

5 If the limiting molar Conductivity forthe cations is +and for the anions , the law of the independent migration of ions states: 0m=v+ ++v (6)wherev+is the number of cations per formula unit,vis the corresponding number of anions, and +and arethe limiting molar conductivities for cations and anions, respectively. For example, for HClv+= 1 andv = 1 butfor MgCl2we havev+= 1 andv= 2. Because weak electrolytes are not fully ionized in solution, the number ofions is not proportional to the concentration of the electrolyte but depends on the degree of dissociation ( ). Theeffectivemolar Conductivity can then be approximated in terms of and the hypothetical molar Conductivity of thefully ionized case ( 0m): m= 0m(7)When a weak acid dissociates in water solution, we have:HA(aq) + H2O(l) H3O+(aq) + A (aq)(8)The effective concentrations in solution are then given by (subscript 0 refers to the initial concentration of the acid):[H3O+]= [HA]0,[A ]= [HA]0, [HA] = (1 ) [HA]0(9)The acidity constant (Ka) can now be written in terms of the ion activities (a):Ka=a(H3O+)a(A )a(HA)a(H2O) = 1 (solvent)=a(H3O+)a(A )a(HA)(10)In order to proceed, we write activity in an alternative form for each species(herei= H3O+, A , HA, H2O).

6 A(i) = ibib (11)2where iis the activity coefficient (dimensionless) for speciesi,biis the molality fori(mol kg 1) andb is the idealsolution molality (constant, 1 mol kg 1). Inserting Eq. (11) into Eq. (10) we get:Ka= H3O+ A HA H3 ObA bHAb =K Kb(12)where notation ofK andKbare used for convenience. Do not confuseKbwith the acidity constant! In dilute solutionsthe mean activity coefficient ( ave= H3O+ A ; geometric mean value) can be calculated using the Debye-Hckellimiting law:log ( ave) = |zH3O+zA |A I(13)wherez s are the ionic charges,Ais a constant (typically for an aqueous solution at 25oC) and I is thedimensionless ionic strength of the solution given by:I=12 Nions i=1z2ibib (14)whereNionsis the number of difference ionic species in the solution.

7 Note that log here denotes a logarithm withbase 10 (ln would denote the natural base logarithm). Since the activity coefficient for the neutral species ( HA) isequal to one and 2ave= H3O+ A , the Eq. (12) gives:Ka= 2ave Kb(15)or by using logarithms:log (Ka) = 2 log ( ave) + log (Kb)(16)In dilute solutions molalities are directly related to concentrations by:[i] =ci=bi (17)whereciis the molar concentration of speciesi(mol L 1; usually denoted by species in brackets) and is the densityof the solution (kg L 1). Inserting Eq. (17) into Eq. (16) we have:log (Ka) = 2 log ( ave) + log([H3O+] [A ] b [HA])(18)Note that the numerical value of b is approximately one, so it is only required for getting correct units. In many casesequilibrium constants are written in terms of concentrations, which tends to lead confusion in units.

8 For example, forEq. (8) we would normally write in terms of concentration:Ka [H3O+] [A ][HA]=:Kc(19)while this gives the correct magnitude, it gives wrong units as the equilibrium constant is dimensionless (see thedefinition in Eq. (10)). VariableKcwas introduced to refer to the equilibrium constant obtained directly fromconcentrations. Thus we can simplify Eq. (18) as:log (Ka) = 2 log ( ave) + log(Kc b )(20)3 From Eqs. (9) and (19)Kccan be obtained as:Kc= 2[HA]01 (21)Next we apply the Debye-H uckel limiting law (Eqs. (13) and (14)) and Eq. (21) to Eq. (20):log (Ka) = 2|zH3O+zA |A I+ log(Kc b )= 2A [HA]0 b + log( 2[HA]0 b (1 ))(22)ORpKa= log (Ka) = 2A [HA]0 b log( 2[HA]0 b (1 ))In practice can be obtained from Eq. (7).

9 However, before Eq. (7) can be applied we must know the limiting molarconductivities for solution consisting of H3O+and A .In order to calculate the limiting Conductivity mentioned above, we must use Eq. (6). Note that we cannot use Eq.(5) since it only applies to strong electrolytes. For the present case Eq. (6) reads (HA = CH3 COOH): 0m(CH3 COOH) = +(H3O+) + (CH3 COOH )(23)Such data is not directly available, but can be calculated, for example, by measuring thelimiting molar conductivitiesof the following strong electrolyte (water) solutions :HCl H3O++ Cl ( 01,m= +(H3O+) + (Cl ))(24)NaCl Na++ Cl ( 02,m= +(Na+) + (Cl ))CH3 COONa CH3 COO + Na+( 03,m= +(Na+) + (CH3 COO ))If the three limiting conductivities can be measured then the limiting Conductivity for Eq.

10 (23) is given by: 0m(CH3 COOH) = 03,m 02,m+ 01,m(25)Thus Eq. (7) now reads: = m(CH3 COOH) 0m(CH3 COOH)= to be measured m(CH3 COOH) 03,m 02,m+ 01,m to be determined separately (26)III. EXPERIMENTALTask overview:Measure conductivities of M, M, M, and M solutions of three strongelectrolytes (NaCl, HCl, CH3CO2Na) and one weak electrolyte (acetic acid; CH3CO2H) using a Conductivity solution:If calibration solution ( KCl in deionized water) is not available, itcan be prepared byweighing g of anhydrous KCl and dissolving it in mL of deionized solutions :Prepare 200 mL electrolyte solutions in volumetric flasks. Use the electrolytes and concentra-tions given in the task overview above. CH3 COOH and HCl are given as M stock solutions whereas the rest ofthe compounds are given as solids.