EXPERIMENT 10: CONDUCTIMETRY

Natorilla, Haziel May C.Mendoza, Enrico

Chem 157.1 - DEMr. Leonardo Dante P. Yambot

Department of Physical Sciences and MathematicsCollege of Arts and Sciences

University of the Philippines Manila=====================================================================

Theoretical Framework

The flow of electricity through a conductor involves the transfer of electrons from a

point of higher negative potential to one of lower negative potential. Transfer can be done

through electronic conductors (conduction takes place via direct migration of electrons

through the conductor), or by electrolytic conductors (conduction takes place via migration of

ions, both positive and negative, towards the electrodes). Upon placing an electric field across

an aqueous solution (that is, by applying a voltage between two parallel plate electrodes), the

cations are attracted towards the negative plate (cathode) and the anions towards the positive

plate (anode).

Electrolyte solutions obey Ohm’s law. If a wire is placed between the two terminals of

a battery and the circuit is closed, current will flow. The amount of current which flows

depends on the resistance of the wire according to Ohm’s law,

V = I R (1)

where V is the voltage, I the current, and R the resistance of the wire.

This resistance will depend on the composition of the wire and on its geometry.

It is possible to characterize a material by measuring an intrinsic property of the

material called resistivity, ρ. The relationship between the resistance of a material and the

geometric independent resistivity is:

r = R A/L (2)

Often, we discuss the conductivity, k, instead of the resistivity. Conductivity is the

inverse of the resistivity,

k= 1/r (3)

The conductance of all electrolytes increases with increasing temperature.

Conductivity, the absolute velocity of an ion moving through a solution (which is dependent

on: type and concentration of ions present, solvent, area of electrode, and distance between

the electrodes) can be measured with a conductivity meter.

If the solvent is pure water, there are only few charged species (ions) present. Kw

=[OH-][H+]=10-14, and the conductivity measured will be low. For instance, the addition of ions

such as in a 0.10 N solution of NaCl (L = 106.8 W -1 cm2 mol-1) will increase the conductivity of

the solution to 106.8 W-1cm2mol-1x 0.01mol/1000 cm3 = 10.68 x 10-6W-1 cm-1 or 1068 x 10-6W-1

m-1.

According to Kohlrausch’s Law of Independent Migration for Infinite Dilution, the

molar conductivity varied as the square root of the concentration for many solutions.

L m = Lmo - k c1/2 (4)

where the limiting molar conductivity at infinite dilution, Lmo , is a constant which

depends on the electrolyte. The constant, k, is more dependent on stoichiometry of the

electrolyte than its nature. The difference in Lo for pairs of salts with a common ion is

expected to be the same regardless of the common ion.

The molar conductivities at infinite dilution can be used to determine conductivity at

infinite dilution for individual ions to yield the law of the independent migration of ions. For

instance, the limiting conductivity of NaCl could be determined by the relationship:

LoNaCl = L o

Na+ + L oCl- (5)

Now, the increase of the equivalent conductance of solutions of strong electrolytes in

the low-concentration range is not associated to an increase in dissociation (because the

dissociation is complete), but rather, it is associated to an increased mobility of the ions. In a

concentrated solution of a highly ionized strong electrolyte, the ions are close enough to one

another so that any one of them in moving is influenced not only by the electrical field

impressed across the electrodes but also by the field of the surrounding ions. The ionic

velocities are, then, dependent upon both forces.

On the other hand, some electrolytes (weak electrolytes) do not dissociate completely

in solution. Instead, equilibrium exists between ions and associated electrolyte. The apparent

equilibrium constant for dissociation may be calculated as

c

cK

)1(

22

(6)

where = degree of dissociation, and c = concentration of the solute.

Conductance measurements can be used to determine the solubility of difficultly

soluble substances, and the ionization constant for a weak electrolyte and for conductimetric

titrations. By using the equation λm = 1000 Ls / C, C being the concentration of substance, its

solubility can be calculated. To find the degree of ionization α, λm and λ0 are required.

According to the Arrhenius theory, the equivalent conductance at any concentration is related

to the degree of dissociation by:

= L/Lo (7)

where L = equivalent conductance at concentration c, and Lo = equivalent

conductance at infinite dilution.

In the case of a weak electrolyte the value of Lo cannot be obtained by the

extrapolation to infinite dilution of results obtained at finite concentration, because L is a

rapidly varying and nonlinear function of Lc. Instead, Lo is obtained by the law of Kohlrausch

(eqn. 5).

The apparent equilibrium constant Ka, is equal to the true equilibrium constant, which

can be expressed in terms of activities only for ideal solutes (because g = 1.0).

HRHR

RRHH

HR

RHa c

ccK

c

ccK

ggg

(8)

where g i is the activity coefficient of species i. Since g i=1.0 for infinitely dilute solutions,

Lim Ka = K

c 0

Arrhenius noted that the molar conductivities of electrolytes decreased with

increasing concentration. He attributed the decrease in conductivity to the decrease in the

degree of ionization of the electrolyte.

The mobility m of an ion is defined as the velocity per unit field strength, or m+ = u+/E

and is m -= u-/E. From Ohm’s law and the definition of conductivity, then k = I/E, where I is the

current across a unit area and E is the electric field. The specific conductivity of the solution is

Ls = ne(m+ + m-) (9)

Hence, molar conductance is then

L = Noe(u+ + u-) (10)

where No is the Avogadro number. Since Noe is the charge on a mole of electrons, or

the Faraday constant F, equation 10 can be written as

Lo = F(m+ + m-) (11)

It is here that Arrhenius made the critical assumption that at infinite dilution ionization

is complete, that is, as m =0, a =1. In the limit,

L = F(m+ + m-) (12)

and by combining previous we get the expression for the degrees of ionization,

= L /Lo (13)

Methodology

Objectives:1. To be able to determine the relative mobilities of some ions in solution2. To be able to determine the molar conductance of different concentrations of

solutions of electrolytes3. To determine the molar conductance at infinite dilution of an electrolyte4. To determine the ionization constant of a weak electrolyte by conductance

measurement5. To be able to determine the concentration of an electrolyte by conductance

measurement

Procedure:

A. Preparation of Conductance Cell

2 holes 1 inch apart in 3” x 3” square cardboard/Styrofoam; Insert electrodes in holes

Put cardboard with electrodes on 50-ml beaker with electrodes1 cm from bottom of beaker

Connect graphite rods in series with 9-V cell and digital multimeter

B. Determination of Relative Ionic Mobility

Prepare: (20 ml of each in 50 ml beaker) 1 M HCl, 1 M NaOH, 1 M NaCl, 1 M NH4Cl and 1 M NaC2H3O2

Dip electrodes in each solution, measure conductance of each solution; Wash probe and dry after use

Tabulate in increasing order solutions containing Cl- ; Determine relative mobility of cations.

Do the same for solutions containing Na+

C. Determination of cell constant

Prepare: (In a 50ml beaker) 25 ml 0.100 m KCl (t = 25˚C)

Measure resistance, calculate cell constant.

D. Variation of molar conductance with concentration

Serial Dilution: From 1 M HCl, 25 ml of the following: 0.50 M, 0.20 M, 0.10 M, 0.050 M, 0.020 M and 0.010

Repeat using NaC2H3O2 and NaCl

Measure resistance of each solution at 25˚C; Calculate L, LS, and λm

E. Determination of Ionization Constant of a Weak Electrolyte by Conductance Measurement

Serial dilution: From 1 M HC2H3O2: 0.20 M, 0.10 M, 0.050 M, 0.020 M, 0.010 M

Measure resistance of each; Calculate LS and λm, α, and Ka of HC2H3O2

F. Determination of Solubility Product Constant by Conductance Measurement

Prepare: (In 50 ml beaker) 12.5 ml 0.05 M Ca(NO3)2 + 12.5 ml 0.10 M Add NaOH dropwise while swirling

Allow precipitate to form and settle at the bottom of beaker (supernatant = clear)

Measure resistance of solution, calc Ls and λm

G. Conductimetric Titration

25 ml analyte, dip probe and take multimeter reading

+ 1 ml increments slowly while mixing.

Stop reading when multimeter reading no longer shows appreciable change in slope

Results and Discussions

A. Determination of Relative Ionic Mobility

Group 3-4 Group 5-6 Group 7-8

Solution I (A)

Hydrochloric acid 0.757 0.172 0.555

Sodium hydroxide 0.763 0.137 0.142

Sodium chloride 0.760 0.118 0.115

Ammonium chloride 0.750 0.131 0.128

Sodium acetate 0.759 0.096 0.099

A2. Relative Mobility of Cations

Group 3-4 Group 5-6 Group 7-8

Cation I (A)

H+ 0.757 0.172 0.555

Na+ 0.760 0.118 0.115

NH4+ 0.750 0.131 0.128

Relative Mobility: H+ > Na+ > NH4+

A3. Relative Mobility of Anions

Group 3-4 Group 5-6 Group 7-8

Anion I (A)

OH- 0.763 0.137 0.142

Cl- 0.760 0.118 0.115

C2H3O2- 0.759 0.096 0.099

Relative Mobility: OH- > Cl- > C2H3O2-

Ionic mobilities depend largely on the size of the molecule. Since we cannot obtain

the absolute mobilities in this experiment (as we cannot measure the mobility of an anion

separate from the cation), we obtained the relative mobilities. Observed that H+ moves fastest

and NH4+ moves the slowest because H+ is a very small particle. For the anions, OH- moves

fastest and acetate ion moves the slowest.

Theoretical values (Absolute Mobilities of Ions at 25°C):

Table -13-10. Maron, Samuel H. and Lando, Jerome B. Fundamentals of Physical Chemistry.

B. Determination of Cell Constant

Group 3-4 Group 5-6 Group 7-8

I (A)

Current of the solution 0.759 0.776 0.076

Resistance of the Solution 11.85770751 Ω 11.59793814 Ω 118.4210526 Ω

Cell Constant 0.152964426 /cm 0.149613402 /cm 1.527631579 /cm

Instead of measuring the resistance, we measured the current, I and solved for K.

Sample Calculations:

Ls KCl = 0.0129 mho/cm at T = 25˚C of 0.100 m KCl

V = I x R, or R = V / I; 9V = 0.776 A x R; R = 11.59793814 Ω

K = R x Ls = 11.59793814 Ω x 0.0129 mho/cm = 0.149613402 /cm

C. Variation of Molar Conductance with Concentration

Sample computations (group 3-4 data):

Resistance, Ω

R = V / I; = 9V / 0.754 A = 11.780010471 Ω

Conductance, L

L=1/R = 1 / 11.780010471 Ω = 1.326259947 Ω-1

Specific conductance, Ls

Ls = K/R or Ls = I (K / V)

Ls = 0.152964426 cm-1 / 11.780010471 Ω = 0.012700617 Ω-1 cm -1

Molar Conductance, λm

λm = Ls/ C

λm = 0.012700617 Ω -1 cm -1 = 0.012700617 Ω -1 cm -1 1 (mol/L) x (1 L / 1000 cm3) 0.001 mol / cm3

λm = 12.53427835 Ω-1 cm2 mole-1

Group 3-4

M of HCl Current

(A)

Resistance (Ω) Conductance (Ω-1) Specific

Conductance

(Ω-1 cm-1)

Molar

Conductance

(Ω-1 cm2 mole-1)

1 0.754

11.780010471 1.326259947 0.012700617

12.53427835

0.5 0.752

11.96808511 1.329787234 0.012501031

25.00206185

0.2 0.748

12.03208556 1.336898396 0.012434536

62.17268039

0.1 0.743

12.11305518 1.34589502 0.012351418

123.5141752

0.05 0.732

12.29508197 1.366120219 0.012168557

243.3711339

0.02 0.721

12.48266297 0.080111111 0.011985696

599.2847936

0.01 0.711

12.65822785 0.079 0.011819459

1181.945876

Group 5-6

M of HCl Current

(A)

Resistance

(Ω)

Conductance (Ω-1) Specific

Conductance

(Ω-1 cm-1)

Molar

Conductance

(Ω-1 cm2 mole-1)

1 0.239

37.6569038 0.02655556 0.00397307

3.973067009

0.5 0.210

42.8571429 0.02333333 0.00349098

6.98195876

0.2 0.170

52.9411765 0.01888889 0.00282603

14.13015463

0.1 0.108

83.3333333 0.012 0.00179536

17.95360824

0.05 0.062

145.16129 0.00688889 0.00103067

20.61340205

0.02 0.023

391.304348 0.00255556 0.00038235

19.11726803

0.01 0.011

818.181818 0.00122222 0.00018286

18.28608247

From the results, it can be seen that molar conductance increased as the solution became more dilute, while the specific conductance decreased as the solution became more dilute. Theoretically, the molar conductance is expected to increase as the solution becomes more dilute, while Ls must decrease as the solution becomes more dilute.

group 5-6: HCl

y = 1.6644x + 0.1137

R2 = 0.118

0

0.05

0.1

0.15

0.2

0.25

0 0.01 0.02 0.03 0.04

square root of C

mo

lar

con

du

ctan

ce

Series1

Linear (Series1)

- all data

group 5-6: HCly = 11.365x + 0.0513

R2 = 0.8885

0

0.05

0.1

0.15

0.2

0.25

0 0.005 0.01 0.015

square root of C

mo

lar

con

du

ctan

ce

Series1

Linear (Series1)

- removed outliers

Y-intercept = 0.0513 = λ0, HCl

HCl vs measured and expected behavior

y = -16.865x + 18.966

R2 = 0.8993

0

5

10

15

20

25

0 0.2 0.4 0.6 0.8 1 1.2

[HCl]

con

du

ctan

ce

measured

Linear (measured)

The linear equation for this graph is the expected behaviour of HCl, because this is a strong acid.

Group 5-6M of NaOAc Current

(A)

Resistance (Ω) Conductance (Ω-1) Specific

Conductance

(Ω-1 cm-1)

Molar

Conductance

(Ω-1 cm2 mole-1)

1 0.178

50.5618 0.019778 0.002959

2.95902062

0.5 0.15

60 0.016667 0.002494

4.9871134

0.2 0.083

108.4337 0.009222 0.00138

6.8988402

0.1 0.047

191.4894 0.005222 0.000781

7.81314433

0.05 0.029

310.3448 0.003222 0.000482

9.64175257

0.02 0.013

692.3077 0.001444 0.000216

10.8054124

0.01 0.0091000 0.001 0.00015

14.9613402

Group 5-6: NaOAc

y = -347.44x + 12.903

R2 = 0.832

0

2

4

6

8

10

12

14

16

0 0.01 0.02 0.03 0.04

square root of c

mo

lar

con

du

ctan

ce

Series1

Linear (Series1)

- all data

Group 5-6: NaOAc

y = -277.24x + 11.328

R2 = 0.9627

0

2

4

6

8

10

12

0 0.01 0.02 0.03 0.04

square root of c

mo

lar

con

du

ctan

ce

Series1

Linear (Series1)

- removed outliersY-intercept = 11.328 = λ0, NaOAc

Group 5-6M of NaCl Current

(A)

Resistance (Ω) Conductance (Ω-1) Specific

Conductance

(Ω-1 cm-1)

Molar

Conductance

(Ω-1 cm2 mole-1)

1 0.178

50.5618 0.019778 0.002959

2.95902062

0.5 0.15

60 0.016667 0.002494

4.9871134

0.2 0.083

108.4337 0.009222 0.00138

6.8988402

0.1 0.047

191.4894 0.005222 0.000781

7.81314433

0.05 0.029

310.3448 0.003222 0.000482

9.64175257

0.02 0.013

692.3077 0.001444 0.000216

10.8054124

0.01 0.0091000 0.001 0.00015

14.9613402

Group 5-6: NaCl

y = -347.44x + 12.903

R2 = 0.832

0

2

4

6

8

10

12

14

16

0 0.01 0.02 0.03 0.04

square root of c

mo

lar

con

du

ctan

ce

Series1

Linear (Series1)

- all data

Group 5-6: NaCl

y = -277.24x + 11.328

R2 = 0.9627

0

2

4

6

8

10

12

0 0.01 0.02 0.03 0.04

square root of c

mo

lar

con

du

ctan

ce

Series1

Linear (Series1)

- removed outliersY-intercept = 11.328 = λ0, NaCl

From the graphs of [HCl vs. conductance and the square root of C vs. molar

conductance, it can be observed that molar conductance decreased as C increased and Ls is

almost directly proportional to C. Theoretically, the conductance must increase gradually until

the equivalence point is reached, then increase sharply afterwards.

D. Determination of Ionization Constant of a Weak Electrolyte by Conductance

Measurement

Sample Computation:

α = λm/λo,HOAc

= λm /(λoNaOAc + λoHCl – λoNaCl)

= 0.00014213/ (11.328 + 0.0513 – 11.328) mho cm2 mol-1

= 2.770565302 x 10-3

HOAc = H+ + OAc-

C-C α C α C α

Ka = (C α2)/(1- α)

= (1(0.00277)2)/(1-0.00277)

= 7.69765 x 10-6

Group 5-6

M HoAc Current

(A)

Resistance

(Ω)

Conductance

(Ω-1)

Specific

Conductance

(Ω-1 cm-1)

Molar

Conductance

(Ω-1 cm2 mole-1)

Degree of

Ionization (α)

Ionization

constant

(Ka)

1 0.00855

1052.632 0.00095 0.000142

0.00014213

0.00277062 7.69765E-060.2 0.00490

1836.735 0.000544 8.15E-05

0.00040728

0.0079392 6.35353E-050.1 0.00380

2368.421 0.000422 6.32E-05

0.0006317

0.01231386 0.000153522

0.05 0.00248

3629.032 0.000276 4.12E-05

0.00082454

0.01607283 0.0002625560.02 0.00171

5263.158 0.00019 2.84E-05

0.00142133

0.02770619 0.0007895070.01 0.00135

6666.667 0.00015 2.24E-05

0.0022442

0.04374661 0.002001317

How does dilution affect the degree of ionization of acetic acid? Ka increases as dilution increases.

Theoretically, as the solution becomes more dilute, the degree of ionization must

decrease because water posses a common ion which suppresses the ionization and

promotes the formation instead of acetic acid.

E. Determination of the Solubility Product Constant by Conductance Measurement

Ca(OH)2 Ca2+ + 2OH-

I = 0.92 A

R, Ca(NO3)2= 9V / 0.92 A = 9.782608696 Ω

L = 1/R = 0.102222222 Ω-1

Ls,soln = K / R = 0.149613402 cm-1 / 9.782608696 Ω = 0.015293814 Ω-1 cm-1

Ls,solute = Ls,soln – Ls,solvent = 0.015293814 Ω-1 cm-1 - 0.58 x 10-7 = 0.015293756 Ω-1 cm-1

Λ, Ca = Ls, solute / C = 0.015293756 Ω -1 cm -1

0.1 mol x 1 Liter

L 1000 cm3

= 152.93756 Ω-1 cm2 mole-1

Λ0,Ca(OH)2 = lCa2+ + 2lOH-

= 2(59.50) + 2(198)

= 515 mho cm2 mole-1

Λ0 = 1000* Ls / C; C = 1000* Ls / Λ0

C = (0.015293756 Scm-1 / 515 Scm2mole-1) x (1000cm3/1L)

C = 0.029696613 M

Ksp, Ca(OH)2 = [C][2C]2 (theoretical = 6.5 x 10-6) = 4 C3 = 4(0.029696613)3 = 1.045764507 x 10-4

Theoretical values (Equivalent Ionic Conductances at Infinite Dilution, 25°C):

Table -13-9. Maron, Samuel H. and Lando, Jerome B. Fundamentals of Physical Chemistry.

Using conductance measurements, the solubility product constant of the difficultly

soluble Ca(OH)2 was determined. The Ls of the solute was determined first. It was equated

with the Ls of the solution because the solution is very dilute because Ca(OH)is very sparingly

soluble. The Ksp that we obtained was slightly off from the theoretical value which is of the

order 10-4. (theoretical = 6.5 x 10-6; experimental = 1.045764507 x 10-4).

F. Determination of Solubility Product Constant by Conductance Measurement

Plot VTitrant vs Conductance

Conductimetric Titration

0

0.01

0.02

0.03

0.04

0.05

0.06

0 10 20 30 40

volume of titrant (ml)

con

du

ctan

ce

Series1

Vtitrant at equivalence point ≈ 14.5 mlConcentration of unknown = 0.041428571 M

M1V1 = M2V2

(0.1 M NaOH)(0.0145L) = M2 (0.0145 L)M2 (acid) = (0.10 moles/L NaOH x 14.5 ml x 1L/1000ml) / (35 ml x 1L /1000ml)M = 0.041428571 M

For the conductimetric titration, the plot showed that the equivalence point was when

Vtitrant ≈ 14.5 ml (extrapolate manually), and the computed Macid = 0.041428571. We can also

deduce from the plot that the unknown analyte is a strong acid because of the V-shaped

graph, which is characteristic of strong electrolytes when titrated with a strong base.

Answers to Questions

1. How does molar conductance vary with concentration?

Based on the equation λm = 1000 Ls/ C, it can be inferred that as concentration of the

solution, C, increases, the molar conductance Λm, decreases.

As the concentration, C, decreases, molar conductance, Λm, increases for both strong

and weak electrolytes, while as C decreases, molar conductance increases.

In the determination of solubility product, Ls is dependent on C – as C increases for

strong electrolytes, Ls increases sharply. If C decreases, Ls decreases. However,

molar conductance still increases because the decrease in Ls is compensated by the

decrease in C.

2. Do the solutions obey Kohlrausch’s Law of independent migration of ions at infinite dilution? Support and explain your answer.

Kohlrausch’s law of independent migration is applicable at infinitely dilute solutions

only. Based on the calculations, it can be deduced that the experimental solutions

used were concentrated since the solutions did not obey Kohlrausch’s law. If the

solution obeys Kohlrausch’s law, λ0 = λm. The two quantities obtained were not

equal, thus proving that the solutions did not obey the law.

3. How does dilution affect the degree of ionization of acetic acid?

According to Kohlrausch and Heydweiller, no matter how long and how carefully

purified, water will still exhibit a definite small conductance. Recall that the degree of

ionization is α = λm/ λ0. Since for strong and weak electrolytes, molar conductance λm

increases with decreasing concentration, diluting a solution of acetic acid (a weak

electrolyte) will increase λm. But then, water possesses a common ion with acetic

acid, H+ - hence the degree of ionization, α, will decrease.

References:

Atkins, Peter and de Paula, Julio. Atkins Physical Chemistry, 7th edition.Laidler. Meisler. Physical Chemistry, 3rd editionMaron, Samuel H. and Lando, Jerome B. Fundamentals of Physical Chemistry.http://www.tau.ac.il/~phchlab/experiments/Conductivity/conductivity.htmhttp://www.tpub.com/neets/book1/chapter1/1p.htmhttp://www.corrosion-doctors.org/Electrochem/conductivity.htm