molar conductivity at infinite dilution of electrolytes lab report

OBJECTIVE

To determine the molar conductivity at infinite dilution (Ʌ0) of sodium chloride, hydrochloric acid,

sodium acetate, and acetic acid at 250c.

INTRODUCTION

Electrolytes are substances which dissolve in water to produce solutions which conduct electrical current.

Such substances produce ions when dissolve in water, and the ions carry the current through solution.

Nonelectrolytes are the substances whose aqueous solution do not contain ions and hence do not conduct

electrical current. Electrolytes are classified as either strong electrolytes or weak electrolytes. Strong

electrolytes when dissolved in water ionize completely to produce ions. For example, when NaCl is

dissolved in water:

NaCl (s) → Na+ (aq) + Cl- (aq)

There are no dissolved NaCl molecule present in the solution. Solutions of strong electrolytes are good

conductor of electricity because they contain a relatively high concentration of ions.

Strong Electrolytes

The electrolytic conductivity (K) (S cm-1) of a solution increases with concentration. However, quantity K

is not a suitable quantity for comparing the conductivities of different solutions. If a solution of one

electrolyte is much more concentrated than another, it may have a higher conductivity simply because it

contains more ions. Instead, molar conductivity (Ʌ) (S cm -1) should be adopted for comparison. It is

defined as (K/concentration). Quantity Ʌ decreases as the concentration increases. Onsager showed

theoretically for strong electrolytes in dilute solution that the effect of ionic attraction reduces the molar

conductivity as in Eq. (1)

Ʌ = Ʌ0 - K√c (1)

Quantities c and Ʌ0 denote concentration of the electrolyte of the molar conductivity at in finite dilution.

Below concentrations of about 0.1 M, a plot of Ʌ against √c result in a straight line. The intersection of

this line with the ordinate is the Ʌ0. The Ʌ0 values are found to be additive. Kohlrausch assumed that in

such a system the molar conductivity at infinite dilution is simply the sum of the independent contribution

of the ions.

Consider a strong electrolytes that yields ions A and B solution:

ApBq → pAz+ + qBz-

Kohlrausch’s law of independent migration of ions proposed:

Ʌ0 = pƛ0+ + qƛ0

- (2)

For a 1-1 electrolyte A+B-

Ʌ0 = ƛ0+ + ƛ0

- (3)

Quantity ƛ0 denotes molar ionic conductivity at infinite dilution. (S cm2 mol-1). This equation has been

written for infinite dilution since it only under such conditions, when ion-ion interactions are at a

minimum that the law strictly holds. It is the applicable to both strong and weak electrolytes.

CHEMICALS

1) 0.1 M sodium chloride, NaCl

2) 0.1 M hydrochloric acid, HCl

3) 0.1 M sodium acetate, NaAc

APPARATUS

1) Digital conductivity meter (1)

2) 100 mL volumetric flask (18)

3) 50 mL burette (3)

4) 100 mL beaker (9)

5) Magnetic stirrer with stirring bar (1)

6) Conductivity probe holder (1)

PROCEDURES

1) A clean burette was filled with 0.1 M NaCl solution.

2) The required volume of 0.1 M NaCl was drained out into each volumetric flask and top up with

DI water to prepared the following molarities, 0.05, 0.01, 0.005, 0.001, 0.0005 and 0.0001 M.

3) The digital conductivity meter was calibrated with conductivity standard of 1413 μS cm-1 or 12.88

mS cm-1 at 250c.

4) Stirring bar and conductivity probe was rinsed by using DI water.

5) The beaker was filled with 50 mL deionized water and stirring bar was placed in.

6) The beaker was placed on the magnetic stirrer.

7) The probe was immersed to a depth approximately 5 cm in the solution. Probe was support with

probe holder.

8) The magnetic stirrer was switch on and the electrolytic conductivity (K) of the DI water at 250c

was recorded.

9) Steps 4-8 were repeated with the diluted NaCl. The NaCl solution was begin with the lowest

concentration first.

10) Steps 1-9 by using HCl and NaAc solutions.

11) All electrolytic conductivity measurements was measured at 250c. If the K values of the solutions

are comparable to the DI water, quantities of K was corrected as (K-Kwater).

RESULTS AND DISCUSSIONS

1) Tabulate the results.

Refer to the appendix

2) Determine the Ʌ (S cm2 mol-1) for each of the strong electrolyte solutions, then plot Ʌ versus √c

(unit of c in mol cm-3). You may include all the three plots (NaCl, HCl, NaAc) 0n the same graph

paper.

NaCl

Concentration = 0.0001 M

Λ (S cm2 mol-1) = K√c

=[ 50.9 x10−6Scm−1

0.0001molL×

1000 L1m3 ×

(10−2 )3m3

cm3 ]= 509 S cm2 mol-1



=[ 103.9 x10−6Scm−1

0.0005molL×

1000L1m3 ×

(10−2 )3m3

cm3 ]= 207.8 S cm2 mol-1



=[ 168.1 x10−6Scm−1

0.001molL×

1000 L1m3 ×

(10−2 )3m3

cm3 ]= 168.1 S cm2 mol-1



=[ 635.0 x10−6Scm−1

0.005molL×

1000L1m3 ×

(10−2 )3m3

cm3 ]= 127 S cm2 mol-1



=[ 1232 x10−6Scm−1

0.01molL×

1000 L1m3 ×

(10−2 )3m3

cm3 ]= 123.2 S cm2 mol-1



=[ 5.71 x10−3Scm−1

0.05molL×

1000L1m3 ×

(10−2 )3m3

cm3 ]= 114.2 S cm2 mol-1

HCl



=[ 199.5 x10−6Scm−1

0.0001molL×

1000 L1m3 ×

(10−2 )3m3

cm3 ]= 1995 S cm2 mol-1



=[ 215.0 x10−6Scm−1

0.0005molL×

1000L1m3 ×

(10−2 )3m3

cm3 ]= 430 S cm2 mol-1



=[ 423.0 x10−6S cm−1

0.001molL×

1000 L1m3 ×

(10−2 )3m3

cm3 ]= 423 S cm2 mol-1



=[ 2.09 x10−3Scm−1

0.005molL×

1000L1m3 ×

(10−2 )3m3

cm3 ]= 418 S cm2 mol-1



=[ 4.12 x10−3Scm−1

0.01molL×

1000 L1m3 ×

(10−2 )3m3

cm3 ]= 412 S cm2 mol-1



=[ 19.73 x10−3Scm−1

0.05molL×

1000L1m3 ×

(10−2 )3m3

cm3 ]= 394.6 S cm2 mol-1

NaAc



=[ 37.1 x10−6Scm−1

0.0001molL×

1000 L1m3 ×

(10−2 )3m3

cm3 ]= 371 S cm2 mol-1



=[ 72.2 x10−6Scm−1

0.0005molL×

1000L1m3 ×

(10−2 )3m3

cm3 ]= 144.4 S cm2 mol-1



=[ 128.1 x10−6Scm−1

0.001molL×

1000 L1m3 ×

(10−2 )3m3

cm3 ]

= 128 S cm2 mol-1



=[ 461.0 x10−6S cm−1

0.005molL×

1000L1m3 ×

(10−2 )3m3

cm3 ]= 92.2 S cm2 mol-1



=[ 857.0 x10−6Scm−1

0.01molL×

1000 L1m3 ×

(10−2 )3m3

cm3 ]= 85.7 S cm2 mol-1



=[ 4.24 x 10−3S cm−1

0.05molL×

1000L1m3 ×

(10−2 )3m3

cm3 ]

= 84.8 S cm2 mol-1

Table of Ʌ versus √c

c (mol L-1) √c (mol cm-3) Ʌ (S cm2 mol-1)

NaCl HCl NaAc

0.0001 3.1623 x 10-4 509 1995 371

0.0005 7.0712 x 10-4 207.8 430 144.4

0.001 1.0000 x 10-3 168.1 423 128

0.005 2.2361 x 10-3 127 418 92.2

0.01 3.1623 x 10-3 123.2 412 85.7

0.05 7.0712 x 10-3 114.2 394.6 84.8

Graph of Ʌ versus √c

Refer to the appendix

3) From the linear regressions, determine the value of Ʌ0 for each of the solution.

From the graph:

Ʌ0 (NaCl) = 230 S cm2 mol-1

Ʌ0 (HCl) = 475 S cm2 mol-1

Ʌ0 (NaAc) = 100 S cm2 mol-1

The value of slope for the graph of Ʌ versus √c are negative. The value of Ʌ0 for each

solution were determined by extended the straight line to the 0.0 M concentration.

4) Report the standard errors in the Ʌ0.

The equivalent conductivity of weak electrolytes increases steeply at very low concentrations

(see the above graph) and hence their limiting values (Λo) cannot be determined by

extrapolating the Λc to zero concentration.

Molar conductivities of weak electrolytes,Λmo can be easily calculated with the help of

Kohlrausch's law.

5) From the Ʌ0 value, determine the value of Ʌ0 for acetic acid (HAc).

Ʌ0 (NaCl) = 230 S cm2 mol-1

Ʌ0 (HCl) = 475 S cm2 mol-1

Ʌ0 (NaAc) = 100 S cm2 mol-1

Ʌ0CH

3COOH = λ0

CH3COO- + λ0

H+

Ʌ0CH

3COONa = λ0

CH3

COO- + λ0Na+

Ʌ0HCl = λ0

H+ + λ0Cl-

Ʌ0NaCl = λ0

Na+ + λ0Cl-

Therefore:

Ʌ0CH

3COOH = Ʌ0

CH3COONa + Ʌ0

HCl - Ʌ0NaCl

= (100 + 475 – 230) S cm2 mol-1

= 345 S cm2 mol-1

6) Using the table of molar ionic conductivities of electrolyte at in finite dilution in aqueous solution

at 250c find the Ʌ0 of NaCl, HCl, NaAc and HAc.

Cations λ+0 /S cm2mol-1 anions λ-

0 /S cm2mol-1

H+ 349.6 OH- 199.1Li+ 38.69 Cl- 76.34Na+ 50.11 Br- 78.4K+ 73.50 I- 76.8

Mg2+ 106.12 SO42- 159.6

Ca2+ 119.00 NO3- 71.4

Ba2+ 127.28 CH3COO- 40.9

HCl

Ʌ0HCl = λ0

H+ + λ0Cl-

= 349.6 + 76.34

= 425.94 S cm2 mol-1

NaCl

Ʌ0NaCl = λ0

Na+ + λ0Cl-

= 50.11 + 76.34

= 126.45 S cm2 mol-1

NaAc

Ʌ0CH

3COONa = λ0

CH3

COO- + λ0Na+

= 40.9 + 50.11

= 91.01 S cm2 mol-1

HAc

Ʌ0CH

3COOH = λ0

CH3COO- + λ0

H+

= 40.9 + 349.6

= 390.5 S cm2 mol-1

7) Compare Ʌ0,exp with the Ʌ0,theory determined from the standard data. Describe possible sources for

systematic errors and degree of their importance.

% error (NaCl) = Ʌ ,exp−Ʌ ,theory

Ʌ , theory x 100

= 230−126.45

126.45 x 100

= 81.89%

% error (HCl) = Ʌ ,exp−Ʌ ,theory

Ʌ , theory x 100

= 475−425.94

425.94 x 100

= 11.51%

% error (NaAc) = Ʌ ,exp−Ʌ ,theory

Ʌ , theory x 100

= 100−91.01

91.01 x 100

= 9.88%

Possible factor that affect the errors:

Interionic attractions: depends on solute-solute interactions.

Solvation of ions: depends on solute-solvent interactions.

Viscosity solvent: depends upon solvent-solvent interactions.

Concentration of solution: Higher concentration of solution less is conduction. If

weak electrolyte, ionization is less & if strong

electrolyte.

Temperature: Electrolytic conduction increase with increase in temp.

CONCLUSION

From the experiment, the value of Ʌ0 at 250c for NaCl is 230 S cm2 mol-1, for HCl is 475 S cm2 mol-1, for

NaAc is 100 S cm2 mol-1 and for HAc is 345 S cm2 mol-1.

REFERENCES

1) http://www.adichemistry.com/physical/electrochemistry/kohlrausch/kohlrausch-law.html

2) http://en.wikipedia.org/wiki/Conductivity_%28electrolytic%29

3) http://www.jesuitnola.org/upload/clark/labs/PerError.htm

http://www.jesuitnola.org/upload/clark/labs/PerError.htm

http://en.wikipedia.org/wiki/Conductivity_(electrolytic)

http://www.adichemistry.com/physical/electrochemistry/kohlrausch/kohlrausch-law.html

molar conductivity at infinite dilution of electrolytes lab report

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