OBJECTIVE To determine the molar conductivity at infinite dilution (Ʌ 0 ) of sodium chloride, hydrochloric acid, sodium acetate, and acetic acid at 25 0 c. 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 = Ʌ 0 - K√c (1) Quantities c and Ʌ 0 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 Ʌ 0 . The Ʌ 0 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.
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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) =
= [ ( ) ]
= 509 S cm2 mol
-1
Concentration = 0.0005 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 207.8 S cm2 mol
-1
Concentration = 0.001 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 168.1 S cm2 mol
-1
Concentration = 0.005 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 127 S cm2 mol
-1
Concentration = 0.01 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 123.2 S cm2 mol
-1
Concentration = 0.05 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 114.2 S cm2 mol
-1
HCl
Concentration = 0.0001 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 1995 S cm2 mol
-1
Concentration = 0.0005 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 430 S cm2 mol
-1
Concentration = 0.001 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 423 S cm2 mol
-1
Concentration = 0.005 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 418 S cm2 mol
-1
Concentration = 0.01 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 412 S cm2 mol
-1
Concentration = 0.05 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 394.6 S cm2 mol
-1
NaAc
Concentration = 0.0001 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 371 S cm2 mol
-1
Concentration = 0.0005 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 144.4 S cm2 mol
-1
Concentration = 0.001 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 128 S cm2 mol
-1
Concentration = 0.005 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 92.2 S cm2 mol
-1
Concentration = 0.01 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 85.7 S cm2 mol
-1
Concentration = 0.05 M
Λ (S cm2 mol
-1) =
= [ ( ) ]
= 84.8 S cm2 mol
-1
Table of Ʌ versus √c
c (mol L-1
) √c (mol cm-3) Ʌ (S cm
2 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.
Related Documents
