LABORATORY REPORT
Course Code: SKF 3023
Semester 2 Session 2017/2018
ID NUMBER AND NAME
1. NUR AMIRA BINTI RAMLI
D20162075606
2. SITI NUR FARIHAH BINTI YUSUF MASRI
D20162076390
LECTURER
3. ANNI ISTI ‘AANAH
D20172079644
DR. NORLAILI BINTI ABU BAKAR
EXPERIMENT NO.
4
TITLE
CONDUCTIVITY OF ELECTROLYTES
DATE & DAY
Monday, 26 March 2018
CHECK LIST (Please tick)
Title
Objective(s)
Methods
Results (Observation, Data,
Calculation, etc.)
Discussions and questions &
answers (if appropriate)
Conclusion(s)
References (at least 2)
TOTAL MARKS
Department of Chemistry
Faculty of Science and Mathematics
UNIVERSITI PENDIDIKAN SULTAN IDRIS
Marks
5
5
10
30
35
10
5
100
TITLE: Conductivity of Electrolytes
OBJECTIVES
Experiment entitled “Conductivity of Electrolytes” has three objectives, there are:
1. To determine the relationship between concentration and conductivity.
2. To determine the conductivity at infinite dilution.
3. To determine the activity coeficients.
APPARATUS
1. Conductivity meter and cell
2. Beaker Volumetric falsk 100 mL
3. Pipette
CHEMICALS
1. Sodium chloride
2. Sodium Sulphate
3. Calcium chloride
4. Potassium nitrate
METHODS
0.1 M of:
 Na2SO4
 NaCl
 KNO3
was prepared as a stock solution.
100 ml of:
 0.05
 0.01
 0.005
 0.001
 0.0005
 0.0001
was prepared from stock solution.
The conductivity of solutions was
measured starting from dilute
solution
RESULTS
Data
Solution
Na2SO4
NaCl
KNO3
Concentration
Conductivity
(M)
(μS/cm)
0.05
8.94
mS/cm
25.5
0.01
2070
μS/cm
25.5
0.005
1110
μS/cm
25.4
0.001
221.2
μS/cm
25.4
0.0005
116.0
μS/cm
25.4
0.0001
33.4
μS/cm
25.4
0.05
5.57
mS/cm
25.2
0.01
1177
μS/cm
25.1
0.005
608
μS/cm
25.4
0.001
133.3
μS/cm
25.3
0.0005
70.2
μS/cm
25.5
0.0001
18.39
μS/cm
25.6
0.05
6.72
mS/cm
25.7
0.01
1245
μS/cm
25.5
0.005
655
μS/cm
25.6
0.001
134.2
μS/cm
25.8
0.0005
62.0
μS/cm
25.8
0.0001
8.54
μS/cm
26
Unit
Calculation
The molar conductivity for each solution.

Concentration in mol cm-3 for 0.05 M
𝐂 = 𝟎. 𝟎𝟓
𝒎𝒐𝒍
𝟏𝒅𝒎𝟑
𝒙
= 𝟓𝒙𝟏𝟎−𝟓
𝒅𝒎𝟑 𝟏𝟎𝟎𝟎 𝒄𝒎𝟑
√𝑪 = √𝟓𝒙𝟏𝟎−𝟓 = 𝟕. 𝟎𝟕𝟏 𝒙𝟏𝟎−𝟑

Concentration in mol cm-3 for 0.01 M
𝒎𝒐𝒍
𝟏𝒅𝒎𝟑
𝐂 = 𝟎. 𝟎𝟏
𝒙
= 𝟏𝒙𝟏𝟎−𝟓
𝒅𝒎𝟑 𝟏𝟎𝟎𝟎 𝒄𝒎𝟑
Temperature
(0C)
√𝑪 = √𝟏𝒙𝟏𝟎−𝟓 = 𝟑. 𝟏𝟔𝟐 𝒙𝟏𝟎−𝟑

Concentration in mol cm-3 for 0.005 M
𝐂 = 𝟎. 𝟎𝟎𝟓
𝒎𝒐𝒍
𝟏𝒅𝒎𝟑
𝒙
= 𝟓𝒙𝟏𝟎−𝟔
𝒅𝒎𝟑 𝟏𝟎𝟎𝟎 𝒄𝒎𝟑
√𝑪 = √𝟓𝒙𝟏𝟎−𝟓 = 𝟐. 𝟐𝟑𝟔 𝒙𝟏𝟎−𝟑

Concentration in mol cm-3 for 0.001 M
𝒎𝒐𝒍
𝟏𝒅𝒎𝟑
𝐂 = 𝟎. 𝟎𝟎𝟏
𝒙
= 𝟏𝒙𝟏𝟎−𝟔
𝒅𝒎𝟑 𝟏𝟎𝟎𝟎 𝒄𝒎𝟑
√𝑪 = √𝟏𝒙𝟏𝟎−𝟔 = 𝟏 𝒙𝟏𝟎−𝟑

Concentration in mol cm-3 for 0.0005 M
𝐂 = 𝟎. 𝟎𝟎𝟎𝟓
𝒎𝒐𝒍
𝟏𝒅𝒎𝟑
𝒙
= 𝟓𝒙𝟏𝟎−𝟕
𝒅𝒎𝟑 𝟏𝟎𝟎𝟎 𝒄𝒎𝟑
√𝑪 = √𝟓𝒙𝟏𝟎−𝟕 = 𝟕. 𝟎𝟕𝟏 𝒙𝟏𝟎−𝟒

Concentration in mol cm-3 for 0.0001 M
𝐂 = 𝟎. 𝟎𝟎𝟎𝟏
𝒎𝒐𝒍
𝟏𝒅𝒎𝟑
𝒙
= 𝟏𝒙𝟏𝟎−𝟕
𝒅𝒎𝟑 𝟏𝟎𝟎𝟎 𝒄𝒎𝟑
√𝑪 = √𝟏𝒙𝟏𝟎−𝟕 = 𝟑. 𝟏𝟔𝟐 𝒙𝟏𝟎−𝟒

Calculation for molar conductivity of each solution.
𝜿
𝚲𝐦 =
𝑪
1. Molar conductivity of Na2SO4

𝚲𝐦 =
𝟖.𝟗𝟒 𝒙𝟏𝟎−𝟑 𝐒 𝒄𝒎−𝟏
𝟓 𝒙 𝟏𝟎−𝟓 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟏𝟕𝟖. 𝟖 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

𝚲𝐦 =
𝟐𝟎𝟕𝟎 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟏 𝒙 𝟏𝟎−𝟓 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟐𝟎𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

𝚲𝐦 =
𝟏𝟏𝟏𝟎 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟓 𝒙 𝟏𝟎−𝟔 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟐𝟐𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

𝚲𝐦 =
𝟐𝟐𝟏.𝟐 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟏𝒙 𝟏𝟎−𝟔 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟐𝟐𝟏. 𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

𝚲𝐦 =
𝟏𝟏𝟔.𝟎 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟓 𝒙 𝟏𝟎−𝟕 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟐𝟑𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

𝚲𝐦 =
𝟑𝟑.𝟒 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟏 𝒙 𝟏𝟎−𝟕 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟑𝟑𝟒 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
2. Molar conductivity of NaCl

𝚲𝐦 =
𝟓.𝟓𝟕 𝐱 𝟏𝟎−𝟑 𝐒 𝒄𝒎−𝟏
𝟓 𝒙 𝟏𝟎−𝟓 𝒎𝒐𝒍 𝒄𝒎−𝟑

𝚲𝐦 =
𝟏𝟏𝟕𝟕 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟏 𝒙 𝟏𝟎−𝟓 𝒎𝒐𝒍 𝒄𝒎−𝟑

𝚲𝐦 =
𝟔𝟎𝟖 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟓 𝒙 𝟏𝟎−𝟔 𝒎𝒐𝒍 𝒄𝒎−𝟑

𝚲𝐦 =
𝟏𝟑𝟑.𝟑 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟏𝒙 𝟏𝟎−𝟔 𝒎𝒐𝒍 𝒄𝒎−𝟑

𝚲𝐦 =
𝟕𝟎.𝟐 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟓 𝒙 𝟏𝟎−𝟕 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟏𝟒𝟎. 𝟒 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

𝚲𝐦 =
𝟏𝟖.𝟑𝟗 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟏 𝒙 𝟏𝟎−𝟕 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟏𝟖𝟑. 𝟗 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏𝟏𝟏. 𝟒 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏𝟏𝟕. 𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏𝟐𝟏. 𝟔 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏𝟑𝟑. 𝟑 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
3. Molar conductivity of KNO3

𝚲𝐦 =
𝟔.𝟕𝟐 𝐱 𝟏𝟎−𝟑 𝐒 𝒄𝒎−𝟏
𝟓 𝒙 𝟏𝟎−𝟓 𝒎𝒐𝒍 𝒄𝒎−𝟑

𝚲𝐦 =
𝟏𝟐𝟒𝟓 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟏 𝒙 𝟏𝟎−𝟓 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟏𝟑𝟒. 𝟒 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏𝟐𝟒. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

𝚲𝐦 =
𝟔𝟓𝟓 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟓 𝒙 𝟏𝟎−𝟔 𝒎𝒐𝒍 𝒄𝒎−𝟑

𝚲𝐦 =
𝟏𝟑𝟒.𝟐 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟏𝒙 𝟏𝟎−𝟔 𝒎𝒐𝒍 𝒄𝒎−𝟑

𝚲𝐦 =
𝟔𝟐.𝟎 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟓 𝒙 𝟏𝟎−𝟕 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟏𝟐𝟒 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

𝚲𝐦 =
𝟖.𝟓𝟒 𝐱 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
𝟏 𝒙 𝟏𝟎−𝟕 𝒎𝒐𝒍 𝒄𝒎−𝟑
= 𝟖𝟓. 𝟒 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
-3
𝐶 (mol cm )
= 𝟏𝟑𝟏 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏𝟑𝟒. 𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝚲𝐦 =
√𝐶 ((mol cm )
-3
𝜿
𝑪
(S cm2 mol-1)
Na2SO4
NaCl
KNO3
5 x 10-5
7.071 x 10-3
178.8
111.4
134.4
-5
1 x 10
-3
3.162 x 10
207
117.7
123.5
5 x 10-6
2.236 x 10-3
222
121.6
131
1 x 10-6
1 x 10-3
221.2
133.3
134.2
5 x 10-7
7.071 x 10-4
232
140.4
124
1 x 10-7
3.162 x 10-4
334
183.9
85.4
Calculation of Λ for each solution
𝚲=
𝜿
𝒏𝒆 𝑪
, where 𝒏𝒆 = 𝒗+ 𝒛+ + 𝒗− 𝒛−
1. Λ for Na2SO4
Na2SO4
2Na+ + SO42-
𝒏𝒆 = (𝟐)(𝟏) + (𝟏)(𝟐) = 𝟒

C = 5 x 10-5 mol cm-3
𝚲=

𝟖. 𝟗𝟒 𝒙 𝟏𝟎−𝟑 𝐒 𝒄𝒎−𝟏
= 𝟒𝟒. 𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟒)(𝟓 𝒙 𝟏𝟎−𝟓 )𝒎𝒐𝒍 𝒄𝒎−𝟑
C = 1 x 10-5 mol cm-3
𝚲=
𝟐𝟎𝟕𝟎 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟓𝟏. 𝟕𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟒)(𝟏 𝒙 𝟏𝟎−𝟓 )𝒎𝒐𝒍 𝒄𝒎−𝟑

C = 5 x 10-6 mol cm-3
𝟏𝟏𝟏𝟎 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟓𝟓. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟒)(𝟓 𝒙 𝟏𝟎−𝟔 )𝒎𝒐𝒍 𝒄𝒎−𝟑
𝚲=

C = 1 x 10-6 mol cm-3
𝟐𝟐𝟏. 𝟐 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟓𝟓. 𝟑 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟒)(𝟏𝒙 𝟏𝟎−𝟔 )𝒎𝒐𝒍 𝒄𝒎−𝟑
𝚲=

C = 5 x 10-7 mol cm-3
𝟏𝟏𝟔 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟓𝟖 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟒)(𝟓𝒙 𝟏𝟎−𝟕 )𝒎𝒐𝒍 𝒄𝒎−𝟑
𝚲=

C = 1 x 10-7 mol cm-3
𝚲=
𝟑𝟑. 𝟒 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟖𝟑. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟒)(𝟏𝒙 𝟏𝟎−𝟕 )𝒎𝒐𝒍 𝒄𝒎−𝟑
2. Λ for NaCl
NaCl
Na+ + Cl-
𝒏𝒆 = (𝟏)(𝟏) + (𝟏)(𝟏) = 𝟐

C = 5 x 10-5 mol cm-3
𝚲=

C = 1 x 10-5 mol cm-3
𝚲=

𝟔𝟎𝟖𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟔𝟎. 𝟖 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟓 𝒙 𝟏𝟎−𝟔 )𝒎𝒐𝒍 𝒄𝒎−𝟑
C = 1 x 10-6 mol cm-3
𝚲=

𝟏𝟏𝟕𝟕 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟓𝟖. 𝟖𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟏 𝒙 𝟏𝟎−𝟓 )𝒎𝒐𝒍 𝒄𝒎−𝟑
C = 5 x 10-6 mol cm-3
𝚲=

𝟓. 𝟓𝟕 𝒙 𝟏𝟎−𝟑 𝐒 𝒄𝒎−𝟏
= 𝟓𝟓. 𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟓 𝒙 𝟏𝟎−𝟓 )𝒎𝒐𝒍 𝒄𝒎−𝟑
𝟏𝟑𝟑. 𝟑 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟔𝟔. 𝟔𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟏𝒙 𝟏𝟎−𝟔 )𝒎𝒐𝒍 𝒄𝒎−𝟑
C = 5 x 10-7 mol cm-3
𝚲=
𝟕𝟎. 𝟐 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟕𝟎. 𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟓𝒙 𝟏𝟎−𝟕 )𝒎𝒐𝒍 𝒄𝒎−𝟑

C = 1 x 10-7 mol cm-3
𝚲=
𝟏𝟖. 𝟑𝟗 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟗𝟏. 𝟗𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟏𝒙 𝟏𝟎−𝟕 )𝒎𝒐𝒍 𝒄𝒎−𝟑
3. Λ for KNO3
KNO3
K+ + NO3-
𝒏𝒆 = (𝟏)(𝟏) + (𝟏)(𝟏) = 𝟐

C = 5 x 10-5 mol cm-3
𝚲=

𝟔. 𝟕𝟐 𝒙 𝟏𝟎−𝟑 𝐒 𝒄𝒎−𝟏
= 𝟔𝟕. 𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟓 𝒙 𝟏𝟎−𝟓 )𝒎𝒐𝒍 𝒄𝒎−𝟑
C = 1 x 10-5 mol cm-3
𝚲=

C = 5 x 10-6 mol cm-3
𝚲=

𝟏𝟑𝟒. 𝟐 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟔𝟕. 𝟏 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟏𝒙 𝟏𝟎−𝟔 )𝒎𝒐𝒍 𝒄𝒎−𝟑
C = 5 x 10-7 mol cm-3
𝚲=

𝟔𝟓𝟓 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟔𝟓. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟓 𝒙 𝟏𝟎−𝟔 )𝒎𝒐𝒍 𝒄𝒎−𝟑
C = 1 x 10-6 mol cm-3
𝚲=

𝟏𝟐𝟒𝟓 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟔𝟐. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟏 𝒙 𝟏𝟎−𝟓 )𝒎𝒐𝒍 𝒄𝒎−𝟑
𝟔𝟐 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟔𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟓𝒙 𝟏𝟎−𝟕 )𝒎𝒐𝒍 𝒄𝒎−𝟑
C = 1 x 10-7 mol cm-3
𝚲=
𝟖. 𝟓𝟒 𝒙 𝟏𝟎−𝟔 𝐒 𝒄𝒎−𝟏
= 𝟒𝟐. 𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
(𝟐)(𝟏𝒙 𝟏𝟎−𝟕 )𝒎𝒐𝒍 𝒄𝒎−𝟑
Calculation of activity coeficient (γ) for all concentrations.
𝜸=
𝚲
𝚲∞
Where;
𝜸
= activity coefficient
𝚲
= equivalent conductivity
𝚲∞
= limiting molar conductivity based on equation 𝚲 = 𝚲∞ − (𝐀 + 𝐁𝚲∞ ) √C
1. Na2SO4
From graph 𝚲 versus √𝑪, the value of 𝚲∞ = 𝟔𝟔. 𝟗𝟖𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

C= 5 x 10-5 mol cm-3 , 𝚲 = 𝟒𝟒. 𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝟒𝟒. 𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=
= 𝟎. 𝟔𝟔𝟕
𝟔𝟔. 𝟗𝟖𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

C= 1 x 10-5 mol cm-3 , 𝚲 = 𝟓𝟏. 𝟕𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝟓𝟏. 𝟕𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=
= 𝟎. 𝟕𝟕𝟑
𝟔𝟔. 𝟗𝟖𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

C= 5 x 10-6 mol cm-3 , 𝚲 = 𝟓𝟓. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝟓𝟓. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=
= 𝟎. 𝟖𝟐𝟗
𝟔𝟔. 𝟗𝟖𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

C= 1 x 10-6 mol cm-3 , 𝚲 = 𝟓𝟓. 𝟑 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=

C= 5 x 10-7 mol cm-3 , 𝚲 = 𝟓𝟖 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=

𝟓𝟓. 𝟑 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟎. 𝟖𝟐𝟔
𝟔𝟔. 𝟗𝟖𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝟓𝟖 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟎. 𝟖𝟔𝟔
𝟔𝟔. 𝟗𝟖𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
C= 1 x 10-7 mol cm-3 , 𝚲 = 𝟖𝟑. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=
𝟖𝟑. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏. 𝟐𝟓𝟕
𝟔𝟔. 𝟗𝟖𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
2. NaCl
From graph 𝚲 versus √𝑪, the value of 𝚲∞ = 𝟕𝟔. 𝟏𝟒𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

C= 5 x 10-5 mol cm-3 , 𝚲 = 𝟓𝟓. 𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=
𝟓𝟓. 𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟎. 𝟕𝟑𝟏
𝟕𝟓. 𝟏𝟒𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

C= 1 x 10-5 mol cm-3 , 𝚲 = 𝟓𝟖. 𝟖 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=

C= 5 x 10-6 mol cm-3 , 𝚲 = 𝟔𝟎. 𝟖 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=

𝟔𝟔. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟎. 𝟖𝟕𝟑
𝟕𝟔. 𝟏𝟒𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
C= 5 x 10-7 mol cm-3 , 𝚲 = 𝟕𝟎. 𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=

𝟔𝟎. 𝟖 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟎. 𝟕𝟗𝟖
𝟕𝟔. 𝟏𝟒𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
C= 1 x 10-6 mol cm-3 , 𝚲 = 𝟔𝟔. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=

𝟓𝟖. 𝟖 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟎. 𝟕𝟕𝟐
𝟕𝟔. 𝟏𝟒𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝟕𝟎. 𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟎. 𝟗𝟐𝟐
𝟕𝟔. 𝟏𝟒𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
C= 1 x 10-7 mol cm-3 , 𝚲 = 𝟗𝟏. 𝟗𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=
𝟗𝟏. 𝟗𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏. 𝟐𝟎𝟖
𝟕𝟔. 𝟏𝟒𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
3. KNO3
From graph 𝚲 versus √𝑪, the value of 𝚲∞ = 𝟓𝟔. 𝟕𝟔𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

C= 5 x 10-5 mol cm-3 , 𝚲 = 𝟔𝟕. 𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=

C= 1 x 10-5 mol cm-3 , 𝚲 = 𝟔𝟐. 𝟐𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=

𝟔𝟕. 𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏. 𝟏𝟖𝟒
𝟓𝟔. 𝟕𝟔𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝟔𝟐. 𝟐𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏. 𝟎𝟗𝟕
𝟓𝟔. 𝟕𝟔𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
C= 5 x 10-6 mol cm-3 , 𝚲 = 𝟔𝟓. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝟔𝟓. 𝟓 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=
= 𝟏. 𝟏𝟓𝟒
𝟓𝟔. 𝟕𝟔𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

C= 1 x 10-6 mol cm-3 , 𝚲 = 𝟔𝟕. 𝟏 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝟔𝟕. 𝟏 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=
= 𝟏. 𝟏𝟖𝟐
𝟔𝟔. 𝟗𝟖𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏

C= 5 x 10-7 mol cm-3 , 𝚲 = 𝟔𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=

𝟔𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟏. 𝟎𝟗𝟐
𝟓𝟔. 𝟕𝟔𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
C= 1 x 10-7 mol cm-3 , 𝚲 = 𝟒𝟐. 𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
𝜸=
𝟒𝟐. 𝟕 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
= 𝟎. 𝟕𝟓𝟐
𝟓𝟔. 𝟕𝟔𝟐 𝐒 𝒄𝒎𝟐 𝒎𝒐𝒍−𝟏
Calculation for ionic strength (I)
𝑰=
𝟏
∑ 𝒄𝒊 𝒛𝒊 𝟐
𝟐
𝒊
Where, I
= ionic strength
𝑐𝑖
= the concentration of the ith ion concentration
𝑧𝑖
= charge of the ith ion
1. Na2SO4

C= 5 x 10-5 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐒𝐎𝟒 −𝟐 ]𝒁𝑺𝑶𝟒 𝟐 )
𝟐
𝟏
𝑰 = (𝟏𝒙 𝟏𝟎−𝟒 (𝟏)𝟐 + 𝟓 𝒙 𝟏𝟎−𝟓 (𝟐)𝟐 )
𝟐
𝑰=
𝑰 = 𝟏. 𝟓 𝐱 𝟏𝟎−𝟒
√𝑰 = 𝟎. 𝟎𝟏𝟐

C= 1 x 10-5 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐒𝐎𝟒 −𝟐 ]𝒁𝑺𝑶𝟒 𝟐 )
𝟐
𝟏
𝑰 = (𝟐𝒙 𝟏𝟎−𝟓 (𝟏)𝟐 + 𝟏 𝒙 𝟏𝟎−𝟓 (𝟐)𝟐 )
𝟐
𝑰=
𝑰 = 𝟑 𝐱 𝟏𝟎−𝟓
√𝑰 = 𝟎. 𝟎𝟎𝟓𝟒𝟖

C= 5 x 10-6 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐒𝐎𝟒 −𝟐 ]𝒁𝑺𝑶𝟒 𝟐 )
𝟐
𝟏
𝑰 = (𝟏𝒙 𝟏𝟎−𝟓 (𝟏)𝟐 + 𝟓 𝒙 𝟏𝟎−𝟔 (𝟐)𝟐 )
𝟐
𝑰=
𝑰 = 𝟏. 𝟓 𝐱 𝟏𝟎−𝟓
√𝑰 = 𝟎. 𝟎𝟎𝟑𝟖𝟕

C= 1 x 10-6 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐒𝐎𝟒 −𝟐 ]𝒁𝑺𝑶𝟒 𝟐 )
𝟐
𝟏
𝑰 = (𝟏𝒙 𝟏𝟎−𝟔 (𝟏)𝟐 + 𝟏 𝒙 𝟏𝟎−𝟔 (𝟐)𝟐 )
𝟐
𝑰=
𝑰 = 𝟐. 𝟓 𝐱 𝟏𝟎−𝟔
√𝑰 = 𝟎. 𝟎𝟎𝟏𝟓𝟖

C= 5 x 10-7 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐒𝐎𝟒 −𝟐 ]𝒁𝑺𝑶𝟒 𝟐 )
𝟐
𝟏
𝑰 = (𝟏𝒙 𝟏𝟎−𝟔 (𝟏)𝟐 + 𝟓 𝒙 𝟏𝟎−𝟕 (𝟐)𝟐 )
𝟐
𝑰=
𝑰 = 𝟏. 𝟓 𝐱 𝟏𝟎−𝟔
√𝑰 = 𝟎. 𝟎𝟎𝟏𝟐𝟐

C= 1 x 10-7 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐒𝐎𝟒 −𝟐 ]𝒁𝑺𝑶𝟒 𝟐 )
𝟐
𝟏
𝑰 = (𝟏𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 + 𝟏 𝒙 𝟏𝟎−𝟕 (𝟐)𝟐 )
𝟐
𝑰=
𝑰 = 𝟐. 𝟓 𝐱 𝟏𝟎−𝟕
√𝑰 = 𝟎. 𝟎𝟎𝟎𝟓
2. NaCl

C= 5 x 10-5 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐂𝐥− ]𝒁𝑪𝒍 𝟐 )
𝟐
𝟏
𝑰 = (𝟓 𝒙 𝟏𝟎−𝟓 (𝟏)𝟐 + 𝟓 𝒙 𝟏𝟎−𝟓 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟓 𝒙 𝟏𝟎−𝟓
√𝑰 = 𝟎. 𝟎𝟎𝟕𝟎𝟕

C= 1 x 10-5 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐂𝐥− ]𝒁𝑪𝒍 𝟐 )
𝟐
𝟏
𝑰 = (𝟏 𝒙 𝟏𝟎−𝟓 (𝟏)𝟐 + 𝟏 𝒙 𝟏𝟎−𝟓 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟏 𝒙 𝟏𝟎−𝟓
√𝑰 = 𝟎. 𝟎𝟎𝟑𝟏𝟔

C= 5 x 10-6 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐂𝐥− ]𝒁𝑪𝒍 𝟐 )
𝟐
𝟏
𝑰 = (𝟓 𝒙 𝟏𝟎−𝟔 (𝟏)𝟐 + 𝟓 𝒙 𝟏𝟎−𝟔 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟓 𝒙 𝟏𝟎−𝟔
√𝑰 = 𝟎. 𝟎𝟎𝟐𝟐𝟑

C= 1 x 10-6 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐂𝐥− ]𝒁𝑪𝒍 𝟐 )
𝟐
𝟏
𝑰 = (𝟏 𝒙 𝟏𝟎−𝟔 (𝟏)𝟐 + 𝟏 𝒙 𝟏𝟎−𝟔 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟏 𝒙 𝟏𝟎−𝟔
√𝑰 = 𝟎. 𝟎𝟎𝟏

C= 5 x 10-7 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐂𝐥− ]𝒁𝑪𝒍 𝟐 )
𝟐
𝟏
𝑰 = (𝟓 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 + 𝟓 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟓 𝒙 𝟏𝟎−𝟕
√𝑰 = 𝟎. 𝟎𝟎𝟎𝟕𝟎𝟕

C= 1 x 10-7 mol cm-3
𝟏
([𝐍𝐚+ ]𝒁𝑵𝒂 𝟐 + [𝐂𝐥− ]𝒁𝑪𝒍 𝟐 )
𝟐
𝟏
𝑰 = (𝟏 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 + 𝟏 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟏 𝒙 𝟏𝟎−𝟕
√𝑰 = 𝟎. 𝟎𝟎𝟎𝟑𝟏𝟔
3. KNO3

C= 5 x 10-5 mol cm-3
𝑰=
𝟏
([𝐊 + ]𝒁𝑲 𝟐 + [𝐍𝐎𝟑 − ]𝒁𝑵𝑶𝟑 𝟐 )
𝟐
𝑰=
𝟏
(𝟓 𝒙 𝟏𝟎−𝟓 (𝟏)𝟐 + 𝟓 𝒙 𝟏𝟎−𝟓 (𝟏)𝟐 )
𝟐
𝑰 = 𝟓 𝒙 𝟏𝟎−𝟓
√𝑰 = 𝟎. 𝟎𝟎𝟕𝟎𝟕

C= 1 x 10-5 mol cm-3
𝟏
([𝐊 + ]𝒁𝑲 𝟐 + [𝐍𝐎𝟑 − ]𝒁𝑵𝑶𝟑 𝟐 )
𝟐
𝟏
𝑰 = (𝟏 𝒙 𝟏𝟎−𝟓 (𝟏)𝟐 + 𝟏 𝒙 𝟏𝟎−𝟓 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟏 𝒙 𝟏𝟎−𝟓
√𝑰 = 𝟎. 𝟎𝟎𝟑𝟏𝟔

C= 5 x 10-6 mol cm-3
𝟏
([𝐊 + ]𝒁𝑲 𝟐 + [𝐍𝐎𝟑 − ]𝒁𝑵𝑶𝟑 𝟐 )
𝟐
𝟏
𝑰 = (𝟓 𝒙 𝟏𝟎−𝟔 (𝟏)𝟐 + 𝟓 𝒙 𝟏𝟎−𝟔 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟓 𝒙 𝟏𝟎−𝟔
√𝑰 = 𝟎. 𝟎𝟎𝟐𝟐𝟑

C= 1 x 10-6 mol cm-3
𝟏
([𝐊 + ]𝒁𝑲 𝟐 + [𝐍𝐎𝟑 − ]𝒁𝑵𝑶𝟑 𝟐 )
𝟐
𝟏
𝑰 = (𝟏 𝒙 𝟏𝟎−𝟔 (𝟏)𝟐 + 𝟏 𝒙 𝟏𝟎−𝟔 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟏 𝒙 𝟏𝟎−𝟔
√𝑰 = 𝟎. 𝟎𝟎𝟏

C= 5 x 10-7 mol cm-3
𝟏
([𝐊 + ]𝒁𝑲 𝟐 + [𝐍𝐎𝟑 − ]𝒁𝑵𝑶𝟑 𝟐 )
𝟐
𝟏
𝑰 = (𝟓 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 + 𝟓 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟓 𝒙 𝟏𝟎−𝟕
√𝑰 = 𝟎. 𝟎𝟎𝟎𝟕𝟎𝟕

C= 1 x 10-7 mol cm-3
𝟏
([𝐊 + ]𝒁𝑲 𝟐 + [𝐍𝐎𝟑 − ]𝒁𝑵𝑶𝟑 𝟐 )
𝟐
𝟏
𝑰 = (𝟏 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 + 𝟏 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟏 𝒙 𝟏𝟎−𝟕
√𝑰 = 𝟎. 𝟎𝟎𝟎𝟑𝟏𝟔

C= 5 x 10-7 mol cm-3
𝟏
([𝐊 + ]𝒁𝑲 𝟐 + [𝐂𝐥− ]𝒁𝑪𝒍 𝟐 )
𝟐
𝟏
𝑰 = (𝟓 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 + 𝟓 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟓 𝒙 𝟏𝟎−𝟕
√𝑰 = 𝟎. 𝟎𝟎𝟎𝟕𝟎𝟕

C= 1 x 10-7 mol cm-3
𝟏
([𝐊 + ]𝒁𝑲 𝟐 + [𝐂𝐥− ]𝒁𝑪𝒍 𝟐 )
𝟐
𝟏
𝑰 = (𝟏 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 + 𝟏 𝒙 𝟏𝟎−𝟕 (𝟏)𝟐 )
𝟐
𝑰=
𝑰 = 𝟏 𝒙 𝟏𝟎−𝟕
√𝑰 = 𝟎. 𝟎𝟎𝟎𝟑𝟏𝟔
Data Processing
1. Plot conductivity versus concentration.
Na2SO4
Graph Conductivity vs Concentration
10000000
Conductivity (μS/cm)

y = 2E+08x - 555015
R² = 0,9624
8000000
6000000
4000000
Ряд1
2000000
Линейная (Ряд1)
0
-2000000
0
0,02
0,04
Concentration (M)
0,06

NaCl
Graph Conductivity vs Concentration
6000000
y = 1E+08x - 345836
R² = 0,9624
Conductivity (μS/cm)
5000000
4000000
3000000
Ряд1
2000000
Линейная (Ряд1)
1000000
0
-1000000 0
0,02
0,04
0,06
Concentration (M)

KNO3
Conductivity (μS/cm)
Graph Conductivity vs Concentration
8000000
7000000
6000000
5000000
4000000
3000000
2000000
1000000
0
-1000000 0
y = 1E+08x - 417299
R² = 0,9624
Ряд1
Линейная (Ряд1)
0,02
0,04
Concentration (M)
0,06
2. Plot molar conductivity (Λm) versus √𝐶 for each salt solution.
Na2SO4
Graph Molar Conductivity (Λm) versus √𝐶
of Na2SO4
400
y = -14667x + 267,93
R² = 0,4829
300
Λm

200
Ряд1
100
Линейная (Ряд1)
0
0
0,002
0,004
√𝐶
0,006
0,008

NaCl
Graph Molar Conductivity (Λm) versus √𝐶
of NaCl
200
y = -7275,6x + 152,29
R² = 0,4835
Λm
150
100
Ряд1
50
Линейная (Ряд1)
0
0
0,002
0,004
0,006
0,008
√𝐶

KNO3
Graph Molar Conductivity (Λm) versus √𝐶
of KNO3
150
y = 3612,5x + 113,52
R² = 0,2379
Λm
100
Ряд1
50
Линейная (Ряд1)
0
0
0,002
0,004
0,006
0,008
√𝐶
3. Compare the molar conductivity at infinite dilution for every solution and
give your comments.
In this case, the molar conductivity at infinite dillution is determined from graph
based on number two. The value of molar conductivity at infinite dilution is the
intercept of graph. The value of molar conductivity at infinite dilution is the
intercept of graph. The molar conductivity at infinite dilution of Na2SO4, NaCl, and
KNO3 are 267.93 S cm2 mol−1 , 152.29 S cm2 mol−1, and 113.52 S cm2 mol−1 .
Those solutions are categorized as strong electrolytes which will undergo
complete ionization and therefore they have higher conductivity. On dilution there
is a regular increase in the molar conductivity of strong electrolyte, due to the
decrease in solute-solute interaction. If we consider the value for each solutions
are different. The molar conductivity at infinite dilution of Na2SO4 is greater than
NaCl, and KNO3. It can be due Na2SO4 has to the number of moles more than
NaCl, and KNO3. While the molar conductivity at infinite dilution should be similar
because they have same number of moles. So, it can be consider that there is
error in experiment. It can be caused by preparation solutions error.
4. Compare the experimental molar conductivity value with reference.
If compare the experimental molar conductivity value with reference there will a
differences. The values anre not same. But it exactly follows same as the formula
equation of molar conductivity. In molar conductivity, the conductivity is inversely
proportional to the concentration. The higher the concentration will give lower
molar conductivity.
5. Determine the Λ for each solution.
𝚲=
√𝐶 ((mol cm )
-3
-3
𝐶 (mol cm )
𝜿
𝒏𝒆 𝑪
(S cm2 mol-1)
Na2SO4
NaCl
KNO3
5 x 10-5
7.071 x 10-3
44.7
55.7
67.2
1 x 10-5
3.162 x 10-3
51.75
58.8
62.25
5 x 10-6
2.236 x 10-3
55.5
60.8
65.5
1 x 10-6
1 x 10-3
55.3
66.5
67.1
-7
-4
5 x 10
7.071 x 10
58
70.2
62
1 x 10-7
3.162 x 10-4
83.5
91.95
42.7
6. Plot Λ vs √𝐶.
 Na2SO4
Graph Λ versus √𝐶 of Na2SO4
100
80
y = -3666,9x + 66,982
R² = 0,4829
Λ
60
Ряд1
40
Линейная (Ряд1)
20
0
0
0,002
0,004
√𝐶
0,006
0,008
 NaCl
Graph Λ versus √𝐶 of NaCl
100
80
y = -3637,8x + 76,145
R² = 0,4835
Λ
60
Ряд1
40
Линейная (Ряд1)
20
0
0
0,002
0,004
0,006
0,008
√𝐶
 KNO3
Λ
Graph Λ versus √𝐶 of KNO3
80
70
60
50
40
30
20
10
0
y = 1806,2x + 56,762
R² = 0,2379
Ряд1
Линейная (Ряд1)
0
0,002
0,004
0,006
0,008
√𝐶
7. Determine the activity coefficient (γ) for all concentrations.
γ
𝐶 (mol
-3
log γ
cm )
Na2SO4
NaCl
KNO3
Na2SO4
NaCl
KNO3
5 x 10-5
0.667
0.731
1.184
-0,176
-0,136
0,073
1 x 10-5
0.773
0.772
1.097
-0,112
-0,112
0,040
5 x 10-6
0.829
0.798
1.154
-0,081
-0,098
0,062
1 x 10-6
0.826
0.873
1.182
-0,083
-0,059
0,073
5 x 10-7
0.866
0.922
1.092
-0,062
-0,035
0,038
1 x 10-7
1.257
1.208
0.752
0,099
0,082
-0,123
8. Plot log γ vs (√𝐼) (I = ionic strength) and give your comments.
 Na2SO
Graph ln γ vs √𝐼 for Na2SO4
y = -37,311x - 0,0059
R² = 0,5749
0,4
ln γ
0,2
0
-0,2 0
0,005
0,01
Ряд1
0,015
Линейная (Ряд1)
-0,4
-0,6
√𝐼
 NaCl
Graph ln γ vs √𝐼 for NaCl
y = -53,035x - 0,0096
R² = 0,5444
0,4
ln γ
0,2
0
-0,2 0
0,002
0,004
0,006
Ряд1
0,008
Линейная (Ряд1)
-0,4
-0,6
√𝐼
 KNO3
Graph ln γ vs √𝐼 for KNO3
0,4
y = 32,978x - 0,0171
R² = 0,2274
Ряд1
ln γ
0,2
0
-0,2
0
-0,4
0,002
0,004
0,006
0,008
Линейная (Ряд1)
√𝐼
Comment:
Graph log γ vs √𝐼 actually follows Deybe Huckle Limiting Law, that is:
𝟏⁄
𝟐
𝒍𝒐𝒈 𝜸 ± = −|𝒛+ 𝒛− |𝑨𝑰
Where:
A = 0.509 / (mol kg-1)1/2 at 298 K
I
= ionic strength
z+ = charge number of cation
z- = charge number of cation
From that equation the the value of log γ should increase in direct proportion to √𝐼 and
give straight line in graph. But the graphs we got are not linear so there are erors that
caused this mistake. The error can be caused by the preparation solution during experiment
such as weighing error, conductivity meter and cell is contamined with other solution when
measure a certain solution.
DISCUSSIONS
Conductivity in aqueous solutions, is a measure of the ability of water to conduct an
electric current. The more ions there are in the solution, the higher its conductivity. Also the
more ions there are in solution, the stronger the electrolyte. The conductivity of an electrolyte
solution depends on concentration of the ionic species and behaves differently for strong
and weak electrolytes.
There is relationship between conductivity and concentration. Conductivity should
increase in direct proportion to concentration. This is same as the data we got from the
experiment. From the graph, it shows its relationship even the graph its not really linear. The
concentration and mobility of the ions are not independent properties. As the concentration
of an ion increases, its mobility decreases. If conductivity were proportional to the
concentration of the electrolyte, the molar conductivity of an electrolyte would be
independent of concentration. Because solutions used in this experiment are strong
electrolyte, so characteristic of a strong electrolyte is that its molar conductivity depends
only slightly on the molar concentration.
From data and calculation, we got vary of molar conductivity for each solutions in
every concentration. Its because the variation of concentration in each solutions. The other
reasons are the number of ions in the solution might not be proportional to the concentration
of the electrolyte and the ions interact strongly with one another, the conductivity of a
solution is not exactly proportional to the number of ions present.
From molar conductivity and square root of concentration can be determined the
value of molar conductivity at infinite dilution by plotting graph. The value of molar
conductivity at infinite dilution is the intercept of graph. The molar conductivity at infinite
dilution of Na2SO4, NaCl, and KNO3 are 267.93 S cm2 mol−1 , 152.29 S cm2 mol−1 , and
113.52 S cm2 mol−1 .
Those solutions are categorized as strong electrolytes which will undergo complete
ionization and therefore they have higher conductivity. On dilution there is a regular increase
in the molar conductivity of strong electrolyte, due to the decrease in solute-solute
interaction. If we consider the value for each solutions are different due to the errors happen
during experimental. If compare the experimental molar conductivity value with reference
there will a differences. The values anre not same. But it exactly follows same as the formula
equation of molar conductivity. In molar conductivity, the conductivity is inversely
proportional to the concentration. The
higher the concentration will give lower molar
conductivity.
In aqueous solutions, with increasing concentration of the solution, the specific
conductance of the electrolytes first increases, reaching a certain maximum and then falls
with further increase of concentration. If the dependent of conductivity towards concentration
were to be studied, we would observe that the conductivity
increases with solution
concentrations due to the increase in the number of ions. However the relationship is not
linear. So the equivalent conductivity approaches to a value called the molar conductivity at
infinite dilution (𝚲∞ ) as the electrolyte becomes more dilute.For a strong electrolyte, the
equivalent conductivity of a dilute solution is given by the Debye-Huckle and Onsanger
equation
𝚲 = 𝚲∞ − (𝐀 + 𝐁𝚲∞ ) √C
Where; A and B = constant
The value of Λ ∞ can be determined by plotting equivalent conductivity (Λ ) versus √𝐶. The
intercept of yhis graph is the velue of Λ ∞ . After these value determined, so it can be used to
determine the activity coefficient (γ) which from the ratio between equivalent conductivity (Λ)
at specific concentration and the conductivity at in infinite dilution (Λ ∞ ). The number of
activity coefficient (γ) for each solution in every concentrations. Each solution have different
mol of ion so there will a different ion mobility and electrophoretic effect that can be reasons
of this differences.
Then the equation used to plot groph between log γ and √𝐼 (ionic strength). Both log γ
and √𝐼 follow Debye Huckle Limiting Law, which the value of log γ should increase in direct
proportion to √𝐼 and give straight line in graph. But the graphs we got are not linear so there
are erors that caused this mistake.
The results we got are far from perfection because there are erros happen during
practical. The error can be caused by the preparation solution during experiment such as
weighing error, conductivity meter and cell is contamined with other solution when measure
a certain solution.
CONCLUSION
From data, calculation, and discussion we can conclude that the relationship
between conductivity and concentration is conductivity should increase in direct proportion to
concentration. The conductivity at infinite dilution can be determine by plotting graph
equivalent conductivity versus √𝐶. The value of molar conductivity at infinite dilution is the
intercept of graph. The molar conductivity at infinite dilution of Na2SO4, NaCl, and KNO3 are
267.93 S cm2 mol−1 , 152.29 S cm2 mol−1 , and 113.52 S cm2 mol−1 . While the activity
coefficients are:
γ
𝐶 (mol
cm-3)
Na2SO4
NaCl
KNO3
5 x 10-5
0.667
0.731
1.184
1 x 10-5
0.773
0.772
1.097
5 x 10-6
0.829
0.798
1.154
-6
1 x 10
0.826
0.873
1.182
5 x 10-7
0.866
0.922
1.092
1 x 10-7
1.257
1.208
0.752
REFERENCES
1) Siyavula, “Electrolytes, Ionisation And Conductivity”,
https://www.siyavula.com/read/science/grade-10/reactions-in-aqueous-solution/18reactions-in-aqueous-solution-03
2) Jack Brubaker, “Conductivity Vs. Concentration”, https://sciencing.com/conductivityvs-concentration-6603418.html
3) Jeloranta, “Conductivity of electrolyte solutions”,
https://www.csun.edu/~jeloranta/CHEM351L/experiment4.pdf
4) Totorost, “Determination of the conductance of strong and weak
electrolyte“,https://www.scribd.com/doc/16348982/Determination-of-theconductance-of-strong-and-weak-electrolyte