The Theory of Conductivity in Solutions
In our study of ions and ionic compounds, we also learned that metals are excellent
conductors of electricity because they contain atoms with loosely held valence
electrons. All conductors contain movable electric charges which move when an
electric potential difference (measured in volts) is applied across a piece of the
conductor (i.e. metal wire). This flow of charge is known as electric current. In most
conductors, the amount of current is proportional to the applied voltage:
V = I R Ohm’s Law
V (voltage, volts); I (current, amperes); R (resistance, ohms)
The ratio between voltage and current is called the resistance (measured in ohms). The
resistance across a standard mass of a material at a given temperature is called the
resistivity of the material. The inverse of resistance and resistivity is conductance and
conductivity.
In the conductivity lab, you measured the change in conductivity as you varied the
number of ions in an electrically conductive solution (distilled water plus dissolved salt),
also known as an electrolyte. An electrolyte is a substance containing free ions that
behaves as an electrically conductive medium. Because they generally consist of ions
in solution, electrolytes are also known as ionic solutions. The conductivity meter you
used to measured conductivity in microsiemens/cm units.
To better understand your data, consider the Theory of Conductivity in Solutions. The
theory states that conductivity is related to ion concentration by the formula:
     kc
1
2
Kohlrausch’s Square Root Law
This formula says that for strong electrolytes with sufficient dilution, the conductivity
 of an electrolyte with concentration c varies with the square root of the molar
concentration. “  ” is the upper case Greek letter lambda (don’t let this throw you, it is
just a variable). Molar means “per mole of substance” (remember the definition of
mole?). Also,
   the value at infinite dilution (distilled water) (published value ≈.000128 μS/(cm)
You still need to solve for “k” and will do that later. But first, there are a few things you
must do to your data to get it in the form above:
1. Calculate concentrations in molarity for each measurement
2. Check to see if your distilled water measurements are close to accepted values
Concentration
Concentration is measured in moles of solute (in our case, ions) per liter of solution
(molarity). If you recall, a mole is a means of counting atoms, ions or molecules
(particles) in a substance:
1 mole = 6.02 x 1023 particles (Avogadro’s number) (also abbreviated mol)
The molar atomic mass of table salt NaCl is the sum of the atomic masses of one mole
of Na atoms (22.99 g/mol) and one mole of Cl atoms (35.45 g/mol):
molar atomic mass of NaCl: 22.99 g + 35.45 g = 58.44 g (1 mol of NaCl weighs 58.44 g)
A 1.0 M NaCl salt solution contains one mole of NaCl dissolved in one liter of distilled
water. You used two different NaCl concentrations (1.0 M and 0.5 M) for this experiment.
To calculate the concentration of dissolved NaCl in distilled water for each of your
measurements, you need to find the amount of NaCl in each drop of water. There are
about 140 drops in 5ml of water, or:
5.000ml
ml
 .0357
140. drops
drop
The amount of NaCl in one liter of 1.0 M salt solution (remember 1000 ml = 1 liter) is:
1.0M 
1 mole of NaCl
58.44g
58.44g
X

1000ml
1 mole of NaCl 1000ml
The calculate the amount of NaCl in each drop of 1.0 M NaCl solution, just multiply the
above by the volume of one drop:
.0357
ml 58.44g
g
x
 0.002087
drop 1000ml
drop
We must convert this mass back to moles. Remember that there are 58.44 g in one
mole of NaCl, or:
1 mole
1
58.44g
0.002087
g
1 mole
moles
x
 3.571x 10 -5
drop 58.44g
drop
This value of moles/drop can be used to calculate the concentration as we add drops.
Starting with 50 ml of distilled water, the concentration is:
Drops of 1.0M
NaCl solution
0
1
2
3
‫׃‬
10
Moles of NaCl
(mol)
0
3.571 x 10-5
7.141 x 10-5
10.711 x 10-5
‫׃‬
35.71 x 10-5
Additional
liquid added
0
0.0357 ml
0.0714 ml
0.1071 ml
‫׃‬
0.357 ml
Total Liquid
(liter)
0.050 L
0.0500357 L
0.0500714 L
0.050171 L
‫׃‬
0.050357 L
Concentration =
moles/total liquid (c)
0M
7.135 x 10-4 M
1.426 x 10-3 M
2.135 x 10-3 M
‫׃‬
7.091 x 10-3 M
You need to do the calculations for drops 4-9, and for drops of the 0.5 M solution.
We are still not done!
Conductivity
Conductivity is also measured in microseimens/cm = μsei/cm
The equation
Drops of 1.0M
NaCl solution
0
1
2
3
‫׃‬
10
   0  kc
“Λ”
Conductivity
(μS/cm)
0
:
:
:
:
≈205
1
2
Concentration =
Moles/Total Liquid
(c = mol/L)
0M
7.135 x 10-4 M
1.426 x 10-3 M
2.135 x 10-3 M
:
7.091 x 10-3 M
Graphing
Now graph the concentration on the x-axis and “Λ” on the y-axis. Use the scatter plot
on MS-EXCEL. The data is represented by the blue dots. This is where you would end
your lab report. A discussion of the red line follows below.
S = 3.56465373
r = 0.95387658
12
50.
Conductivity
29
43.
45
36.
62
29.
79
22.
95
15.
2
9.1 0.1
1.4
2.6
3.9
5.2
6.5
7.7
Concentration
Checking to see if your data fits the theory
We said earlier that we did not know the value of the “k” in the equation:
1
2
     kc , where    0.000128 (μS/cm) for NaCl from published data.
Using a computer program to model the collected data the red line above is obtained.
This red line gives the following equation:
  307.  .271c
1
20
which is not very close to the model. This is the best fit that could be found.
Leaving the power of the concentration to
1
1
rather than the
obtained from the
20
2
model, the following graph is obtained which doesn’t look much different:
S = 4.83561055
r = 0.90026095
12
50.
Y Axis (units)
29
43.
45
36.
62
29.
79
22.
95
15.
2
9.1 0.1
1.4
2.6
3.9
X Axis (units)
The equation of the red line is:
  50.5  .0161c
1
2
5.2
6.5
7.7