Determining the Solubility Product Constant for Calcium Sulfate:
The Effect of Ionic Strengths of Electrolyte Solutions
David Liu
Chemistry 102
May 19, 2015

Abstract
The method of this experiment was done by filtering the solids of calcium sulfate out
from its diluted solution. Once the solid calcium sulfate has been filtered out of the solution,
ammonia buffer at pH of 10 is added into the remaining solution, also adding approximately 2
drops of eriochrome black 1 indicator to determine the endpoint. The final part of the preparation
consists of preparing a burette filled with EDTA to titrate into the solution until it turns from red
to blue. Once data has been collected for three trials, three analyses in different methodologies
will be performed, to calculate the solubility product constant of calcium sulfate. The different
methods used for calculation is using molarities at room temperature, activities at room
temperature, and Gibbs Free Energy. By employing the method molarity yielded a result of
6.22x10
-5
M, 1.17x10
-5
M, and 4.48x10
-5
M respectively. Further by using the activities of the
ions, the calculations resulted in 1.19x10
-5
M, 1.12x10
-5
M, and 1.20x10
-5
M. Lastly the method
of using Gibbs free energy were used and determined to be 4.48x10
-5
. These numbers are then
compared to the literature value for the solubility product constant (K
sp
) of calcium sulfate,
determined to be at 7.10x10
-5
M.
Introduction
Electrical conductivity in solutions has played a major role in the modern day era, from
any industry using electricity to every residence living on electricity. Understanding the method
and strength of how electricity is conducted, also known as electrolytes in solutions is vital for
these applications. When a solid dissolves into water, the solid dissociates into ions, which would
then allow electricity to be conducted through the solution, which are also known as electrolytes.
The equation to be used for calculating equilibrium constant is known to chemists as
K =
a
C
c
a
B
b
/
a
A
a
a
B
b
(1)
for any generalized chemical reaction
aA
(aq)
+ bB
(aq)
cC
(aq)
+ dD
(aq)
(2)

where
a
is the activity of a given chemical species involved in the equilibrium, and the activity of
the species is used as one of the method to calculate the solubility product constants, with
a
being calculated with the expression
a
=
f
M
(3)
This all relates to one of the method of calculating solubility constants, and the calculations for
obtaining
f
(0<f<1), the activity coefficient will be mentioned later in the section. However, if
were to allow
f
to be 1, then a more familiar equilibrium constant equation surfaces,
K = [C]
c
[D]
d
/[A]
a
[B]
b
(4)
Another common method to calculate equilibrium constant is to include Gibbs free energy
change, since it is a thermodynamic quantity, by using the expression
ΔG
o
= -RTlnK
(5)
where ΔG
o
is the change in Gibbs free energy, R being the constant 8.314J/mol·K, and T would