1. The Application of
Common Ion to Purify
Hard Water
Chemistry- Experiment
Rolla Tyas Amalia
Grade 11 Science

2. Aim: To vanish the hardness of natural water.
Materials:
1. 4 Pippets 11. 2grm Ca(OH)2
2. 8 tube reactions 12. NaOH 0,025 M and 0,1, 50 ml
3. 8 conical flasks 13. 25 ml Hcl, 0,1 M
4. A funnel 14. Distilled water
5. A stative
6. A tube rack
7. Paper Filter
8. A beaker glass 250 ml
9. A gratitude glass
10. A burette
Procedures:
1. Prepare 4 conical flasks, put 2 grm of Ca(OH)2 powder into each flask.
2. Label the flasks with number 1-4.
3. Add 50 ml water into flask 1; 50 ml NaOH 0,025 M into flask 2; 50 ml
NaOH 0,05 M into flask 3; and 50 ml NaOH 0,1 M into flask 4.
4. Plug flask in, then shake and let it for one night.
5. After one night, separate the solutions at each flask so you can get the
filtrate.
6. Put 10 ml of filtrate 1 into conical flask and add 3 drops of methyl
indicator. Titrate the solution with HCL 0,1 M until there is change in
solution color! Make a note on the volume of used HCL.
7. Determine the concentration of OH- in the filtrate using equation V1 M1 =
V2 M2
8. Record the observation in a table.
9. Do a calculation and record the data in the following table:
Saturated Used (OH-) in (OH-) in (Ca2+) in Ksp= (Ca2+)
solution of volume of solution Ca(OH)2 solution (OH-)
Ca(OH)2 HCL 0.1 M
8.7 ml 0.087 M 0.174 M 0.087 M 2.6 x 10^-3
In water
13.2 ml 0.132 M 0.264 M 0.132 M 9.1 x 10^-3
In NaOH 0.025 M
7.3 ml 0.073 M 0.146 M 0.073 M 7.78 x 10^-3
In NaOH 0.05 M
15.3 ml 0.153 M 0.292 M 0.153 M 0.024
In NaOH 0.1 M

3. Evaluating Data
From the data of the experiment, constanta of solubility of Ca(OH)2 in
water is smaller than the constanta of solubility of Ca(OH)2 in NaOH with 0.025
M and 0.05 M. This means that Ca(OH)2 is more soluble in water than in NaOH
with 0.025 M and 0.05 M.
The effect of NaOH solution in the solubility of Ca(OH)2 is reducing the
solubility of insoluble substances, that we known as the common ion effect. We
could see that the solubility of Ca(OH)2 in NaOH is decreasing as the
concentration of NaOH is increasing.
My conclusion is Ca(OH)2 is more soluble in water than NaOH. The NaOH
with bigger concentration will give the common ion effect to the solubility of
Ca(OH)2 in NaOH. The bigger the concentration of NaOH, the smaller the
solubility of the Ca(OH)2, which make the Ca(OH)2 can be more soluble.
Communicating
The effect of common ion to the solubility is reducing the solubility of the
insoluble substance, so that it might be soluble. The effect of common ion to the
value of Ksp is decreasing the value of the Ksp. From the data of the experiment,
the Ksp of Ca(OH)2 in NaOH with 0.1 M is smaller than the Ksp of Ca(OH)2 in
water. This means that salt is more soluble in a solution with common ion than
in water.