Solubility Lab Report: Calcium Hydroxide and Precipitation Analysis

This lab report investigates the solubility of calcium hydroxide (lime) in water, focusing on the formation of precipitates through dilution and double decomposition reactions. The experiment involves serial dilutions of calcium nitrate solution and the subsequent addition of NaOH to observe precipitation patterns. The report details the materials used, experimental methodology, and calculations of the solubility product constant (Ksp). The results show a decrease in precipitate formation with increasing dilution, demonstrating the impact of concentration on solubility. The report also calculates the Ksp of calcium hydroxide and compares it to literature values. The discussion section analyzes the results and explains the chemical reactions involved, concluding that the concentration of a product or a solution continues to decrease with dilution with water. The report includes a bibliography of relevant sources.

SOLUBILITY 1
Laboratory Report
Name
Name of the Class
Instructor
Institution
City and State
Date

SOLUBILITY 2
Abstract
This report uses five drops of deionised water and 5 drops of calcium hydroxide solution
from the proceeding wells (1-12) to determine its solubility and the formation of a precipitate
(an insoluble salt).The results show that a precipitate is formed in the first wells as the
concentration continues to decrease moving from left to right due to dilution of the solution.
Ultimately, it can be concluded that the concentration of a product or a solution continues to
decrease with dilution with water.
Introduction
Calcium hydroxide or lime is solid that is partially soluble in water. It dissolves in water to
give calcium hydroxide solution that dissociates to form Ca and OH ions (Rojas 2014, p.255).
When dissolved in water, Ca(OH)2 reaches a saturation point or equilibrium. However, the
equilibrium constant is referred to as solubility constant and is expressed by Ksp (Bullard &
Flatt, 2010). Therefore the Ksp of dissolving lime is water is = [Ca2+ (aq)][OH- (aq)]2.
The rationale of this report is to calculate the solubility constant of lime when dissolved in
water and when mixed with a soluble salt.
Materials used
 Dropper
 Well plates, 12
 Deionised water
 0.1M Ca(No3)2
 0.1M NaOH
 Dust coat & Eyeglasses for protection
Methodology

SOLUBILITY 3
Step 1
1. For the first experiment, 5 drops of calcium nitrate solution were put in well 1, and 5
drops of deionised water each in the reaming wells
2. 5 drops of the solution were added in well 2 and the solution mixed with a dropper.
3. A dropper was used to add 5 drops of the mixture in well 2 to well 3 and the mixture
mixed. Continue the serial dilution until to the last well. The concentration was then
calculated.
Step 2
1. 5 drops of 0.1M NaOH were put in the 12 wells. An empty dropper was used to
mix each of the solutions
Step 3
1. 3-4 minutes were allowed for the precipitates to form and the patterns of the
precipitation observed.
2. The concentration of ions in the well was calculated and the solution used to
determine the solubility product of lime.
Results
2. Lime or calcium hydroxide dissolves in water as shown as shown below
Ca(OH) 2 (s) ⇌ Ca2+ (aq) + 2OH- (aq)
3. Solubility product constant expression of calcium hydroxide
Ksp = [Ca2+ (aq)] [OH- (aq)]2
4. Calculate Ksp , we need molarity of OH- and Ca2+
Since the molarity of Ca(OH)2 is 0.05M, we can use the mole ratio to find the molarity of
combining ions
Mole ratio Ca(OH)2 : Ca2+ is 1:1 and

SOLUBILITY 4
Ca(OH)2: OH- is 1:2
Molarity of OH- ; (0.05M*2) = 0.1M
Molarity of Ca2+ = 0.05M
Ksp of Ca(OH)2 = (0.05)(0.1)2
=0.0005 or 5.0x10 -4
The well number is 2.
5. The reaction taking place here is double decomposition or precipitation, adding the two
salts, would give a precipitate and a soluble salt (NaNo3)
Equation is
Ca(OH)2 (aq) +NaOH (aq) Ca(OH)2 (s) +NaNo3(aq)……..Not balanced
The spectator ions are Na+ and NO3, so the remaining ions will form
Ca2+ (aq) + 2OH- (aq) Ca(OH)2 (s)
Since the molarity of NaoH is 0.1M
We can use the mole ratio to get molarity of Ca ions of 1: 2
0.1
2 = 0.05M
Therefore, Ksp = [Ca2+ (aq)] [OH- (aq)]2
(0.05)(0.1)2 = 5.0 x104; the well number is 2
6. Molar solubility of calcium hydroxide based on literature Ksp value
The actual Ksp for Ca(OH)2 is 5.5x10-6
Who equation is Ca(OH)2(s) ⇌Ca2+(aq) +2OH- (aq)
Ksp = = [Ca2+ (aq)] [OH- (aq)]2, where we can let the ionic values take the form t
Hence Ksp = t x (2t)2 = 4t3 = 5.5 x10-6
T3 = 1.375 x 10-6 ; t = 3
√ (1.375 x 10−6) = 1.1 x 10-2
Solubility = 0.011M
Discussion

SOLUBILITY 5
As it can be noted from the experiments, the concentration of lime continued to decrease
from well 1 to well 12. This is due to the dilution of the solution with water in the preceding
wells. Therefore, it is clearly evident that, there was little or no formation of a precipitate in
the last wells than in the first three wells. Ultimately, this experiments shows that salts
(solutions) can be formed in different methods such as reacting a salt with water or through
double decomposition, where two soluble salts react together to form a precipitate and a
soluble salt, as shown in experiment 2.

SOLUBILITY 6
Bibliography List
Abidi, S.S.A. and Murtaza, Q., 2014. Synthesis and characterization of nano-hydroxyapatite
powder using wet chemical precipitation reaction. Journal of Materials Science &
Technology, 30(4), pp.307-310.
Bullard, J.W. and Flatt, R.J., 2010. New insights into the effect of calcium hydroxide
precipitation on the kinetics of tricalcium silicate hydration. Journal of the American
Ceramic Society, 93(7), pp.1894-1903.
Chelazzi, D., Poggi, G., Jaidar, Y., Toccafondi, N., Giorgi, R. and Baglioni, P., 2013.
Hydroxide nanoparticles for cultural heritage: consolidation and protection of wall paintings
and carbonate materials. Journal of colloid and interface science, 392, pp.42-49.
Rojas, R., 2014. Copper, lead and cadmium removal by Ca Al layered double
hydroxides. Applied Clay Science, 87, pp.254-259.