Solubility of Calcium Hydroxide – Titration Method
Prediction: Give the balanced chemical equation of calcium hydroxide with HCl (aq).
Ca(OH)2 (aq) + 2 HCl (aq)  CaCl2 (aq) + 2 H2O (l)
Find the Ksp of calcium hydroxide from the reference sheet and determine the molar
solubility of calcium hydroxide:
Ca(OH) 2  Ca 2
ICE
x
Ksp  [Ca 2 ][OH - ]2 
(x)(2x) 2 
4x 3 
x 


2 OH 2x
6.5 x 10-5
6.5 x 10-5
6.5 x 10-5
2.5 x 10-2
Observations: A student titrates saturated calcium hydroxide with 0.0525 M HCl (aq) and
acquires the following results when titrating 50.00 mL of calcium hydroxide with HCl
(aq) in a buret. Complete the table:
Final Volume of Buret with HCl (aq)
Initial Volume of Buret with HCl (aq)
Volume of HCl (aq)
21.65 mL
0.15 mL
23.05 mL
2.25 mL
22.80 mL
1.10 mL
21.50 mL
20.80 mL
21.70 mL
Average Volume of HCl (aq)
Colour of Phenolphalein
21.33
Purple - Pink
Light-Pink
Purple-Pink
Analysis: Using the average volume of HCl (aq) and the balanced equation above,
calculate the experimental concentration of the calcium hydroxide.
1 mol Ca(OH)2
1
[Ca(OH) 2 ]  (0.0525 M HCl x 0.02133 L) x
x
2 mol HCl
0.05000 L
-2
 1.12 x 10 M
Calculate the concentration of the [OH-] and [Ca2+] ions (1):

Ca(OH) 2  Ca 2  2 OH1.12 x 10-2
1.12 x 10-2 2(1.12 x 10-2 )  2.24 x 10-2
Calculate the Ksp value for calcium hydroxide given the experimental data:
Ksp = [Ca2+] [OH-]2
= 4(0.0112)3
= 5.62 x 10-6
What is the percentage error between the theoretical value for Ksp and the experimental
value?
5.62 x 10-6  6.5 x 10-6
% difference 
x 100  14 %
-6
6.5 x 10

Solubility of Calcium Hydroxide – Precipitation Method
Observations: A student pipets 25.00 mL of distilled water into a clean beaker and adds an amount of
calcium hydroxide according to the chart below. The excess calcium hydroxide is filtered from solution
and the data summarized in the chart.
Quantity
Mass of Weigh Boat with Ca(OH)2
Mass of Weigh Boat
Mass of Ca(OH)2
Mass of Filter Paper with Precipitate
Mass of Filter Paper
Mass of Precipitate
Mass (grams)
1.490 g
1.210 g
0.280 g
1.347 g
1.089 g
0.258 g
Calculations:
Calculate the mass of Ca(OH)2 in solution.
mCa(OH)2 (aq)  mCa(OH)2 (added) - mCa(OH)2 (ppt)
= 0.280 g - 0.258 g
= 0.022 g
Calculate the solubility of Ca(OH)2 at room temperature in grams per 100 mL.
mCa(OH) 2
x

Vsol' n
100 mL
0.022 g
x

25.00 mL 100 mL
0.022 g x 100 mL
x
25.00 mL
 0.088 g
solubility Ca(OH) 2  0.088 g/100 mL
Determine the experimental molar solubility of calcium hydroxide.
[Ca(OH) 2 ] 
0.088 g
mol
100 mL
x
x
100 mL 74.10 g
0.1L
= 1.2 x 10- 2 mol/L
Determine the experimental Ksp for calcium hydroxide.
Ca(OH) 2  Ca 2 
x
ICE
 2OH - ; Ksp  [Ca 2  ][OH- ]2
 2x
x 1.2 x 10 - 2 mol/L
Ksp  (x)(2x)2
 4x 3
 4(1.2 x 10- 2 )3
 6.9 x 10 -6
Determine the percentage error for the experiment.
% error 
6.9 x 10-6 - 6.5 x 10-6
 6.2 %
6.5 x 10-6
x 100