Measuring the Solubility Product of Ca OH 2

Topics: Chemistry, Solubility, Chlorine Pages: 3 (554 words) Published: April 9, 2015
Measuring the Solubility Product of Ca(OH)­2
Purpose: The purpose of this investigation is to find the solubility product (Ksp) of Ca(OH)2 by titrating the hydrochloric acid with calcium hydroxide and using their entities to find the concentration of Ca­2+ and OH- ions. Materials: Refer to lab sheet “Measuring the Solubility of Ca(OH)2” (handout) Method (Procedure): Refer to lab sheet “Measuring the Solubility of Ca(OH)2” (handout) Observations

Trial 1
Trial 2
Initial burette reading
0mL
17.75mL
Final burette reading
17.75mL
24.75mL
Volume of Ca(OH)2 used
17.75mL
7.25mL
** average volume from two trials is 21.125mL
Questions
1. Calculation of [OH-] from titration data (using average of all trials) C=0.0500M
V=0.01L
n= CxV
n= (0.0500M)(0.01L)
n= 0.0005mol

2HCl(aq) + Ca(OH)2(aq)  CaCl2(s) + H2O(l)

0.0005 mol of HCl X 1 mol of Ca(OH)2 = 0.00025 mol of Ca(OH)2 2 mol of HCl
Therefore, nCa(OH)2 = 0.00025 mol
Ca(OH)2(aq)  Ca2+ + 2OH-

0.00025mol Ca(OH)2 X 2 mol OH- = 0.0005 mol
1 mol Ca(OH)2
Concentration of OH- :
n= 0.0005 mol
V= 0.02125L
C= n/V [OH-] = 2.35X10-2M = 0.0005mol/0.02125L
= 2.35X10-2M

2. Calculating the [Ca2+] from the balanced equation.

Since:
nCa(OH)2 = 0.00025mols
Ca(OH)2(aq)  Ca2+ + 2OH-

0.00025mol Ca(OH)2 X 1 mol Ca2+ = 0.00025 mol
1 mol Ca(OH)2
Concentration of Ca2+ :
n= 0.00025 mol
V= 0.02125L
C= n/V [Ca2+] = 1.18X10-2M = 0.00025mol/0.02125L
= 1.18X10-2M

3. Using the Ksp expression to calculate the value of Ksp
According to the balanced equation: Ca(OH)2(aq)  Ca2+ + 2OH- Ksp = [Ca2+][OH-]2
= (1.18X10-2)( 2.37X10-2)2
= 6.51X10-6
Therefore, the experimental value of Ksp (solubility product) of Ca(OH)2 is 6.51X10-6.

Experimental error
The...
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