Determination of the Solubility Product Constant of Calcium Hydroxide

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Determination of the Solubility Product Constant of Calcium Hydroxide

Introduction

The equilibrium constant for the solubility equilibrium between an ionic solid and its ions is called solubility constant [1] , Ksp of the solute. For example, the solubility product is defined by

MxAy(s) ⇋xM(aq)y++ yA(aq)x- (1)
Where M is the metal cation, A is the anion, x and y are the corresponding charges of the ions. The equilibrium expression is
Ksp=[MY+]x[AX-]Y (2)

In the example, MxAy(s) does not appear in the equilibrium constant expression since its activity is 1 and does not appear in the equation.
The expression for the reaction quotient of
the solid MxAy(s) is
Qsp=[MY+]x[AX-]Y(3)

If Qsp<Ksp, there is no precipitate formed and ions are dissociated in the solution. Likewise, if Qsp>Ksp, ions tend to form precipitate and eventually appear in the solution.

The ionic strength is another factor that affects the Ksp value. It depends on the concentration of the ions present in the solution and their charges. The ionic strength is expressed as
μ=12cizi2 (4)

Where μ is the ionic strength of the solute in the solution is, ci is the molar concentration of each ion and zi is the charge of the constituent ions.
In the experiment, the solute Ca(OH)2 is taken into consideration. The Ksp of Ca(OH)2 is expressed as
Ksp=[Ca2+][OH-]2 (5)

And was alsp determined by calculating the hydroxide ion concentration from solutions saturated with Ca(OH)2. The effect of diverse and common ions on the solubility was also examined in the experiment.

Methodology

A precipitate was prepared by mixing 10 mL of 1.0 M Ca(NO3)2 and 20 mL of 1.0 M NaOH. The class was divided into five groups with each group assigned to prepare a calcium hydroxide suspension, each with a different media: distilled water, 1.0 M KCl, 0.5 M KCl, 0.1 M KCl, 0.01 M KCl, 0.005 M KCl, and 0.001 M KCl. The 1.0 M KCl was prepared using solid KCL, while the lower concentrations were prepared by diluting the 1.0 M KCl solutions. The assigned medium was then isolated to 100 mL for each group and then Ca(OH)2 was gradually added to the solution until it is saturated

C. Common-Ion Effect:

100 mL of 0.1 M Ca(NO3)2 in a 250 mL beaker was added with Ca(OH)2 solid to the assigned media media until it was saturated. 50 mL of the solution was then filtered and an aliquot of 25 mL of the solution with 3 drops of phenolphthalein, giving the solution a dark pink color, was titrated with 0.1 M HCl until the dark pink color of the phenolphthalein vanished.

Results and Discussion

The hydroxide ion concentrations can be solved using the relationship MH+VH+= MOH-Valiquot (6)
where MH+ and MOH- are the concentrations of H+ and OH- ions, respectively.

Table 1: 0.1 M HCl Volumes Used to Neutralize Saturated Ca(OH)2 Solutions in Different Media Medium| HCl Volume (mL)|
Distilled Water| 9.7|
0.001 M KCl| 8.3|
0.005 M KCl| 8.8|
0.010 M KCl| 9.0|
0.050 M KCl| 8.3|
0.100 M KCl| 6.3|
0.500 M KCl| 12.3|
0.100 M Ca2+| 9.7|

Table 1 shows the volume by which hydroxide ions saturated in the different solutions by hydronium ions present in 0.10 M HCl.

Table 2: Ca2+ and OH- Concentrations
Medium| [OH-]| [Ca2+]|
Distilled Water| 0.0388| 0.0194|
0.001 M KCl| 0.0332| 0.0166|
0.005 M KCl| 0.0352| 0.0176|
0.010 M KCl| 0.036| 0.018|
0.050 M KCl| 0.0332| 0.0166|
0.100 M KCl| 0.0252| 0.0126|
0.500 M KCl| 0.0492| 0.0246|
0.100 M Ca2+| 0.0388| 0.0194|

Table 2 shows the calcium and hydroxide ion concentrations obtained at different media. Based from the table, as the concentration of KCl increases, the Ca2+ and OH- concentrations also increase. This is due to the diverse non-common ion effect. The ions of KCl tend to surround...
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