Determination of the Solubility Product Constant of Calcium Hydroxide
Topics: Chemistry, Solubility, Chemical equilibrium, Ion, Concentration / Pages: 5 (1103 words) / Published: Feb 11th, 2013

Determination of the Solubility Product Constant of Calcium Hydroxide


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.


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,

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