A CAPE Chemistry student sought to determine the solubility product, Ksp of calcium hydroxide by titrating saturated solution of Ca(OH)2 against standardized hydrochloric acid solution, HCl. The saturated Ca(OH)2 solution was filtered, to remove excess solute, and 25mL aliquots were titrated against 0.10M HCl solution using bromothymol blue indicator.Â
Ca(OH)2(s)  ↔  Ca2+(aq)  +  2OH-(aq)Â
A)Calculate the concentration of Calcium Hydroxide.
B)Use the equilibrium expression for the dissolution of Ca(OH)2 to determine the concentration of Ca2+ and OH–.
 A)
1mole of Ca(OH)2Â reacts with 2mole of HCl so if we consider same concentration of Ca(OH)2Â and HCl then moles of Ca(OH)2Â = (25 * 0.1 / 1000 ) = 0.0025 moles ..
so , Moles of HCl = 2 * 0.0025 = 0.005 moles .
so, volume of HCl used for this titration = ( 0.005 / 0.1 ) * 1000 = 50 ml . answer...
B). Moles of HCl required For This Titration = 0.005 Moles.
C). Ca(OH)2Â + 2HCl ------- CaCl2Â + 2H2O ;
D). number of moles of calcium hydroxide that reacted. = Half mole of HCl reacted = 0.005/2 = 0.0025 moles . answer.
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