Answer to Question #343617 in Inorganic Chemistry for tom

Question #343617

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.

State the colour change the student observed as the end-point was reached.

Calculate the concentration of Calcium Hydroxide.

Write the Ksp expression for Ca(OH)2.

Use the equilibrium expression for the dissolution of Ca(OH)2 to determine the concentration of Ca2+ and OH–.

Calculate the Ksp for calcium hydroxide (state the units).

The value the student calculated deviated significantly from the accepted value. The reason given was

‘some of the excess solute got into the filtrate during filtration of the saturated calcium hydroxide’.

State how this could have affected the Ksp value.

State one other factor that could have affected the Ksp value.

Expert's answer

1mole of Ca(OH)₂ reacts with 2mole of HCl so if we consider same concentration of Ca(OH)₂ and HCl then moles of Ca(OH)₂ = (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)₂ + 2HCl ------- CaCl₂ + 2H₂O ;

D). number of moles of calcium hydroxide that reacted. = Half mole of HCl reacted = 0.005/2 = 0.0025 moles . answer.