Question

A sample of solid Ca(OH)₂ was stirred in water at a certain temperature until the solution contained as much dissolved Ca(OH)₂ as it could hold. A 50.7-mL sample of this solution was withdrawn and titrated with 0.0621 M HBr. It required 66.7 mL of the acid solution for neutralization.

(a) What was the molarity of the Ca(OH)₂ solution?

________ M



(b) What is the solubility of Ca(OH)₂ in water, at the experimental temperature, in grams of Ca(OH)₂ per 100 mL of solution?

_______  g/100mL

Answer 1

Molarity of HBr = 0.0621 M

Volume of HBr solution = 66.7 mL = 0.0667 L

Moles = Molarity x Volume (L)

=> Moles of HBr = 0.0621 x 0.0667 = 0.00414 moles

The reaction is,

2HBr + Ca(OH)2 ----> Ca(Br)2 + 2H2O

According to the stoichiometry of the reaction 1 mole of Ca(OH)2 is required for complete neutralization of 2 moles of HBr

Moles of HBr = 0.00414

=> Moles of Ca(OH)2 = 0.00414 / 2 = 0.00207 moles

Volume of Ca(OH)2 solution = 50.7 mL = 0.0507 L

Molarity = Moles / Volume (L)

a) Molarity of Ca(OH)2 solution = 0.00207 / 0.0507 = 0.0408 M

Molar Mass of Ca(OH)2 = 74.093 g / mol

b)

Solubility = 0.0408 x 74.093 = 3.027 g / L

=> Solubility = 0.3027 g / 100 mL