Answer to Question #82051 in General Chemistry for Annie

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 A 92.7-mL sample of this solution was withdrawn and titrated with 0.0518 M HBr. It required 98.7 mL of the acid solution for neutralization.

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

(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?

Answer 1

Answer:

Moles HBr used:

0.0987L*0.0518 mol/L = 5.11*10^-3 mol HBr

Moles Ca(OH) in solution:

5.11*10^-3 mol HBr * (1 mol Ca(OH)2 / 2 mol HBr) = 2.555*10^-3 mol Ca(OH)2

Concentration Ca(OH)2:

2.555*10^-3 mol / 0.0927 L = 0.0276 M Ca(OH)2

b) 0.0276 mol/ L * 74.1 g/mol = 0.204 g Ca(OH)2 /100 mL