Answer to Question #165638 in General Chemistry for Maddy Brown

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 A 73.4 mL sample of this solution was withdrawn and titrated with 0.0945 M HBr. It required 92.8 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

Balanced equation:

2HBr + Ca(OH)2 → CaBr2 + 2H2O

1mol HBr reacts with 2mol Ca(OH)2

Use equation:

M1V1 = M2V2 *2

0.05 * 48.8 = M2*100*2

M2 = 0.05*48.8/(100*2)

M2 = 0.0122M Ca(OH)2 solution. answer to first question.

Molar mass Ca(OH)2 = 74.0932 g/mol

0.0122M solution contains: 74.0932*0.0122 *100/1000 = 0.09g/100ml

Solubility of Ca(OH)2 at 30°C = 0.09g / 100ml solution.