Question

A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution...

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 A 93.5-mL sample of this solution was withdrawn and titrated with 0.0700 M HBr. It required 71.5 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

Homework Answers

Answer #1

a)

M1 = ?

V1 = 93.5 mL

M2 = 0.0700M

V2 = 71.5 mL

M1V1 = M2V2

M1 = M2V2/V1 = 0.0700 M x 71.5 mL / 93.5 mL = 0.0535 M

Conc. of saturated soln. of Ca(OH)2 is equal to 0.0535M

The molarity of the Ca(OH)2 solution is 0.0535M

b) Conc. of saturated soln. of Ca(OH)2 is equal to 0.0535M which is equal to 0.0535 mol/L

Molecular weight of Ca(OH)2 = 74.093 g/mol

so, 0.0535 mol corresponds to (0.0535 x 74.093 g/mol) 3.963 g/L or 0.396 g/100 mL of water

The solubility of Ca(OH)2 in water, at the experimental temperature, in grams of Ca(OH)2 per 100 mL of solution is 0.396g or 396 mg

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