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 A 58.6-mL sample of this solution was withdrawn and titrated with 0.0872 M HBr. It required 79.0 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

a) Ca(OH)2 solution is titrated with HBr, So let us write the balanced reaction for this.

Ca(OH)₂(aq) + 2HBr(aq) ----------> CaBr₂(aq) + 2H₂O(l)

step 1.

Finding the moles of HBr used for neutralization reaction

Molarity of HBr =  0.0872 M

Volume of HBr solution = 79.0 mL = 0.0790 L

Moles of HBr reacted = 0.0872 M x 0.079 L = 6.90x10^-3 mol

Step 2. FInding the moles Ca(OH)2

Moles of Ca(OH)2 reacted =

=3.44x10^-3 mol

Moles of Ca(OH)2 = 3.44x10^-3 mol

Volume of the solution = 58.6 mL + 79.0 mL = 137.6 mL = 0.1376 L

Molarity of Ca(OH)2 =

Molarity of Ca(OH)2 =0.025 M

b)

Molarity of Ca(OH)2 =0.025 M

Moles of Ca(OH)2 in 1 L = 0.025 mol

Mass of Ca(OH)2 in 1 L solution = moles * molar mass =0.025 mol x 74.1 g/mol

=1.85 g/1000 mL

Mass of Ca(OH)2 in 100 mL solution = 0.185 g/100 mL