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 A 93.5-mL sample of this solution was withdrawn and titrated with 0.0700 M HBr.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