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 A 59.6-mL sample of this solution was withdrawn and titrated with 0.0543 M HBr. It required 67.1mL of the acid solution for neutralization.

(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

Answer 1

Sol:-

Solubility :- Number of moles of substance dissolved in 1 litre volume of saturated solution .

Unit of Solubility = mol/L

Because the unit of Solubility is same as the unit of Molarity therefore

solubility of Ca(OH)₂ = Molarity of Ca(OH)₂.

Given

Volume of Ca(OH)₂ i.e V₁ = 59.6 mL = 0.0596 L

Molarity of Ca(OH)₂ i.e M₁= ?

also Valency factor (v.f₁) of Ca(OH)₂ = 2 (i.e number of OH^- given by Ca(OH)₂ ).

and Molarity of HBr (M₂) = 0.0543 M

Volume of HBr (V₂) = 67.1mL = 0.0671 L

also Valency factor (v.f₂) of HBr = 1 ( i.e number of H⁺ given by HBr )

For Neutralization

gram equivalent of Ca(OH)₂ i.e E_Ca(OH)2= gram equivalent of HBr i.e E_HBr .

N₁V₁ (Ca(OH)₂) = N₂V₂ ( HBr)       ,[ because Normailty (N) = gram equivalent of solute / volume of solution in litre]

M₁ x v.f₁ x V₁ ( Ca(OH)₂ ) = M₂ x v.f₂ x V₂ (HBr)   , [because Normality (N) = Molarity (M) x valency factor (v.f) ]

M₁ x 2 x 0.0596 L = 0.0543 M x 1 x 0.0671 L

M₁ = 0.0543M x 0.0671L / 2 x 0.0596 L

M₁ = 0.0306 M

Hence Molarity of Ca(OH)₂ = solubility of Ca(OH)₂ = 0.0306 M = 0.0306 mol/L

because Number of moles = mass of substance in g / gram molar mass of substance

and gram molar mass of Ca(OH)₂ = 74.093 g/mol

Now Solubility of Ca(OH)₂ =0.0306 mol/L

                                           =0.0306 mol x 74.093g/mol / 1000mL

                                           = 2.27 g / 1000 mL

                                           = 0.227 g / 100 mL

Hence Solubility of Ca(OH)₂ = 0.227g / 100 mL