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 A 78.7-mL sample of this solution was withdrawn and titrated with 0.0696 M HBr. It required 68.0 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

Answer 1

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

1 mole       2 moles

Ca(OH)2                                                                                       HBr

molarity M1 =                                                                              molarity M2 = 0.0696M

volume V1 = 78.7 ml                                                                  volume V2 = 68ml

no of moles n1 = 1                                                                    no of moles n2 = 2

M1V1/n1     = M2V2/n2

M1   = M2V2n1/n2V1

      = 0.0696*68*1/2*78.7 = 0.032 M

solubility of Ca(OH)2 = 0.032moles/L

molar mass of Ca(OH)2   = 74g/mole

solubility of Ca(OH)2 = 0.032*74/1000ml

                                  = 0.2368 g/100ml