Question

50.00ml of a saturated solution of calcium hydroxide is titrated with (5.00 x 10^-2 M) HNO₃. The endpoint is reached when 24.40ml of the acid is added. What is the molarity and solubility (grams/liter of solution) of the Ca(OH)₂ solution?

Thank you

Answer 1

first we can write balenced equation

Ca(OH)₂ + 2 HNO₃ -------->     Ca(NO₃)₂ + 2H₂O

from the reaction it is very clear that for 1 mol of Ca(OH)₂ nuetralistion required 2 moles of HNO3

now calculate the moles of HNO3 required to complete the reaction

moles = molarity x V in liters

moles = 5.00 x 10^-2 M x 0.0244 L

moles = 0.00122 moles HNO₃

from this we can calulate the moles of Ca(OH)₂

2 moles of HNO3 required to nuetralise one mole of Ca(OH)₂

0.00122 moles of HNO3 is sufficient for how many moles of Ca(OH)₂

Ca(OH)₂ moles = 0.00122 / 2

Ca(OH)₂ moles = 0.00061

now we know moles of Ca(OH)₂ and volume of Ca(OH)₂

from this we can calculate the molarity of Ca(OH)₂

Molarity = 0.00061 / 0.05

Molarity = 0.0122 M Ca(OH)₂

= 1.22 x 10^-2 M

this is called molar solubility

Ca(OH)₂ ----> Ca⁺² + 2OH^-

Ksp = [ Ca⁺²] [OH^-]²

now we have to calculate the molarity of the each ion

i mol of Ca(OH)₂ will give one mole of Ca⁺² so

concentration of  [ Ca⁺²] = concentration of Ca(OH)₂

[ Ca⁺²] = 1.22 x 10^-2 M

but one mole of Ca(OH)₂ will give 2 mole of OH^- so

concentration of [OH^-] = 2 x concentration of Ca(OH)₂

= 2 x 1.22 x 10^-2 M

= 2.44 x 10^-2 M

now put these values in above equation

Ksp = [1.22 x 10^-2 M] x [2.44 x 10^-2 M]²

Ksp = 7.3 x 10^-6