Question

the solubility of slaked lime, Ca(OH)2, in water is 0.185 g/100 mL. what volume of 0.00300 M HCl is needed to neutralize 13.0 mL of a saturated Ca(OH)2 solution?

Answer 1

mass of Ca(OH)2 = solubility x volume

so

mass of Ca(OH)2 present = 0.185 x 13 / 100

mass of Ca(OH)2 present = 0.02405

now

moles = mass / molar mass

so

moles of Ca(OH)2 = 0.02405 / 74

moles of Ca(OH)2 = 3.25 x 10-4

now

the reaction is

Ca(OH)2 + 2HCl --> CaCl2 + 2H20

we can see that

moles of HCl required = 2 x moles of Ca(OH)2

so

moles of HCl = 2 x 3.25 x 10-4

moles of HCl = 6.5 x 10-4

now

we know that

volume (ml) = moles x 1000 / molarity

so

volume of HCl (ml) = 6.5 x 10-4 x 1000 / 0.003

volume of HCl (ml) = 216.666

so

216.666 ml of HCl is required