Question

A 0.4400-g sample of impure Ca(OH)2 is dissolved in enough water to make 60.30 mL of solution. 20.00 mL of the resulting solution is then titrated with 0.1461-M HCl. What is the percent purity of the calcium hydroxide if the titration requires 18.33 mL of the acid to reach the endpoint?

Answer 1

The balanced chemical equation for the reaction is

Ca(OH)₂ + 2 HCl -------> CaCl₂ + 2 H₂O

As per the stoichiometric equation,

1 mole Ca(OH)₂ = 2 mole HCl.

Millimoles of HCl added = (18.33 mL)*(0.1461 M)*(1 mol/L/1 M) = 2.678013 mmole.

Millimoles of Ca(OH)₂ titrated = (2.678013 mmole HCl)*(1 mole Ca(OH)₂/2 mole HCl) = 1.3390065 mmole.

The said amount of Ca(OH)₂ is in 20.00 mL of the prepared solution; the volume of the original solution is 60.30 mL. Therefore , the millimoles of Ca(OH)₂ in 60.30 mL of the prepared solution is (1.3390065 mmole)*(60.30 mL/20.00 mL) = 4.037104598 mmole.

Molar mass of Ca(OH)₂ = (1*40.078 + 2*15.9994 + 2*1.008) g/mol = 74.0928 g/mol.

Mass of Ca(OH)₂ corresponding toe 1.3390065 mmole Ca(OH)₂ = (1.3390065 mmole)*(1 mole/1000 mmole)*(74.0928 g/mol) = 0.09921 g ≈ 0.0992 g.

Percent Ca(OH)₂ in the impure sample = (0.0992 g)/(0.4400 g)*100% = 22.5454% (ans).