Question

1.5 g of caco3 (M.W 100.09g/mol) are transferred to a 500 ml volumetric flask and 1:1 Hcl is added dropwise until effervescence ceases and the solution is clear. everything is diluted with water in the mark. 30 ml of that solution are titrated with EDTA and 15.5 ml are spent to reach the endpoint. Calculate the molarity of EDTA

Answer 1

CaCO₃ + HCl -----------> CaCl₂ + CO₂ + H₂O

Amount of CaCO₃ = 1.5 g

Molecular weight = 100.09 g/mol

Moles of CaCO₃ = 1.5 g / 100.09 gmol^-1 = 0.0149 mol = 14.9 mmol

14.9 mmoles of CaCO₃ produced 14.9 mmoles of CaCl₂

Volume of the flask= 500 mL

Molarity = Moles /Volume = 14.9 mmol / 500 mL = 0.0298 M

Volume of EDTA consumed = 15.5 mL

Molarity of EDTA = ?

Molarity of CaCl₂ = 0.0298

Volume of of CaCl₂ = 30 mL

Using M₁V₁ = M₂V₂,

0.0298 x 30 mL = M₂ x 15.5 mL

M₂ = 0.0576 M

So the molarity of EDTA = 0.0576 M