Question

1.5 g of caco3 (M.W 100.09g/mol) are transferred to a 500 ml volumetric flask and 1:1...

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

Homework Answers

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

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