Question

a) A standard ZnCl2 solution is prepared by dissolving 0.6328 g of Zn in an HCl solution and diluting to volume in a 1.00 L volumetric flask. An EDTA solution is standardized by titrating a 10.00 mL aliquot of the ZnCl2 solution, which requires 10.84 mL of EDTA solution to reach the end point. Determine the concentration of the EDTA solution.

b) A 1.4927 g sample of powdered milk is dissolved and the solution titrated with the EDTA solution prepared in part (a). The titration requires 11.93 mL of the EDTA solution to reach the end point. Determine the concentration of calcium in the powered milk in ppm Ca.

Answer 1

a)

Mass of Zn = 0.6328 g

Atomic mass of Zn = 65.38 g/mol

Moles of Zn = 0.6328/65.38 = 0.009679 mol

0.009679 mol of Zn will produce 0.009679 mol of ZnCl₂

Molarity of the ZnCl₂ solution (C₁) = 0.009679 mol/1L = 0.009679 M

Volume of the aliquot of the ZnCl₂ solution (V₁) = 10.00 mL

Volume of EDTA solution required (V₂) = 10.84 mL

Let us say that the concentration of EDTA = C₂

Now equating the moles of the ZnCl₂ and the EDTA,

V₂ x C₂ = V₁ x C₁

or, C₂ = (10.00x 0.009679)/10.84

           = 0.008929 M

Therefore the concentration of EDTA = 0.008929 M

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b)

11.93 mL of 0.008929 M EDTA = 11.93 x 0.008929 = 0.1065 mmol = 0.1065 x 10^-3 mol

0.1065 x 10^-3 mol of EDTA will form complex with 0.1065 x 10^-3 mol of Ca²⁺

Atomic mass of Ca = 40.078 g/mol

Therefore, mass of calcium in powdered milk = 40.078 x 0.1065 x 10^-3 = 0.004268 g

Thus, 1.4927 g sample of powdered milk contains 0.004268 g = 4268 x 10^-6 g calcium

Therefore, the concentration of Ca in powdered milk = (4268 x 10^-6)/1.4927 = 2859 ppm