Question

Analysis of Aluminum in a Mixture by Solvent Extraction and EDTA Titration.

*the amount of EDTA, which is equivalent to the number of moles of zinc and the number of moles of aluminium.*

A sample required 9.73 mL of zinc nitrate solution to titrate the EDTA released from aluminum through the addition of fluoride. If the molarity of the zinc solution was 0.025M and EDTA was 0.05M, what was the total weight of the aluminum present in the initial sample?

Answer 1

amount of EDTA = no of moles of Zn²⁺ = no of moles of Al³⁺

no of moles of Al³⁺ = volume of Zn²⁺ used to titrate X Molarity of Zn²⁺ = amount of EDTA

no of moles of Al³⁺ = 9.73 ml X 0.025 M = 0.243 Mol L^-1 ml ( M = Mol L^-1 and 1000 ml = 1L ) = 0.243 mMol

Weight of Aluminium in the initial sample = no of mol X Molar mass

Weight of Aluminium in the initial sample = 0.243 mMol X 26.98 g Mol^-1

= 6.556 mg