Titration with EDTA question?

Question

25.00 Ml sample of a solution of Al_2(SO4)3 and NiSO4 was diluted to 500.0 mL and the Al^3+ and Ni^2+ in a 25.00 mL aliquat were complexed by addition of 40.00 mL of 0.01175 M EDTA in a pH 4.8 buffer.

The excess EDTA was then back titrated with 10.07 mL of 0.00993 M Cu^2+. an excess of F- was then added to the solution to displace the EDTA that was bouned to the Al^3+. This liberated EDTA consumed 26.30 mL of 0.00993 M Cu^2+.

Calculate the mg/ml of Al2(SO4)3 (342.17 g/mol) and NiSO4 (154.75 g/mol) in the sample

Answer 1

((a) Calculate moles of EDTA added

(b) Calcuate moles of excess EDTA

(c) Calculate moles of EDTA that reacted with Al3+ and Ni2+ by subtration: (a) minus (b)

(d) Calculate the moles of EDTA displace form Al3+ = moles of Al3+

(e) Calculate the moles of Ni2+: (c) minus (d)

(f) Multiply (d) and (e) with the appropriate molar mass to get total mass of each compound in grams

(g) Convert masses in (f) to mg and divide by 25 mL