Analytical Chem istry - DePauw University

Chapter 9 Titrimetric Methods477of which 1.524×10 –3 mol are used to titrate Ni and 5.42×10 –4 mol areused to titrate Fe. This leaves 8.50×10 –4 mol of EDTA to react with Cuand Cr. The amount of EDTA reacting with Cu is2+0.06316 molCu2+× 0.00621 LCuL1 molEDTA−× = 392 . × 10 4 molEDTAmolCu2+leaving 4.58×10 –4 mol of EDTA to react with Cr. The sample, therefore,contains 4.58×10 –4 mol of Cr.Having determined the moles of Ni, Fe, and Cr in a 50.00-mL portionof the dissolved alloy, we can calculate the %w/w of each analyte in thealloy.–31.524×10 molNi58. 69 g Ni× 250.0 mL×50.00 mLmolNi= 0.44720.4472 gNi× 100 = 62. 32%w/wNi0.7176 gsample–45.42×10 molFe55. 847 g Fe× 250.0 mL×50.00 mLmolFe0.151 gFe× 100 = 21.%0 w/wFe0.7176 gsample–44.58×10 molCr51. 996 g Cr× 250.0 mL×50.00 mLmolCr0.119 gCr× 100 = 16.%6 w/wFe0.7176 gsamplegNi= 0.151 gFe= 0.119gCrPractice Exercise 9.16A indirect complexation titration with EDTA can be used to determinethe concentration of sulfate, SO 2– 4 , in a sample. A 0.1557-g sample isdissolved in water, any sulfate present is precipitated as BaSO 4 by addingBa(NO 3 ) 2 . After filtering and rinsing the precipitate, it is dissolvedin 25.00 mL of 0.02011 M EDTA. The excess EDTA is then titratedw iith 0.01113 M Mg 2+ , requiring 4.23 mL to reach the end point.Calculate the %w/w Na 2 SO 4 in the sample.Click here to review your answer to this exercise.