Question

A 1.000-mL sample of unknown containing Co2+ and Ni2+ was treated with 25.00 mL of 0.03872 M EDTA. Back titration with 0.02127 M Zn2+ at pH 5 required 23.54 mL to reach the xylenol orange endpoint. A 2.000-mL sample of unknown was passed through an ion exchange column that retards Co2+ more than Ni2+. The Ni2+ that passed through the column was treated with 25.00 mL of 0.03872 M EDTA and required 25.63 mL of 0.02127 M Zn2+ for back titration. The Co2+ emerged from the column later. It, too, was treated with 25.00 mL of 0.03872 M EDTA. How many milliliters of 0.02127 M Zn2+ will be required for back titration?

Answer 1

In 1.00 ml sample of Ni2+ and Co2+

Total EDTA added initially = 0.03872 M x 25 ml = 0.968 mmol

Total unused EDTA remained is back titrated with Zn2+ = 0.02127 M x 23.54 ml = 0.5007 mmol

Total moles of EDTA reacted with Co2+ and Ni2+ = 0.968 - 0.5007 = 0.4673 mmol

Now, 2.00 ml unknown sample passed through column

So total Ni2+ and Co2+ present = 2 x 0.4673 = 0.9346 mmol

Total EDTA added in first step = 0.03872 M x 25 ml = 0.968 mmol

Total unused EDTA back titrated with Zn2+ = 0.02127 M x 23.63 ml = 0.5026 mmol

Actual moles of EDTA reacted with Ni2+ = 0.968 - 0.5026 = 0.4654 mmol

Total EDTA added in second step = 0.03872 M x 25 ml = 0.968 mmol

moles of Co2+ to be present in 2.00 ml unknown = 0.9346 - 0.4654 = 0.4692 mmol

Milliliters of Zn2+ needed for back titration = 0.4692 mmol/0.02127 M = 22.06 ml