An unknown solution was analyzed for Ni by an EDTA titration. A 50.00 mL sample of...
An unknown solution was analyzed for Ni by an EDTA titration. A 50.00 mL sample of the unknown was treated with 25.00 mL of 0.2404 M EDTA. The excess EDTA was then back titrated with 8.52 mL of 0.0694 M Zn2+ to reach the equivalence point. What was the concentration of Ni (in unit of M) in the 50.00 mL sample? Please keep your answer to three decimal places.
Homework Answers
The added EDTA moles and the required Zn + 2 moles are calculated:
n EDTA = M * V = 0.2404 M * 0.025 L = 6.01E-3 mol
n Zn + 2 = 0.0694 M * 0.00852 L = 5.91E-4 mol
The moles of Ni are calculated by difference from the previous moles:
n Ni = 6.01E-3 - 5.91E-4 = 5.42E-3 mol
The molarity is calculated:
[Ni] = n / V = 5.42E-3 mol / 0.05 L = 0.108 M
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