Nickel by EDTA back titration

Worked example

Ni²⁺ reacts too slowly with EDTA for a direct titration, so 25.00 cm³ of the nickel solution was treated with 50.00 cm³ of 0.0200 mol dm⁻³ EDTA, and the excess was back-titrated with 0.0200 mol dm⁻³ zinc sulfate (19.00 cm³ needed). Find [Ni²⁺].

1. 1. Concentration from moles and volume: solve for n(EDTA) added

   $n = cV$

   = 0.001000 mol

2. 2. Concentration from moles and volume: solve for n(EDTA) left = n(Zn²⁺)

   $n = cV$

   = 3.800 × 10⁻⁴ mol

3. 3. Subtract n(EDTA) added − n(EDTA) left = n(Zn²⁺) → n(Ni²⁺) = n(EDTA) bound

   = 6.200 × 10⁻⁴ mol

4. 4. Concentration from moles and volume: solve for c(Ni²⁺)

   $c = \dfrac{n}{V}$

   = 0.02480 mol dm⁻³

Answer: c(Ni²⁺) = 0.0248 mol dm⁻³

Method

All complexes are 1 : 1. n(EDTA) added = cV; the Zn²⁺ titre gives the EDTA left over; the difference was bound by Ni²⁺. A back titration is used because [Ni(EDTA)]²⁻ forms slowly, so the end point of a direct titration would drift.

Common mistakes

• The zinc titre measures the EDTA left over, not the nickel. Ni²⁺ = EDTA added − EDTA left over.

Formulas used

• Concentration from moles and volume

  $c = \dfrac{n}{V}$

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