Copper(II) by iodometric titration

Worked example

In an iodometric titration, 25.00 cm³ of Cu²⁺(aq) was treated with excess KI, and the I₂ formed was titrated with 0.100 mol dm⁻³ Na₂S₂O₃, needing 31.00 cm³. Find [Cu²⁺] in the original solution.

1. 1. Concentration from moles and volume: solve for n(S₂O₃²⁻)

   $n = cV$

   = 0.003100 mol

2. 2. Mole ratio from the balanced equation: solve for n(I₂)

   $n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$

   = 0.001550 mol

3. 3. Mole ratio from the balanced equation: solve for n(Cu²⁺)

   $n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$

   = 0.003100 mol

4. 4. Concentration from moles and volume: solve for c(Cu²⁺)

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

   = 0.1240 mol dm⁻³

Answer: c(Cu²⁺) = 0.124 mol dm⁻³

Method

n(S₂O₃²⁻) = cV. Each I₂ needs two thiosulfate, and each I₂ came from two Cu²⁺, so the twos cancel: n(Cu²⁺) = n(S₂O₃²⁻). Divide by the volume of copper solution.

Common mistakes

• Two Cu²⁺ make one I₂ (2Cu²⁺ + 4I⁻ → 2CuI + I₂), so n(Cu²⁺) = 2n(I₂) = n(S₂O₃²⁻).

Formulas used

• Concentration from moles and volume

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

• Mole ratio from the balanced equation

  $n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$

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