Question

A 60 mL sample solution of KI was electrolysed for 579 seconds using a 5 amp constant current. The $I_2$ produced completely reacted with 0.20 M sodium thiosulphate solution and unreacted KI solution required 30 mL of 0.4 M faintly alkaline solution of $KMnO_4$. Calculate the molarity of original KI solution?

Answer 1

The molarity of the original KI solution is 0.90 M.

Answer 2

1. Electrolysis Step

   • Total charge passed, $$Q = I \times t = 5\,\text{A} \times 579\,\text{s} = 2895\,\text{C}.$$
   • Moles of electrons transferred, $$n_{e^-} = \frac{Q}{F} = \frac{2895}{96485} \approx 0.03\,\text{mol}.$$
   • In the oxidation at the anode, the reaction is $$2I^- \rightarrow I_2 + 2e^-.$$ Hence, moles of $I_2$ produced: $$n_{I_2} = \frac{n_{e^-}}{2} = \frac{0.03}{2} = 0.015\,\text{mol}.$$

2. Sodium Thiosulfate Titration

   • Iodine reacts with thiosulfate as: $$I_2 + 2\,S_2O_3^{2-} \rightarrow 2\,I^- + S_4O_6^{2-}.$$ Thus, 1 mole $I_2$ requires 2 moles $S_2O_3^{2-}$. The moles of $I_2$ produced (0.015 mol) would have consumed $$2 \times 0.015 = 0.03\,\text{mol}\,I^- \text{ (via oxidation)}.$$

3. KMnO₄ Titration

   • Unreacted KI (i.e. excess $I^-$) is titrated with 0.4 M alkaline KMnO₄. In a faintly alkaline medium, the reaction is: $$\mathrm{MnO_4^-} + 2\,OH^- + 2\,I^- \longrightarrow \mathrm{MnO_4^{2-}} + I_2 + H_2O.$$ Here, 1 mole of $\mathrm{MnO_4^-}$ reacts with 2 moles of $I^-$.
   • Moles of $\mathrm{KMnO_4}$ used = Volume × Molarity $$= 0.030\,\text{L} \times 0.4\,\text{M} = 0.012\,\text{mol}.$$
   • Therefore, moles of $I^-$ remaining = $$2 \times 0.012 = 0.024\,\text{mol}.$$

4. Total Iodide in the Original KI Solution

   • The iodide originally present is consumed partly in electrolysis (to form $I_2$) and partly remains unreacted.
   • Iodide consumed to form $I_2$ = $2 \times n_{I_2} = 2 \times 0.015 = 0.03\,\text{mol}$.
   • Total moles $I^-$ originally = $0.03 + 0.024 = 0.054\,\text{mol}$.
   • The original volume is $60\,\text{mL} = 0.06\,\text{L}$. Thus, the molarity of the KI solution is: $$M = \frac{0.054}{0.06} = 0.90\,\text{M}.$$