A student attempts a permanganometry titration in near-neutral solution instead of strongly acidic solution. What goes wrong?
AThe permanganate oxidizes the solvent instead of the analyte, consuming extra titrant
BMnO₄⁻ is reduced to MnO₂ (a brown precipitate) rather than Mn²⁺, so the endpoint color change is obscured and the stoichiometry changes
CThe reaction proceeds too quickly in neutral solution, making it impossible to detect the endpoint
DNothing changes — the endpoint is equally well-defined in neutral and acidic solution
Permanganometry is self-indicating only in strongly acidic solution because only there is the MnO₄⁻ → Mn²⁺ reduction (5 electrons per Mn) thermodynamically favored and fast. In neutral or basic solution, MnO₄⁻ is reduced to MnO₂ (brown precipitate, 3 electrons per Mn), which clouds the solution and makes the persistent-pink endpoint undetectable. The different stoichiometry also invalidates any calculation based on the acidic-solution reaction. Acid concentration is not a minor procedural detail — it determines which reduction product forms and therefore whether the titration is valid.
Question 2 Multiple Choice
In an indirect iodometric titration, the analyte oxidizes iodide (I⁻) to iodine (I₂), which is then back-titrated with sodium thiosulfate. Why is starch indicator added near the endpoint rather than at the beginning?
AStarch destroys thiosulfate if present from the start, preventing accurate measurement
BStarch forms an intensely blue complex with I₂; added at the start when I₂ concentration is high, the blue color is so dark it masks the gradual color change, making the endpoint hard to detect
CThe starch–I₂ complex is irreversible, so if added early it permanently sequesters iodine from the titration
DStarch reacts with the analyte rather than with iodine if it is present before the back-titration begins
Starch binds I₂ (as I₃⁻) to form a deep blue complex — this is the detection chemistry. But at high I₂ concentrations (early in the titration), the intense blue color is so dark that tracking the gradual lightening is difficult, and the bound complex is slow to release I₂ to react with thiosulfate, which can cause a premature apparent endpoint. Near the endpoint when I₂ concentration is already low (the solution is pale yellow-straw), adding starch gives a clean blue → colorless transition that is easy to detect precisely. Timing of starch addition is a practical precision issue.
Question 3 True / False
Before the equivalence point in a redox titration, the solution potential is governed by the redox couple of the analyte rather than the titrant.
TTrue
FFalse
Answer: True
The Nernst equation governs potential through the ratio of oxidized to reduced species. Before the equivalence point, excess analyte remains — for example, a mixture of Fe²⁺ and Fe³⁺ if iron is being titrated. The potential is determined by the Fe³⁺/Fe²⁺ ratio as titrant converts Fe²⁺ to Fe³⁺. Only after the equivalence point, when excess titrant (e.g., MnO₄⁻/Mn²⁺) dominates, does the titrant's redox couple govern the potential. At the equivalence point itself, the potential is intermediate and both couples contribute — and it jumps sharply, defining the endpoint.
Question 4 True / False
In iodometric titrations, the pH of the solution has little effect on the accuracy of the result.
TTrue
FFalse
Answer: False
pH is critical in iodometric methods. In alkaline conditions, iodine disproportionates: I₂ + 2OH⁻ → I⁻ + IO⁻ + H₂O. This converts I₂ to iodate (IO₃⁻) and iodide, meaning the I₂ that was supposed to be the measured species is consumed by a side reaction, causing low results. Acidic pH is required to suppress this disproportionation. Additionally, some analytes (like dichromate) generate different products at different pH values. pH control is as important in redox titrations as in acid–base work.
Question 5 Short Answer
Why is it often necessary to pre-treat an analyte sample before performing a redox titration, and what would go wrong if this step were skipped?
Think about your answer, then reveal below.
Model answer: Many real samples contain the analyte in mixed oxidation states — for example, an iron ore sample may contain both Fe²⁺ and Fe³⁺. A redox titrant reacts with only one oxidation state (e.g., KMnO₄ oxidizes Fe²⁺ but not Fe³⁺). If pre-reduction is skipped, the titration only consumes the portion of iron already in the lower state, giving a result that is systematically low and irreproducible depending on sample preparation. Pre-treatment (e.g., reducing all iron to Fe²⁺ with SnCl₂) ensures every mole of the analyte element reacts with the titrant, making the stoichiometric calculation valid.
The excess reductant from pre-treatment must also be destroyed before titration begins — otherwise it reacts with the titrant and consumes extra equivalents, giving a high result. This step (adding HgCl₂ to oxidize excess Sn²⁺, for example) is easy to overlook but essential. The concept generalizes: pre-oxidation may be needed instead if the analyte must be brought to a higher state before titration with a reducing titrant.