= Solution
\b[<Potassium iodide> is prepared by converting <iodine> into <iodide>, then obtaining a pure crystalline <potassium> salt.] One chemical route uses <potassium hydroxide>; an alternative first forms <iron(II) iodide>. The account here describes the chemical operations and their balances, without assigning a later pharmacopoeial formula to 1866.
In the alkaline route, <iodine> reacts with hot <potassium hydroxide> to give a mixture of <potassium iodide> and <potassium iodate>:
$$
\boxed{3\mathrm I_2+6\mathrm{KOH}\longrightarrow5\mathrm{KI}+\mathrm{KIO}_3+3\mathrm H_2\mathrm O.}
$$
This is <disproportionation>: <iodine> goes from <oxidation state> $0$ to both $-1$ and $+5$. Merely evaporating the solution would leave the unwanted <iodate> with the <iodide>. Heating the dried salts with <charcoal> reduces the <iodate> to <iodide>. A limiting balance with <carbon monoxide> as the carbon oxide product is
$$
\boxed{\mathrm{KIO}_3+3\mathrm C\longrightarrow\mathrm{KI}+3\mathrm{CO}.}
$$
The carbon oxides can depend on the reaction conditions; the indispensable chemical point is removal of the iodate's <oxygen> without losing its <iodine>. Extraction with <water>, filtration of insoluble residue, concentration and crystallization then recover <potassium iodide>. A modern USDA technical evaluation explicitly describes this same reaction sequence; it is corroboration of the chemistry, not evidence that its modern document was the 1866 official recipe.<Potassium Iodide technical evaluation by United States Department of Agriculture>[2011 report, printed pp. 5–6, Evaluation Question 1, method (a)]{ref}
In the alternative iron route,
$$
\mathrm{Fe}+\mathrm I_2\longrightarrow\mathrm{FeI}_2,
$$
followed by
$$
\mathrm{FeI}_2+\mathrm K_2\mathrm{CO}_3\longrightarrow2\mathrm{KI}+\mathrm{FeCO}_3\downarrow.
$$
The insoluble iron-containing material is separated and the <potassium iodide> recovered from the liquid. Oxidation of the ferrous precipitate during working can alter its final form, so the displayed <chemical equation> is the ideal metathesis balance rather than a guarantee that every collected particle remains pure <iron(II) carbonate>.
Pure <potassium iodide> is a colourless or white crystalline ionic <solid>, commonly in cubic crystals, readily soluble in <water> and less soluble in <ethanol>. It has no intrinsic brown iodine colour. Its aqueous solution contains <potassium> <ions> and <iodide> <ions>. Free <iodine> dissolves in the presence of <iodide> through <triiodide> formation,
$$
\mathrm I_2+\mathrm I^-\rightleftharpoons\mathrm I_3^-.
$$
With <silver nitrate>, <iodide> gives yellow <silver iodide>, sparingly soluble in <water> and resistant to ordinary aqueous <Ammonia>:
$$
\mathrm{Ag}^++\mathrm I^-\longrightarrow\mathrm{AgI}\downarrow.
$$
A little <chlorine> liberates <iodine>, which gives the characteristic colour with <starch>:
$$
\mathrm{Cl}_2+2\mathrm I^-\longrightarrow2\mathrm{Cl}^-+\mathrm I_2.
$$
By contrast, <iodate> contamination can liberate <iodine> simply on acidification in the presence of <iodide>:
$$
\mathrm{IO}_3^-+5\mathrm I^-+6\mathrm H^+\longrightarrow3\mathrm I_2+3\mathrm H_2\mathrm O.
$$
That distinction explains why removal of the <iodate> is part of preparing a well-defined product, rather than an optional cosmetic step. These are chemical properties and recognition principles, not advice to administer the salt.
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