Question

Which of the following solutions will turn violet when a drop of lime juice is added to it?
(A) A solution of ${\text{NaI}}$
(B) A solution mixture of ${\text{KI}}$ and ${\text{NaI}}{{\text{O}}_{\text{3}}}$
(C) A solution mixture of ${\text{NaI}}$ and ${\text{KI}}$
(D) A solution mixture of ${\text{KI}}{{\text{O}}_{\text{3}}}$and ${\text{NaI}}{{\text{O}}_{\text{3}}}$

Answer 1

Hint: Lime juice is acidic in nature as it contains citric acid. The citric acid present in lime juice will liberate iodine from the iodide which will give purple or violet coloration to the solution.

Complete step by step answer: When the hydrogen ions liberated from an acid reacts with an iodide ion $\left( {{{\text{I}}^ - }} \right)$ and an iodate ion $\left( {{\text{I}}{{\text{O}}_{\text{3}}}^{\text{ - }}} \right)$ , the following reaction takes place.
${{\text{I}}^{\text{ - }}}{\text{ + I}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{ + }}{{\text{H}}^{\text{ + }}} \to {{\text{I}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
In this reaction between the hydrogen ions and iodide and iodate ions, violet coloration appears due to formation of iodine $\left( {{{\text{I}}_{\text{2}}}} \right)$ . Thus, in acidic medium, the iodate ion in which iodine is in $ + 5$ oxidation state will oxidize the iodide ion in which iodine is in $ - 1$ state to iodine in which it is in zero oxidation state.
Now, consider the first option which is a solution of sodium iodide. Sodium iodide has the chemical formula ${\text{NaI}}$ and so the solution will contain only iodide ions. There is no source of iodate ions in this solution and so the hydrogen ions from the acidic lime juice will not be able to liberate iodine accompanied by purple coloration. Thus, option A is incorrect.
Now, the second option is a solution mixture of ${\text{KI}}$ and ${\text{NaI}}{{\text{O}}_{\text{3}}}$. Thus, this solution mixture contains iodide ions from ${\text{KI}}$ and iodate ions from ${\text{NaI}}{{\text{O}}_{\text{3}}}$. So, the hydrogen ions from the acidic lime juice will be able to liberate iodine accompanied by purple coloration.
${\text{NaI}}{{\text{O}}_{\text{3}}}{\text{ + 5KI + 6}}{{\text{H}}^{\text{ + }}} \to {\text{3}}{{\text{H}}_{\text{2}}}{\text{O + 3}}{{\text{I}}_{\text{2}}}{\text{ + 5}}{{\text{K}}^{\text{ + }}}{\text{ + N}}{{\text{a}}^{\text{ + }}}$
Thus, option B is correct.
The third option is a solution mixture of ${\text{NaI}}$ and ${\text{KI}}$. Thus, this solution mixture contains only iodide ions from ${\text{KI}}$and ${\text{NaI}}$ and no iodate ions. So iodine and purple coloration will not appear in addition to lime juice. Thus, option C is incorrect.
The fourth option is a solution mixture of ${\text{KI}}{{\text{O}}_{\text{3}}}$ and ${\text{NaI}}{{\text{O}}_{\text{3}}}$. Thus, this solution mixture contains only iodate ions from ${\text{KI}}{{\text{O}}_{\text{3}}}$ and ${\text{NaI}}{{\text{O}}_{\text{3}}}$ and no iodide ions. So iodine and purple coloration will not appear in addition to lime juice. Thus, option D is incorrect.

Note: Potassium iodide is a major component of iodised table salt alongwith sodium iodide, sodium iodate and potassium iodate. Potassium iodide is used as a dietary supplement. This is done for the prevention of thyroid problems due to lack of iodine in our normal diet. Thus, addition of lemon juice to table salt will also give a bluish color.