Question

Tincture of Iodine has antiseptic properties. This solution is made by dissolving :

• A. Iodine in Potassium iodide
• B. Iodine in vaseline
• C. Iodine in water
• D. Iodine in alcohol

Answer 1

The correct option is D

Iodine in alcohol

The explanation for the correct option:

Option (D):

In the Tincture of Iodine, the solvent is Alcohol and Water $\left({\mathrm{H}}_2\mathrm{O}\right)$ while the solute is Iodine $\left({\mathrm{I}}_2\right)$.

1. A Tincture of Iodine is also known as a weak Iodine solution or Iodine tincture.
2. It is usually $2$ to $7\%$ elemental Iodine $\left({\mathrm{I}}_2\right)$, along with Potassium iodide $\left(\mathrm{KI}\right)$ or Sodium iodide $\left(\mathrm{NaI}\right)$ dissolved in a mixture of Alcohol (Ethanol, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$) and Water $\left({\mathrm{H}}_2\mathrm{O}\right)$.
3. Tincture solutions are characterized by the presence of Alcohol since Alcohol is a good solvent.
4. Iodine tincture is an antiseptic used to disinfect wounds.

Therefore, the Tincture of Iodine has antiseptic properties. This solution is made by dissolving Iodine $\left({\mathrm{I}}_2\right)$ in Alcohol.

The explanation for the incorrect option:

Option (A):

Iodine $\left({\mathrm{I}}_2\right)$ does not dissolve in Potassium iodide $\left(\mathrm{KI}\right)$ because we know that when Iodine $\left({\mathrm{I}}_2\right)$ is dissolved in aqueous Potassium iodide $\left(\mathrm{KI}\right)$ they generally form a compound that is Potsaaium triiodide $\left({\mathrm{KI}}_3\right)$.

Option (B):

Iodine $\left({\mathrm{I}}_2\right)$ does not dissolve in Vaseline because Vaseline is used for the treatment of dry and damaged skin not as an antiseptic.

Option (C):

Iodine $\left({\mathrm{I}}_2\right)$ does not dissolve in Water $\left({\mathrm{H}}_2\mathrm{O}\right)$ because Water $\left({\mathrm{H}}_2\mathrm{O}\right)$ is an extremely polar molecule, while Iodine $\left({\mathrm{I}}_2\right)$ exists in the diatomic form of ${\mathrm{I}}_2,$ and is therefore non-polar, and will not dissolve in water $\left({\mathrm{H}}_2\mathrm{O}\right)$.

Hence, option (D) is correct.