A student adds lead (II) nitrate (Pb(NO₃)₂) to an unknown solution. A yellow precipitate forms. What could the unknown solution be?

The correct option is Potassium iodide (KI).

Step-by-Step Explanation:

1. Identify the Reactant: The student is adding lead(II) nitrate, ${\mathrm{Pb(NO}}_3{)}_2$, which is a soluble salt in aqueous solution containing lead ions (${\mathrm{Pb}}^{2+}$) and nitrate ions (${\mathrm{NO}}_3^{-}$).

2. Observe the Reaction: A yellow precipitate forms upon adding the unknown solution. This indicates a precipitation reaction where one of the products is an insoluble solid with a characteristic yellow color.

3. Analyze the Options:
- Potassium iodide (KI): When potassium iodide reacts with lead(II) nitrate, double displacement occurs:
${\mathrm{Pb(NO}}_3{)}_2\left(\mathrm{aq}\right)+2\mathrm{KI}\left(\mathrm{aq}\right)\to {\mathrm{PbI}}_2\left(s\right)+2{\mathrm{KNO}}_3\left(\mathrm{aq}\right)$
Lead(II) iodide (${\mathrm{PbI}}_2$) is well-known as a bright yellow precipitate.
- Copper sulphate (CuSO₄): This would form lead(II) sulphate (${\mathrm{PbSO}}_4$), which is a white precipitate.
- Sodium chloride (NaCl): This would form lead(II) chloride (${\mathrm{PbCl}}_2$), which is a white precipitate.
- Barium sulphate (BaSO₄): Barium sulphate is highly insoluble in water and would not react significantly in a double-displacement solution with lead(II) nitrate to yield a yellow precipitate.

Therefore, the unknown solution must contain iodide ions, which react with lead(II) ions to form the yellow precipitate of lead(II) iodide. This confirms the unknown solution is potassium iodide (KI).