In the double displacement reaction between aqueous Potassium iodide and aqueous Lead nitrate, a yellow precipitate of Lead iodide is formed. While performing the activity if lead Nitrate is not available, which of the following can be used in place of Lead nitrate?

The correct option is B

Lead acetate

Explanation for correct answer:

Option B Lead acetate

• When two compounds react, positive ions (cation) and the negative ions (anion) of the two reactants switch places, forming two new compounds or products is known as a double displacement reaction.
• A double replacement reaction is represented by the general equation, $\mathrm{AB}+\mathrm{CD}\to \mathrm{AD}+\mathrm{CB}$
• In the double displacement reaction, Potassium Iodide$\left(\mathrm{KI}\right)$ and Lead nitrate$(\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2)$ dissociate in their aqueous states to form ions.
• Thus Lead$\left({\mathrm{Pb}}^{2+}\right)$ ions combine with the Iodide$\left({\mathrm{I}}^{-}\right)$ ions to form precipitates of Lead iodide.
• The Lead acetate dissociates in the aqueous state to form ${\mathrm{Pb}}^{2+}$ ions and ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ ions.
• Thus, Potassium iodide combines with Lead acetate to form precipitates of Lead iodide and

$\underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{aq}\right)}+\underset{\mathrm{Lead}\mathrm{acetate}}{\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2\left(\mathrm{aq}\right)}\to \underset{\mathrm{Lead}\mathrm{iodide}}{{\mathrm{PbI}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Potassium}\mathrm{acetate}}{2{\mathrm{CH}}_3\mathrm{COOK}\left(\mathrm{s}\right)}$

• Hence Option B is correct.

Explanation for incorrect answer:

Option A Lead sulphate (insoluble)

• If Lead sulphate is used in place of lead nitrate, lead iodide won't get precipitated, since the lead sulphate is insoluble in water and it does not give ${\mathrm{Pb}}^{2+}$ ions which can combine with ${\mathrm{I}}^{-}$ to form ${\mathrm{PbI}}_2$.
• Hence Option A is incorrect.

Option C Ammonium nitrate

• If Ammonium nitrate is used in place of Ammonium nitrate, lead iodide won't get precipitated, since the Ammonium nitrate and Potassium iodide does not give ${\mathrm{Pb}}^{2+}$ ions which can combine with ${\mathrm{I}}^{-}$ to form ${\mathrm{PbI}}_2$.
• Hence Option C is incorrect.

Option D Potassium sulphate

• If Potassium sulphate is used in place of Potassium sulphate, lead iodide won't get precipitated, since the Potassium sulphate and Potassium iodide does not give ${\mathrm{Pb}}^{2+}$ ions which can combine with ${\mathrm{I}}^{-}$ to form ${\mathrm{PbI}}_2$.
• Hence Option D is incorrect.