Question

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?

• A. Lead sulphate (insoluble)
• B. Lead acetate
• C. Ammonium nitrate
• D. Potassium sulphate

Answer 1

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.