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

Answer: (B)

Lead acetate

$2KI\left(aq\right)+Pb\left(N{O}_3{)}_2\right(aq)\to Pb{I}_2(s)+2KN{O}_3$
In the above double displacement reaction, potassium Iodide(KI) and lead nitrate ($Pb(N{O}_3{)}_2$) dissociate in their aqueous states to form ions. The lead ($P{b}^{2+}$) ions combine with the iodide ($I^{-}$) ions to form precipitates of lead iodide ($Pb{I}_2$). If Lead sulphate is used in place of lead nitrate, no precipitates of lead iodide will be formed, because lead sulphate being insoluble in water does not give $P{b}^{2+}$ ions which can combine with $I^{-}$ to form $Pb{I}_2$. On the other hand, lead acetate dissociates in aqueous state to give $P{b}^{2+}$ ions and $C{H}_{3}CO{O}^{-}$ ions . Therefore potassium iodide combines with lead acetate to form precipitates of lead iodide and potassium acetate.
$2KI\left(aq\right)+Pb\left(C{H}_{3}COO{)}_2\right(aq)\to Pb{I}_2(s)+2C{H}_{3}COOK$