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(D) The reaction between potassium iodide $(KI)$ and lead nitrate $(Pb(NO_3)_2)$ is a double displacement reaction that produces a yellow precipitate

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Lead acetate

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(D) The reaction between potassium iodide $(KI)$ and lead nitrate $(Pb(NO_3)_2)$ is a double displacement reaction that produces a yellow precipitate of lead iodide $(PbI_2)$.$2KI(aq) + Pb(NO_3)_2(aq) \rightarrow PbI_2(s) + 2KNO_3(aq)$To obtain the yellow precipitate of lead iodide,we need a soluble lead salt that can provide lead ions $(Pb^{2+})$ in the aqueous solution.Lead sulphate is insoluble in water,so it cannot be used.Potassium sulphate and ammonium nitrate do not contain lead ions.Lead acetate $(Pb(CH_3COO)_2)$ is a soluble lead salt that dissociates to provide $Pb^{2+}$ ions,which react with iodide ions to form the yellow precipitate of lead iodide. Therefore,lead acetate is the correct substitute.

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?

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