Question

what are the products of the precipitation reaction between aqueous potassium iodide, ki, and aqueous lead(ii) nitrate, pb(no₃)₂?
○ pbi₂ and kno₃
○ pbi and k(no₃)₂
○ pbk and ino₃
○ pbi and kno₃

Explanation:

Step1: Identify Reaction Type

This is a double - displacement (precipitation) reaction. In double - displacement reactions, the general form is \(AB + CD
ightarrow AD+CB\), where the cations and anions of the two reactants switch places. For the reaction between \(KI\) (where \(A = K^+\), \(B=I^-\)) and \(Pb(NO_3)_2\) (where \(C = Pb^{2+}\), \(D = NO_3^-\)).

Step2: Determine Products

The cations \(K^+\) and \(Pb^{2+}\) will switch anions. The anion of \(KI\) is \(I^-\) and the anion of \(Pb(NO_3)_2\) is \(NO_3^-\).

• The product formed from \(Pb^{2+}\) and \(I^-\): The charge of \(Pb\) is \(+ 2\) and the charge of \(I\) is \(-1\). To form a neutral compound, we need 2 \(I^-\) ions for each \(Pb^{2+}\) ion. So the formula is \(PbI_2\) (since \(Pb^{2+}+2I^-

ightarrow PbI_2\)).

• The product formed from \(K^+\) and \(NO_3^-\): The charge of \(K\) is \(+1\) and the charge of \(NO_3^-\) is \(- 1\). So the formula is \(KNO_3\) (since \(K^++NO_3^-

ightarrow KNO_3\)).

So the products of the reaction between \(KI\) and \(Pb(NO_3)_2\) are \(PbI_2\) and \(KNO_3\).

Answer:

A. \(PbI_2\) and \(KNO_3\)