Question

When a solution of ammonium chloride, NH4Cl, is added to a solution of lead (II) nitrate , Pb(NO3)2, a white precipitate, PbCl2, forms. Which of the following is the total ionic equation for this reaction?
A) 2 NH4Cl(aq)+Pb(NO3)2(aq)-->PbCl2(s)+2 NH4NO3 (aq)
B) 2 NH4+(aq)+2Cl-(aq)+Pb2+(aq)+2NO3-(aq)-->PbCl2(s)+2NH4+(aq)+2NO3-(aq)
C) Pb2+(aq)+2Cl-(aq)-->PbCl2(s)
D) Pb2+(aq)+Cl2-(aq)-->PbCl2(s)

Answer 1

Answer:

B) $2NH_4^+_{(aq)}+2Cl^-_{(aq)}+Pb^{2+}_{(aq)}+2NO_3^-_{(aq)}\rightarrow PbCl_2_{(s)}+2NH^{4+}_{(aq)}+2NO_3^-_{(aq)}$

Explanation:

The total ionic equation contains all the species in the reaction in the ionic form.

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

Spectator ions are also written in total ionic equations.

The chemical equation is:

$2NH_4Cl_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbCl_2_{(s)}+2NH_4NO_3_{(aq)}$

The total ionic equation is -

$2NH_4^+_{(aq)}+2Cl^-_{(aq)}+Pb^{2+}_{(aq)}+2NO_3^-_{(aq)}\rightarrow PbCl_2_{(s)}+2NH^{4+}_{(aq)}+2NO_3^-_{(aq)}$