Question

What is the net ionic equation of the reaction of ${\text{MgS}}{{\text{O}}_4}$ with ${\text{Pb(N}}{{\text{O}}_3}{)_2}$ ?

Answer 1

The net ionic equation is the equation after the spectator ions have been canceled. Spectator ions are species that are not directly involved in the production of a gas, water, or insoluble product in a chemical reaction. The reaction will not take place if one of these three products is not produced.

Complete step by step solution:
We know that the given reaction is a precipitation reaction, in which two soluble aqueous salts are combined to create an insoluble salt known as a precipitate, which can then be filtered out and collected. So, we expect the compounds to exchange their ions to form new compounds.
We know that magnesium sulphate ( ${\text{MgS}}{{\text{O}}_4}$ ) when reacted with lead nitrate ( ${\text{Pb(N}}{{\text{O}}_3}{)_2}$ ) gives magnesium nitrate ( ${\text{Mg}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}$ ) and lead sulphate ( ${\text{PbS}}{{\text{O}}_4}$ ).
We will write the balanced equation first, to get the net ionic equation.
The balanced chemical equation is $MgS{O_4}\left( {aq} \right) + Pb{\left( {N{O_3}} \right)_2}\left( {aq} \right) \to Mg{\left( {N{O_3}} \right)_2}\left( {aq} \right) + PbS{O_4}(s) \downarrow$ .
Now we will split the compounds into ions to get the complete ionic equation, which is,
$M{g^{2 + }}\left( {aq} \right) + SO_4^{2 + }\left( {aq} \right) + P{b^{2 + }}\left( {aq} \right) + 2NO_3^ - \left( {aq} \right) \to M{g^{2 + }}\left( {aq} \right) + 2NO_3^ - \left( {aq} \right) + PbS{O_4}\left( s \right) \downarrow$
Note that we can not split the precipitate as it is no longer in aqueous state and has become a solid.
We have to now cancel the spectator ions. They can be identified as ions that are present on both sides of the equation.
${{M{g^{2 + }}\left( {aq} \right)}} + SO_4^{2 + }\left( {aq} \right) + P{b^{2 + }}\left( {aq} \right) + {{2NO_3^ - \left( {aq} \right)}} \to {{M{g^{2 + }}\left( {aq} \right)}} + {{2NO_3^ - \left( {aq} \right)}} + PbS{O_4}\left( s \right) \downarrow$
$\Rightarrow SO_4^{2 + }\left( {aq} \right) + P{b^{2 + }}\left( {aq} \right) \to PbS{O_4}\left( s \right) \downarrow$ .

Note:
The product lead (II) sulphate is an insoluble salt in white color. Note that precipitation reactions are always a double displacement reaction. But the converse double displacement reaction forms a precipitate need not be true.