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