Strontium nitrate and sulfuric acid equation

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Discussion Overview

The discussion revolves around the chemical reaction between strontium nitrate and sulfuric acid, specifically focusing on deriving the molecular, total ionic, and net ionic equations for the reaction. Participants explore the details of the equations and the cancellation of ions, as well as the concept of spectator ions in the context of double replacement reactions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a proposed molecular equation: Sr(NO3)2 + H2SO4 -> 2HNO3 + SrSO4, and seeks validation on the correctness of their total and net ionic equations.
  • Another participant emphasizes the importance of including charges and states in the ionic equations to identify aqueous species and precipitates, suggesting that if no state change occurs, the net ionic equation could indicate "No reaction."
  • Several participants discuss the cancellation of ions in the net ionic equation, with one asserting that the two hydrogen ions and two nitrate ions should cancel out, while another questions how the hydrogen ions should be treated in the cancellation process.
  • One participant clarifies that the net ionic equation should reflect the actual reaction, which involves the formation of a precipitate, and notes that spectator ions should be excluded from the final equation.
  • There is a discussion about the charge of the sulfate ion, with one participant correcting another's assumption about its charge in the context of the reaction.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of hydrogen ions in the net ionic equation and whether certain ions should be considered spectator ions. The discussion remains unresolved regarding the final form of the net ionic equation and the treatment of specific ions.

Contextual Notes

There are unresolved assumptions regarding the charges of ions and the definitions of spectator ions, as well as the specific conditions under which the net ionic equation is derived.

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Find the molecular, total ionic, and net ionic equation for the aqueous reaction between strontium nitrate and sulfuric acid.

I did like this, please check this out...

Sr(NO3)2 + H2SO4 -> 2HNO3 + SrSO4 is molecular equation.

Sr+2(aq) + 2NO3-1(aq) + H+1(aq) + HSO4-1(aq) -> 2H+1(aq) + 2NO3-1(aq) + SrSO4(s) is total ionic equation.

Sr+2(aq) + HSO4-1(aq) -> H+1(aq) + SrSO4(s) is net ionic equation.

Are they correct? I'm really confusing the last one. The hydrogen can be canceled out or not?

Please help me out...! Thanks...
 
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You should write the charges and states of each ion in the Net and Total ionic equations. You do this because you want to know what is aqueous and what is not. In the end equation there you should end up with one of those being a precipitate (solid) and then the finished equation would have only the two ions on the reactant side and the non aqueous product on the product side. If all of the molecules in the double replacement reaction are aqueous and no state change took place, then all of the ions would be considered spectator ions and thus the Net Ionic equation would actually be "No reaction" since all of the ions cancel out.
 
QuarkCharmer said:
You should write the charges and states of each ion in the Net and Total ionic equations. You do this because you want to know what is aqueous and what is not. In the end equation there you should end up with one of those being a precipitate (solid) and then the finished equation would have only the two ions on the reactant side and the non aqueous product on the product side. If all of the molecules in the double replacement reaction are aqueous and no state change took place, then all of the ions would be considered spectator ions and thus the Net Ionic equation would actually be "No reaction" since all of the ions cancel out.


=>>> I already know that part. I just want to check the final ionic equation. Anyway I edited.
 
The 2 hydrogen ions and 2 nitrate ions on both side of the equation should cancel out. So you should have only the ions that took place in the reacion in the form of a synthesis.

Like: X+ + Y- -> XY

On the left side of your net equation you have Hydrogen sulfate (aq), since it is aqueous, it should be a hydrogen ion and a sulfate ion, then the hydrogen would cancel the one on the product side.
 
QuarkCharmer said:
You should write the charges and states of each ion in the Net and Total ionic equations. You do this because you want to know what is aqueous and what is not. In the end equation there you should end up with one of those being a precipitate (solid) and then the finished equation would have only the two ions on the reactant side and the non aqueous product on the product side. If all of the molecules in the double replacement reaction are aqueous and no state change took place, then all of the ions would be considered spectator ions and thus the Net Ionic equation would actually be "No reaction" since all of the ions cancel out.

QuarkCharmer said:
The 2 hydrogen ions and 2 nitrate ions on both side of the equation should cancel out. So you should have only the ions that took place in the reacion in the form of a synthesis.

Like: X+ + Y- -> XY

=>>> I think H2SO4(aq) is dissociate like H+(aq) + HSO4-(aq).
How 2 hydrogens should be canceled out? And I canceled out 2NO3 each side.
 
Let's take a look at what exactly is happening here.

You have Sr(NO3)2 (aq) and you pour in H2(SO4)(aq). The double replacement rxn creates 2H(NO3)(aq) and Sr(SO4)(s).

The Net ionic equation (or end result) is that an Sr(aq) ion combines with an SO4(aq) ion to create Sr(SO4)(s).

Anything that is not reacting to produce the precipitate is considered a spectator and canceled out.

In this case, the 2 Hydrogen ions on the left (H2 is really 2 H atoms right?) are not taking place in the reaction at all, and therefor, they are present on both sides in aqueous form in the Total Ionic Equation. The nitrate (NO3) is also not taking place in the reaction to produce the precipitate, and as expected, in the Total Ionic Equation, there are the two nitrates on both sides. What you are left with is a solid of Strontium Sulfate and a bunch of Hydrogen and Nitrate ions floating about. But the purpose of the Net ionic is to show only those ions that were involved in the reaction to make the precipitate.

In your net ionic you have:

Sr+2(aq) + 2HSO4-1(aq) -> H+1(aq) + SrSO4(s)

The red is what is actually taking place. The polyatomic ion SO4 does not have a charge of -1, otherwise how could it initially be bonded to the diatomic molecule of Hydrogen (+2)?
 
Last edited:
QuarkCharmer said:
Let's take a look at what exactly is happening here.

You have Sr(NO3)2 (aq) and you pour in H2(SO4)(aq). The double replacement rxn creates 2H(NO3)(aq) and Sr(SO4)(s).

The Net ionic equation (or end result) is that an Sr(aq) ion combines with an SO4(aq) ion to create Sr(SO4)(s).

Anything that is not reacting to produce the precipitate is considered a spectator and canceled out.

In this case, the 2 Hydrogen ions on the left (H2 is really 2 H atoms right?) are not taking place in the reaction at all, and therefor, they are present on both sides in aqueous form in the Total Ionic Equation. The nitrate (NO3) is also not taking place in the reaction to produce the precipitate, and as expected, in the Total Ionic Equation, there are the two nitrates on both sides. What you are left with is a solid of Strontium Sulfate and a bunch of Hydrogen and Nitrate ions floating about. But the purpose of the Net ionic is to show only those ions that were involved in the reaction to make the precipitate.

In your net ionic you have:

Sr+2(aq) + 2HSO4-1(aq) -> H+1(aq) + SrSO4(s)

The red is what is actually taking place. The polyatomic ion SO4 does not have a charge of -1, otherwise how could it initially be bonded to the diatomic molecule of Hydrogen (+2)?

=>>> Thanks a lot... you helped me so much... : ) but... I don't get the first problem about light. Do you remember the question about spectrum?
 
I don't see a first problem about light?
 

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