Net Ionic Equation

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Caroline C 1G
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Net Ionic Equation

Postby Caroline C 1G » Wed Oct 04, 2017 3:45 pm

Fundamentals M
M.9 Copper (II) nitrate reacts with sodium hydroxide to produce a precipitate of light blue copper (II) hydroxide. a. Write the net ionic equation for the reaction.

I know we haven't done net ionic equations in class yet, but this question came up in one of the fundamental sections. Does anyone know how to get a net ionic equation?

Ryan Sydney Beyer 2B
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Re: Net Ionic Equation

Postby Ryan Sydney Beyer 2B » Wed Oct 04, 2017 5:17 pm

So the net ionic equation from my understanding is basically getting rid of anything that didn't participate in the reaction. A net ionic equation gets rid of all spectator ions.

So our equation is :: Cu(NO3)2 (aq) + 2 NaOH (aq) --> Cu(OH)2 (s) + 2NaNO3 (aq)

We then write out the total ionic equation which is where the electrolytes dissolved in aqueous solution are written as ions and will basically show us what didn't participate in the reaction.

This can be done like this :: Cu2+ (aq) + 2NO3- (aq) + 2Na+ (aq) + 2OH- (aq) --> Cu(OH)2 (s) + 2Na+ (aq) + 2NO3- (aq)

We can then see that nitrate doesn't participate in the reaction as it is in the aqueous state on the reactants side and the products side. The same is true for sodium. I think of it as a matching game and you match up and eliminate all the pairs that have the same signs, charges and states. So now we take out the nitrate and sodium on both sides of the equation.

This leaves us with our net ionic equation :: Cu2+ (aq) + 2OH- (aq) --> Cu(OH)2 (s)

Morgan Baxter 1E
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Re: Net Ionic Equation

Postby Morgan Baxter 1E » Wed Oct 04, 2017 6:05 pm

Yes I completely agree. The response above was stated very well. When thinking about net ionic equations, an important part is to remember the solubility rules. If you know your solubility rules (outlined within the fundamentals), then you will be able to determine which of the compounds in the products is a solid. The elements involved in the solid product will not be able to be eliminated on the reactants side of the equation because those elements play an active role in the reaction.

Ava Harvey 2B
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Re: Net Ionic Equation

Postby Ava Harvey 2B » Thu Oct 05, 2017 5:44 pm

Can someone please elaborate a little more on what happens to the sodium and nitrate ions? I'm still a little bit confused because in the net ionic equation the sodium and nitrate ions are not there, but they're still used in the calculations and mole ratios. So I'm not sure what ratios to use if they're not present in the equation. Thanks so much!

Guillermo Nuno 1J
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Re: Net Ionic Equation

Postby Guillermo Nuno 1J » Thu Oct 05, 2017 6:00 pm

Finding the Net Ionic Equation is essentially recognizing what changed in the reaction and what didn't. You eliminate what didn't change and keep what DID change. So an example would be:

NaCl (aq) + AgNO3 (aq) -> NaNO3 (aq) + AgCl (s)

We then break it up into a Total Ionic Equation

Na+ (aq) + Cl- (aq) + Ag+ 9 (aq) + NO3- (aq) ---> Na+ (aq) + NO3- (aq) + AgCl (s)

We see that both Na+ (aq) and NO3- (aq) have stayed the same on both sides of the equations so they can be crossed out.
All that we have left here would be Cl- (aq) + Ag+ 9 (aq) -> AgCl (s) because they went from (aq) to (s).
Like a matching game, as stated before.

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