M9 Part A

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Nancy Le - 1F
Posts: 52
Joined: Fri Sep 29, 2017 7:07 am

M9 Part A

Postby Nancy Le - 1F » Fri Oct 06, 2017 12:32 am

How do you find the net ionic equation for the reaction?

deeksha1I
Posts: 21
Joined: Fri Sep 29, 2017 7:04 am

Re: M9 Part A

Postby deeksha1I » Fri Oct 06, 2017 12:40 am

We haven't covered this topic in class yet. But from my understanding, net ionic equations are basically two charged elements that undergo a chemical reaction. So in this case, Cu has a charge of 2+ and OH has a charge of 1-, so when Cu(OH)subscript2 forms, there has to be 2 OH's to balance out the charge. And since there are 2 OH's on the right, you need two OH's on the left.
Hope this helps!

Nehal Banik
Posts: 64
Joined: Thu Jul 13, 2017 3:00 am

Re: M9 Part A

Postby Nehal Banik » Fri Oct 06, 2017 2:18 am

To find the net ionic, you have to first understand what product is going to become a solid and which reactants are soluble meaning they became aqueous, so, the precipitate which is blue copper hydroxide is actually a solid and the reactants are soluble. The copper nitrate is soluble and so is the sodium hydroxide because hydroxides don't dissolve however sodium hydroxide will dissolve, therefore it becomes this
Cu^(2+)+2NO3+2Na+2OH-->Cu(OH)solid+2Na+2NO3

Cancel out the reactants and products that are the same on each side, these are called spectator ions, and you should have Cu^(2+)aq+2OHaq-->Cu(OH)2

Hope this help! I'm trying to explain this to the best of my ability.


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