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Help with M9

Posted: Tue Oct 03, 2017 11:28 am
by 304922790
Why is the Na canceled out in the equation (according to the solutions manual)? There is a similar questions posted previously but the answer was not very detailed and I am still confused.

Thanks for your help in advance! :)

Re: Help with M9

Posted: Tue Oct 03, 2017 1:36 pm
by Hannah Chew 2A
I actually missed this too, but I realize now that the question asks for the net ionic equation. This means that spectator ions like sodium wouldn't be part of the equation unless it was asking for the complete ionic equation.

Re: Help with M9

Posted: Tue Oct 03, 2017 8:52 pm
by 304922790
Hmmm... I only took high school chemistry so I don't really know/ remember what spectator ions are. Can you please expand that for me please? :)

Re: Help with M9

Posted: Tue Oct 03, 2017 9:24 pm
by Kourtney Nham 1L
Hi! Spectator ions are ions whose forms aren't changed during the reaction. In the question, the original equation would be Cu(NO3)2 + 2NaOH --> Cu(OH)2 + 2NaNO3. Based on solubility rules, the equation in an ionic form would look like Cu2+ + 2NO3- + 2Na+ + 2OH- --> Cu(OH)2 + 2Na+ + 2NO3-. As you can see, both the sodium ions and nitrate ions are unchanged on both sides of the reaction while the copper and hydroxide ions form the precipitate. Net ionic equations are meant to show which ions are actually reacting in the reaction, so you can "cross out" the sodium ions and nitrate ions which leaves you with the answer.

Re: Help with M9

Posted: Wed Oct 04, 2017 11:44 pm
by Nina Gautam 1K
In ionic equations, how do you know when to split up molecules (as Kourtney did for both reactant molecules) vs keeping molecules together? (Such as keeping Cu(OH)2 together in the products but splitting up 2NaNO3 to 2Na+ + 2NO3-?

Re: Help with M9

Posted: Thu Oct 05, 2017 10:58 am
by Charles Ang 1E
Nina Gautam 1L wrote:In ionic equations, how do you know when to split up molecules (as Kourtney did for both reactant molecules) vs keeping molecules together? (Such as keeping Cu(OH)2 together in the products but splitting up 2NaNO3 to 2Na+ + 2NO3-?


When in an aqueous solution, solubility rules apply. 2NaNO3 is soluble (most molecules with NO3 are), therefore it is split up when writing the ionic equation.