# Homework Help: Net Ionic Equation

1. Jul 18, 2006

### leebongemail

Problem: When a solution of sodium hydroxide is added to a solution of ammonium carbonate, and the solution is heated, ammonia gas, (NH3) is released. Write a net ionic equation for this reaction. Hint: both NaOH and (NH4)2CO3 exist as dissociated ions in aqueous solution.

Can someone help me through the steps to find the net ionic equation?
I need help to begin writing the balanced equation for this system.

2. Jul 18, 2006

### bubbles

First, you need to write a balanced molecular equation.
Then, write an ionic equation that includes all the ions in the reaction.
Third, cross out the spectator ions on both sides of your equation and you get the net ionic equation.
Remember to check solubility rules.

Hint: It is a double displacement reaction.

Last edited: Jul 18, 2006
3. Jul 18, 2006

### leebongemail

Hm, well for the first step i got

NaOH+(NH4)2CO3 <---->NaCO3+NH3+H20

is that correct?

4. Jul 18, 2006

### bubbles

The sodium ion has a 1+ charge whereas the carbonate ion has a 2- charge, so your equation is wrong. (It's Na2CO3)

Last edited: Jul 18, 2006
5. Jul 18, 2006

### leebongemail

so the equation is NaOH+(NH4)2CO3 <---> Na2CO3+NH3+H2O?

6. Jul 18, 2006

### bubbles

Yes, but you have to balance the equation.

7. Jul 18, 2006

### leebongemail

2NaOH+(NH4)2CO3 <---> 2Na2CO3+2NH3+2H2O

is this balanced equation correct?

Last edited: Jul 18, 2006
8. Jul 18, 2006

### PPonte

Hint: Count the number of $$Na^+$$ that you have in both members.

:tongue:

If you have difficulties balancing chemical equations, you may want to see-
I always use this method when I can't figure out the stoichiometric coefficients by trial-error.

http://en.wikipedia.org/wiki/Chemical_equations
http://www.studyworksonline.com/cda/content/article/0,,EXP1315_NAV2-100_SAR1316,00.shtml [Broken]

Last edited by a moderator: May 2, 2017
9. Jul 19, 2006

### leebongemail

where did K come from?

10. Jul 19, 2006

### sdekivit

First you need to know what ions your solution contains. Ammoniumcarbonate solution contains:

$$NH_{4} ^{+}$$ and $$CO_{3} ^{2-}$$

Sodium hydroxide solution contains:

$$Na^{+}$$ and $$OH^{-}$$

We already know that $$NH_{3}(g)$$ will be formed and so you know that the ammonium ions will react with the hydroxide ions. The sodium carobonate that will be formed is soluble in water and thus don't need to be taken up in the ionic equation. Thus the net ionic reaction occuring here is:

$$NH_{4} ^{+} + OH^{-} \rightarrow NH_{3} + H_{2}O$$

remember that you added 2 ion conataining solution with each other, thus the ions already exist. When solid ammonium carbonate is added to a sodium hydroxide solution, you refer to $$(NH_{4})_{2}CO_{3}$$ in the equation, because it needs to be dissolved first.

Last edited: Jul 19, 2006
11. Jul 19, 2006

### leebongemail

is there a website that has the common or many ions for an element? I'm taking a summer introduction to chem class at this community college, and it goes by really fast. does someone recommend a site? i can sure use it. I have a test tomorrow on 'Acids and Bases' and 'Reaction Rates and Chemical Equilibrium'. I'm sure to have questions later in the day!:tongue:

12. Jul 19, 2006

### bubbles

Last edited by a moderator: Apr 22, 2017