If you wrote it correctly you don't have any charges. Nitric acid is HNO3. That is a compound and has no charge. Same for Ca(OH)2, Ca(NO3)2, and H2O.
This is getting really confusing for me. My teacher said that if its an ionic compound then it must have a charge in. On this sheet she gave me of common ions, hydroxide has a negative 1 charge. I also went back to some of the problems she did as an example in class, and she put charges on them. I have no idea what to do with these charges now. Like for Ca(OH)2 she put 2 charges on calcium and - 1 charge on OH. Then she put paranthesis around OH to make it 2 to match Ca.
OK. You may be trying to write ionic equations. If so, then the HNO3 would be written as H^+ + NO3^-. The Ca(OH)2 [Ca ion is a +2 charge and OH is a -1 charge which accounts for the formula as it is] is only partially soluble in water and may be written Ca(OH)2(s), Ca(NO3)2 is ionic and in solution dissolves easily so it consists of Ca^+2 and 2NO3^-. The H2O is a liquid, and is largely unionized, so we write it as H2O(l).
So the ionic equation would be written as
H^+ + NO3^- + Ca(OH)2(s) ==> Ca^+2 + 2NO3^- + H2O
I don't know how your teacher treats Ca(OH)2, s/he may have written it as Ca(OH)2(s), or as Ca(OH)2(aq), or as Ca^+2 + 2OH^-.
Is this close to what you are trying to do?
My teacher does not care that we put symbols as the end of our equations like (s) or (aq) but I think we are balancing ionic equations. But she puts +2 charge on Ca and -1 charge on OH
She also puts charge +2 charge on Calcium on the product side and a -1 charge on the NO3 as well.
For the final equation that I wrote (Unbalanced):
HNO3 (H has +1 charge, NO has -1) + Ca(OH)2 (Ca has +2 charge, OH has -1) -------> Ca(NO3)2 (Ca +2 charge, NO -1 charge) + H2O
That is what I wrote, would that be an ionic equation? Does that even look right? Also, I have trouble balanceing it.
The way I have been taught is to go the easy route and write, and balance, the molecular equation. Here is the molecular equation.
2HNO3 + Ca(OH)2 ==> Ca(NO3)2 + 2H2O
The we make that into the ionic equation.
The rules are to write those compounds that are ionic as ions, write solids, and unionized compounds as the molecule. Here we would have,
2H^+ + 2NO3^- + Ca^+2 + 2OH^- ==> Ca^+2 + 2NO3^- + 2H2O.
THEN, the next step is change this into a NET IONIC EQUATION. We do that by canceling ions common to both sides. In the above equation, We have Ca^+2 on both sides so cancel those. We have 2 NO3^- on both sides so cancel those. What's left?
2H^+ + 2OH^- ==> 2H2O and that is the net ionic equation. That may be where your teacher is going in the end. To answer an earlier question, you put the charges on the ions when you move from the molecular equation to the ionic equation.
Thanks bob. I think im going to ask my teacher about this. She probably made us put charges on them to help us because we just started doing it.
I agree with DrBob's approach. There is a need to add in the phase symbols as this does enable the equation to make more sense.
This experiment is usually done by dissolving solid Ca(OH)2 in nitric acid, other wise there is nothing to see.
So I would start from
2HNO3(aq) + Ca(OH)2(s) ==> Ca(NO3)2(aq) + 2H2O(l)
then the ions for the (aq) species
2H^+(aq) + 2NO3^-(aq) + Ca(OH)2(s)==> Ca^+2(aq) + 2NO3^-(aq) + 2H2O(l)
losing the common ions
2H^+(aq) + Ca(OH)2(s)==> Ca^+2(aq) + 2H2O(l)
which is what would be expected for dissolving Ca(OH)2 in an acid.
Does this help?