I answered this at an earlier post. The second arrow should ber a + sign.
Ca(OH)2 + Na2CO3 ==> CaCO3 + 2NaOH
Is the answer you gave the ionic and net ionic equation....
No. My response only addresses why there is a second arrow. The second arrow shouldn't be there. It should be a + sign.
What do you not understand about ionic and net ionic equations. If we can get to your troubler spot you will be able to work these yourself.
i do not know how to break down the ions when i know that they are soluble.
for example...Ca(HO)2 is soluable but i dont know what to do with the two or what superscripts to use...
If they are soluble (note the spelling), you break them apart and give each the charge they possess. For example, Ca(OH)2 will break into Ca^+2 and 2OH^-.
These are just the valences (or oxidation number) of each. If the material is a precipitate (not soluble) we write it as CaCO3(s) or if it is a weak electrolyte (such as acetic acid, HC2H3O2, or water, H2O, we write them as the molecule. Gases are written as the molecule, also.)
Ca(OH)2(aq) + Na2CO3(aq)==>CaCO3(s) + 2NaOH(aq).
Then Ca(OH)2 becomes Ca^+2(aq) and 2OH^-(aq), Na2CO3 becomes 2Na^+(aq) and CO3^-2(aq), CaCO3 is written as CaCO3(s) and 2NaOH becomes 2Na^+(aq) + 2OH^-(aq).
To change this ionic equation to the net ionic equation you simply cancel ions common to both sides. For example, there are 2OH^- on each side so they cancel. There are two 2Na^+ on each side so they cancel. What is left is the net ionic equation. That is
Ca^+2(aq) + CO3^-2(aq) ==> CaCO3(s)
The ions that cancel are called spectator ions, as Bob Pursley wrote in an earlier post.
Thank you so much for giving me a step-by-step! I am homeschooled so i have no one to ask. I have had trouble with this and you are the only thing that has clearly told me how to do this. I only asked bob pursley b/c he gave me the answer last time and i was fustrated and looking for an easy way out. kudos to you!!!! :D
If you want to do KOH + HBr, we shall be glad to check it for you.
K^+1 + OH^-1 + H^+1 + Br(not sure which ionization number to put here)-> K^+1 + Br + H20 (l)
also, i am not sure what the other ions are.... (aq)? is my guess
Then the net ionic equation is...
OH^-1 + H^+1 -> H2O (l)
The net ionic equation you wrote is correct. Good work.
The complete equation is
KOH + HBr ==> KBr + H2O
(You would do well to write the complete molecular equation as a first step; otherwise, there is not a smooth flow from molecular to ionic to net ionic equation AND you must try to remember all of those ions since they aren't written anywhere.)
You may put the aq if you wish.
KOH(aq) + HBr(aq) ==> KBr(aq) + H2O
K^+ + OH^- + H^+ + Br^- (you know Br is -1 because H^+ is +1 and all compounds are zero charged (HBr is zero). ==> K^+(aq) + Br^-(aq)
Tben the K^+ and Br^- cancel.
(Also, Br is in group 17 (or VIIA depending upon the system you are using) and that entire column of F, Cl, Br, I, in column 17 are -1 charged ions in simple compounds such as NaCl, KBr, KI, etc.
two more questions....
1. so when you use the superscript ^-
that is the same as saying ^-1 ?
2. what are OH^-1 and H^+1 in the equations? (liquid , or gas)
Yes, any time a number is not written, the number 1 is understood; therefore, OH^- means OH^-1 and Na^+ means Na^+1. Likewise, we may write Ca^+2 or Ca^++ which would look like this, Ca+2 or Ca++.
If OH^- comes from KOH(aq) then OH^- is OH^-(aq). H^+ comes from HBr; therefore, H^+ is H^+(aq). So the net ionic equation for that KOH + HBr reaction ==> etc would be
H^+(aq) + (OH^-)(aq) ==> H2O(l) where (l) is "ell" (for liquid although on my screen it looks like a number 1)
Thank you once again for helping me!!!!