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Does anyone know how to solve net ionic equations for chemistry???
I desperateley need to figure out how to solve net ionic equations for my chemistry class... i have a big test on it tomorrow and if i don't pass this test i will fail chemistry I have absolouteley no clue how to solve these problems except for the fact that the aqueous compounds need to be divided and the solid ones do not. I slso have NO idea how to figure out if a precipitate forms or not so if you know that please let me know... I have a practice worksheet with a few problems so i'll put a few on here for examples!! Thank You soooo much for you help
INSTRUCTIONS: Write a balanced net ionic equation for each reaction. If a precipitate forms identifiy it. If no precipitate forms then write no precipitate. (the numbers on the end of compounds should be smaller my computer will not let me make them the way they should be)
Pb(ClO3)2 (aq) + NaI (aq) ---->
Ca(OH)2 (aq) + H3 PO4 (aq)--->
Fe(NO3)3 (aq) + NaOH (aq)-->
NH4Cl (aq) + Pb(NO3)2 (aq)-->
2 Answers
- 1 decade agoFavorite Answer
You must separate each chemical into its constituent cations and anions to make this easier to see. ie, NaI---> Na+ and I- . Then pair off the anions and cations with equal charges that will react, and, using the rules of solubility, determine if the result is a precipitate. ie, lead is very insoluble. When paired with other anions, it is generally a precipitate. Hope this helps.
- 5 years ago
Molecular Equation: FeCl3 (aq) + Na3PO4 (aq) -----> FePO4 (s) + 3NaCl (aq) Total Ionic Equation: Fe3+ (aq) + 3Cl- (aq) + 3Na+ (aq) + PO43- (aq) ------> FePO4 (s) + 3Na+ (aq) + 3Cl- (aq) Net Ionic Equation: Fe3+ (aq) + PO43- (aq) ------> FePO4 (s) The only substances that you break up into ions are the soluble (aq) substances. Be sure to include the charge on each ion in the total and net ionic equations. Substances that are not broken apart are (s), (l), and (g). The net ionic equation includes only the ions involved in forming the (s), (l), or (g). Ions that are the same on both sides of the equation (spectator ions) are omitted from the net ionic equation.
