what is the net ionic for these following balanced equations:
2(NH4)3PO4(aq) + 3Ba(OH)2(aq) -> Ba(PO4)2(s) + 6H2O(l) + 6NH3(g)
2HNO3(aq) + K2CO3(aq) -> H2O(l) + CO2(g) +2KNO3
NaF(aq) + HCl(aq) -> NaCl(aq) + HF(aq)
Reactions go to completion for one of just a few reasons.
1. A ppt (insoluble material) is formed. You need to know the solubility rules to know what is soluble and what isn't. Here is a simplified table.
http://en.wikipedia.org/wiki/Solubility_chart
2. A gas is formed. (H2, O2, Cl2, F2, N2, CO2, SO2, NO2 are few common ones).
3. A slightly ionized substance (weak acid/weak base/water) are formed.
4. Redox reaction.
If one of these doesn't happen generally there is no reaction. To write the net ionic equation, remember that if it is a gas, insoluble, or weak electrolyte formed, you write that as the molecule. If none of those, the material isn't listed. For example, for #1, Ba3(PO4)2 is insoluble (rule 1) above and the net ionic equation is
2PO4^3-(aq) + 3Ba^2+(aq) => Ba3(PO4)2(s)
#3 is rule 3 above.
F^-(aq) + H^+(aq) ==> HF(aq)
I will leave 2 for you but note CO2.
To find the net ionic equation for each reaction, you need to first write out the complete balanced equations and then use the solubility rules to determine which ions dissociate (break apart) into their individual components in the reaction.
1. 2(NH4)3PO4(aq) + 3Ba(OH)2(aq) -> Ba(PO4)2(s) + 6H2O(l) + 6NH3(g)
The dissociation of the compounds are as follows:
(NH4)3PO4(aq) -> 3NH4+(aq) + PO4^3-(aq)
Ba(OH)2(aq) -> Ba^2+(aq) + 2OH-(aq)
Using the solubility rules, we know that Ba3(PO4)2 is insoluble and precipitates out as a solid.
Thus, the net ionic equation is:
3NH4+(aq) + PO4^3-(aq) + 3Ba^2+(aq) + 6OH-(aq) -> Ba3(PO4)2(s) + 6H2O(l)
2. 2HNO3(aq) + K2CO3(aq) -> H2O(l) + CO2(g) + 2KNO3
The dissociation of the compounds are as follows:
HNO3(aq) -> H+(aq) + NO3-(aq)
K2CO3(aq) -> 2K+(aq) + CO3^2-(aq)
We can see that H+(aq) and CO3^2-(aq) react and form water (H2O(l)) and carbon dioxide gas (CO2(g)). The other ions, K+ and NO3-, remain unchanged.
Thus, the net ionic equation is:
H+(aq) + CO3^2-(aq) -> H2O(l) + CO2(g)
3. NaF(aq) + HCl(aq) -> NaCl(aq) + HF(aq)
The dissociation of the compounds are as follows:
NaF(aq) -> Na+(aq) + F-(aq)
HCl(aq) -> H+(aq) + Cl-(aq)
In this case, the Na+(aq) and Cl-(aq) ions do not participate in any reactions with each other. However, the F-(aq) and H+(aq) ions react to form the molecular compound HF(aq).
Thus, the net ionic equation is:
F-(aq) + H+(aq) -> HF(aq)
To find the net ionic equation for each balanced equation, follow these steps:
Step 1: Write out the balanced chemical equation.
Step 2: Write the complete ionic equation by breaking down all the soluble ionic compounds into their individual ions.
Step 3: Identify the spectator ions, which are the ions that appear on both sides of the equation and do not participate in the chemical reaction.
Step 4: Write the net ionic equation by removing the spectator ions from the complete ionic equation.
Let's apply these steps to the given balanced equations:
1. 2(NH4)3PO4(aq) + 3Ba(OH)2(aq) -> Ba(PO4)2(s) + 6H2O(l) + 6NH3(g)
Step 1: Balanced equation is already provided.
Step 2: Complete ionic equation:
2(NH4)3PO4(aq) + 3Ba(OH)2(aq) -> 2Ba(PO4)(s) + 6NH4OH(aq)
Step 3: Spectator ions: NH4+ and OH-
Step 4: Net ionic equation:
Ba(OH)2(aq) -> Ba(PO4)(s) + 2H2O(l)
2. 2HNO3(aq) + K2CO3(aq) -> H2O(l) + CO2(g) + 2KNO3
Step 1: Balanced equation is already provided.
Step 2: Complete ionic equation:
2H+(aq) + 2NO3-(aq) + 2K+(aq) + CO3^2-(aq) -> 2H2O(l) + CO2(g) + 2K+(aq) + 2NO3-(aq)
Step 3: Spectator ions: 2K+ and 2NO3-
Step 4: Net ionic equation:
2H+(aq) + CO3^2-(aq) -> H2O(l) + CO2(g)
3. NaF(aq) + HCl(aq) -> NaCl(aq) + HF(aq)
Step 1: Balanced equation is already provided.
Step 2: Complete ionic equation:
Na+(aq) + F-(aq) + H+(aq) + Cl-(aq) -> Na+(aq) + Cl-(aq) + H+(aq) + F-(aq)
Step 3: Spectator ions: Na+ and Cl-
Step 4: Net ionic equation:
F-(aq) + H+(aq) -> HF(aq)
Therefore, the net ionic equations for the given balanced equations are:
1. Ba(OH)2(aq) -> Ba(PO4)(s) + 2H2O(l)
2. 2H+(aq) + CO3^2-(aq) -> H2O(l) + CO2(g)
3. F-(aq) + H+(aq) -> HF(aq)