Write the complete and net ionic equations for the following reactions. Be sure to indicate the states of the reaction products.
1) AgNO3(aq) + NaCl(aq) ---->
2) CaCl2(aq) + Na2SO4(aq) ---->
3) Pb(ClO4)2(aq) + NaOH(ag) ---->
4) (NH4)2SO4(aq) + BaCl2(aq) ---->
I'll do #1.
Step 1. Write the molecular equation.
AgNO3(aq) + NaCl(aq) = AgCl + NaNO3
Step 2. If a reaction occurs, it is because
a. a precipitate forms (an insoluble material forms). For this you need to know the solubilities of the salts. Here is a simplified chart for solubility.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
b. A gas is formed.
c. A slightly ionized material is formed (usually water).
Step 3. Now you can take the products and add their state.
AgNO3(aq) + NaCl(aq) ==> AgCl(s) + NaNO3(aq)
You will note from the solubility chart that AgCl is not soluble and NaNO3 is; therefore, I place a (s) by AgCl and (aq) by NaNO3.
Step 4. Next you convert to the FULL ionic equation. (Anything NOT in step 2a,b, or c gets converted to ions.)
Ag^+ + NO3^- + Na^+ + Cl^- ==> AgCl(s) + Na^+ + NO3^-
Note: All of the soluble materials should have (aq) by them but that clutters the screen and I have omitted them. You need to write these in for your instructor. For example, part of the equation would look like this
Ag^+(aq) + NO3^-(aq) + Na^+(aq) + Cl^-(aq) etc etc. AgCl does NOT have (aq) since it is insoluble but the NaNO3 does.
Step 5. Cancel ions common to both sides. I see Na^+ on both sides and I see NO3^- on both sides. Cancel those.
Step 6. What is left is the NET ionic equation.
Ag^+(aq) + Cl^-(aq) ==> AgCl(s)
Students often ask what is the purpose of a net ionic equation. This last equation shows you what actually happens in the solution. The Ag^+(aq) reacts with Cl^(aq) to form insoluble AgCl(s). What about the other ions? It doesn't matter what they are. They are just spectator ions. They don't enter into the reaction. So AgNO3 + NaCl gives the same net ionic equation as AgNO3 and KCl or LiCl or any other soluble chloride.
THANK YOU!!!!
1) AgNO3(aq) + NaCl(aq) ----> AgCl(s) + NaNO3(aq)
Complete Ionic Equation:
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) ----> AgCl(s) + Na+(aq) + NO3-(aq)
Net Ionic Equation:
Ag+(aq) + Cl-(aq) ----> AgCl(s)
2) CaCl2(aq) + Na2SO4(aq) ----> CaSO4(s) + 2NaCl(aq)
Complete Ionic Equation:
Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) + SO4^2-(aq) ----> CaSO4(s) + 2Na+(aq) + 2Cl-(aq)
Net Ionic Equation:
Ca2+(aq) + SO4^2-(aq) ----> CaSO4(s)
3) Pb(ClO4)2(aq) + 2NaOH(aq) ----> Pb(OH)2(s) + 2NaClO4(aq)
Complete Ionic Equation:
Pb2+(aq) + 2ClO4-(aq) + 2Na+(aq) + 2OH-(aq) ----> Pb(OH)2(s) + 2Na+(aq) + 2ClO4-(aq)
Net Ionic Equation:
Pb2+(aq) + 2OH-(aq) ----> Pb(OH)2(s)
4) (NH4)2SO4(aq) + BaCl2(aq) ----> BaSO4(s) + 2NH4Cl(aq)
Complete Ionic Equation:
2NH4+(aq) + SO4^2-(aq) + Ba2+(aq) + 2Cl-(aq) ----> BaSO4(s) + 2NH4+(aq) + 2Cl-(aq)
Net Ionic Equation:
SO4^2-(aq) + Ba2+(aq) ----> BaSO4(s)
To write the complete and net ionic equations for these reactions, we need to identify the reactants, determine their chemical formulas, and write the products formed after the reaction. Let's go step by step:
1) AgNO3(aq) + NaCl(aq) ---->
First, let's identify the reactants:
- AgNO3(aq): This is silver nitrate, which dissociates in water to form Ag+ and NO3- ions.
- NaCl(aq): This is sodium chloride, which also dissociates in water to form Na+ and Cl- ions.
Now, let's write the products:
- AgNO3(aq) + NaCl(aq) --> AgCl(s) + NaNO3(aq)
The complete ionic equation shows all the ions involved in the reaction:
- Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) --> AgCl(s) + Na+(aq) + NO3-(aq)
The net ionic equation removes the spectator ions (the ones that appear unchanged on both sides of the equation):
- Ag+(aq) + Cl-(aq) --> AgCl(s)
2) CaCl2(aq) + Na2SO4(aq) ---->
For this reaction, let's identify the reactants:
- CaCl2(aq): This is calcium chloride, which dissociates in water to form Ca2+ and Cl- ions.
- Na2SO4(aq): This is sodium sulfate, which dissociates in water to form Na+ and SO4(2-) ions.
Now, let's write the products:
- CaCl2(aq) + Na2SO4(aq) --> CaSO4(s) + 2NaCl(aq)
The complete ionic equation shows all the ions involved in the reaction:
- Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) + SO4(2-)(aq) --> CaSO4(s) + 2Na+(aq) + 2Cl-(aq)
The net ionic equation removes the spectator ions:
- Ca2+(aq) + SO4(2-)(aq) --> CaSO4(s)
3) Pb(ClO4)2(aq) + NaOH(aq) ---->
Let's identify the reactants:
- Pb(ClO4)2(aq): This is lead(II) perchlorate, which dissociates in water to form Pb2+ and ClO4- ions.
- NaOH(aq): This is sodium hydroxide, which dissociates in water to form Na+ and OH- ions.
Now, let's write the products:
- Pb(ClO4)2(aq) + 2NaOH(aq) --> Pb(OH)2(s) + 2NaClO4(aq)
The complete ionic equation shows all the ions involved in the reaction:
- Pb2+(aq) + 2ClO4-(aq) + 2Na+(aq) + 2OH-(aq) --> Pb(OH)2(s) + 2Na+(aq) + 2ClO4-(aq)
The net ionic equation removes the spectator ions:
- Pb2+(aq) + 2OH-(aq) --> Pb(OH)2(s)
4) (NH4)2SO4(aq) + BaCl2(aq) ---->
Let's identify the reactants:
- (NH4)2SO4(aq): This is ammonium sulfate, which dissociates in water to form 2NH4+ and SO4(2-) ions.
- BaCl2(aq): This is barium chloride, which dissociates in water to form Ba2+ and 2Cl- ions.
Now, let's write the products:
- (NH4)2SO4(aq) + BaCl2(aq) --> BaSO4(s) + 2NH4Cl(aq)
The complete ionic equation shows all the ions involved in the reaction:
- 2NH4+(aq) + SO4(2-)(aq) + Ba2+(aq) + 2Cl-(aq) --> BaSO4(s) + 2NH4+(aq) + 2Cl-(aq)
The net ionic equation removes the spectator ions:
- SO4(2-)(aq) + Ba2+(aq) --> BaSO4(s)
Remember to indicate the states of the reaction products (s for solid, aq for aqueous) as shown in the final equations above.