write the molecular, ionic and net ionic equation for the following
Ph(NO3)2 + NaOH
Pb(NO3)2 + Na2CO3
Ni(NO3)2 + NaOH
Ni(NO3)2 + Na2Co3
recognize what is going to precipate, that takes those ions in solution out.
For instance on first.
Pb(NO3)2+NaOH>> Pb(OH)2 precipates, so net ionic is ...
Pb++ + OH- >> Pb(OH)2
To write the molecular, ionic, and net ionic equations for the given reactions, we first need to identify the formulas of the reactants and products involved.
1. Ph(NO3)2 + NaOH:
The reactants are Ph(NO3)2 (phenyl nitrate) and NaOH (sodium hydroxide). To write the equation, we need to determine their formulas.
- Ph(NO3)2: The compound contains the Ph (phenyl) cation and the NO3 (nitrate) anion. The formula for phenyl nitrate is C6H5NO3.
- NaOH: This compound consists of the Na (sodium) cation and OH (hydroxide) anion. The formula for sodium hydroxide is NaOH.
The molecular equation for the reaction is:
C6H5NO3 + NaOH → C6H5NO3 + NaOH
Now, let's break the molecular equation into its ionic form:
C6H5NO3 + NaOH → C6H5NO3 + NaOH
Na+ + OH- → Na+ + OH-
The net ionic equation removes the spectator ions (those that appear on both sides of the equation):
OH- → OH-
2. Pb(NO3)2 + Na2CO3:
The reactants are Pb(NO3)2 (lead(II) nitrate) and Na2CO3 (sodium carbonate). Let's determine their formulas:
- Pb(NO3)2: The compound consists of the Pb2+ (lead) cation and the NO3- (nitrate) anion. The formula for lead(II) nitrate is Pb(NO3)2.
- Na2CO3: This compound contains Na+ (sodium) cation and CO3^2- (carbonate) anion. The formula for sodium carbonate is Na2CO3.
The molecular equation for the reaction is:
Pb(NO3)2 + Na2CO3 → Pb(NO3)2 + Na2CO3
Next, let's break it down into its ionic form:
Pb2+ + 2NO3- + 2Na+ + CO3^2- → Pb2+ + 2NO3- + 2Na+ + CO3^2-
The net ionic equation eliminates the spectator ions:
CO3^2- → CO3^2-
3. Ni(NO3)2 + NaOH:
The reactants are Ni(NO3)2 (nickel(II) nitrate) and NaOH (sodium hydroxide). We determine their formulas as follows:
- Ni(NO3)2: The compound contains the Ni2+ (nickel) cation and the NO3- (nitrate) anion. The formula for nickel(II) nitrate is Ni(NO3)2.
- NaOH: This compound has the Na+ (sodium) cation and OH- (hydroxide) anion. The formula for sodium hydroxide is NaOH.
The molecular equation for the reaction is:
Ni(NO3)2 + 2NaOH → Ni(NO3)2 + 2NaOH
Breaking it down into its ionic form:
Ni2+ + 2NO3- + 2Na+ + 2OH- → Ni2+ + 2NO3- + 2Na+ + 2OH-
The net ionic equation removes the spectator ions:
OH- → OH-
I hope this helps you understand how to write the molecular, ionic, and net ionic equations for these reactions!