Write the net ionic equation for any precipitation reaction that occurs when .1M solutions of the following are mixed.

Does the molarity affect the equation at all, or is it just a normal net ionic equation?

If the solutions are too dilute A might not ppt with B to form AB but for most 0.1M solutions I think you can ignore that. Post any questions you have and I'll be glad to look at them to see if a ppt forms.

the questions are
a. zinc nitrate and nickel (II) chloride
b. potassium phosphate and calcium nitrate
c. sodium hydroxide and zinc nitrate
d. iron (III) nitrate and barium hydroxide

would i be able to just ignore the molarity and write the equations as I normally would?

yes

Yes, for precipitation reactions, the molarity of the solutions does not affect the net ionic equation. The net ionic equation only shows the species that are directly involved in the precipitation reaction. The molarity of the solutions affects the rate of reaction and the amount of precipitate formed, but not the overall reaction itself.

Here are the net ionic equations for each of the given reactions:

a. Zinc nitrate and nickel (II) chloride:
Zn2+(aq) + 2Cl-(aq) -> ZnCl2(s)

b. Potassium phosphate and calcium nitrate:
3Ca2+(aq) + 2PO4^3-(aq) -> Ca3(PO4)2(s)

c. Sodium hydroxide and zinc nitrate:
Zn2+(aq) + 2OH-(aq) -> Zn(OH)2(s)

d. Iron (III) nitrate and barium hydroxide:
Fe3+(aq) + 3OH-(aq) -> Fe(OH)3(s)

In general, molarity does not affect the net ionic equations of precipitation reactions. The net ionic equation represents only the species that are directly involved in the formation of the precipitate, and the molarity of the solutions does not play a role in determining these species.

To write the net ionic equation for a precipitation reaction, you need to follow these steps:

1. Write the balanced chemical equation for the overall reaction.
2. Identify the strong electrolytes (ionic compounds that dissociate completely into ions) in the reaction.
3. Write the complete ionic equation, showing all the ions that are present in the reaction.
4. Cancel out the spectator ions, which are ions that do not participate in the formation of the precipitate, on both sides of the equation.
5. Write the final net ionic equation, which includes only the ions that are involved in the formation of the precipitate.

Now, let's go through the questions you provided:

a. Zinc nitrate and nickel (II) chloride:
- Balanced chemical equation: Zn(NO3)2(aq) + 2NiCl2(aq) -> 2Ni(NO3)2(aq) + ZnCl2(aq)
- Complete ionic equation: Zn2+(aq) + 2NO3-(aq) + 2Ni2+(aq) + 2Cl-(aq) -> 2Ni2+(aq) + 2NO3-(aq) + Zn2+(aq) + 2Cl-(aq)
- Net ionic equation: Zn2+(aq) + 2Cl-(aq) -> ZnCl2(s)

b. Potassium phosphate and calcium nitrate:
- Balanced chemical equation: 3K3PO4(aq) + 2Ca(NO3)2(aq) -> Ca3(PO4)2(s) + 6KNO3(aq)
- Complete ionic equation: 3K+(aq) + 3PO4^3-(aq) + 2Ca2+(aq) + 2NO3-(aq) -> Ca3(PO4)2(s) + 6K+(aq) + 6NO3-(aq)
- Net ionic equation: 3Ca2+(aq) + 2PO4^3-(aq) -> Ca3(PO4)2(s)

c. Sodium hydroxide and zinc nitrate:
- Balanced chemical equation: 2NaOH(aq) + Zn(NO3)2(aq) -> 2NaNO3(aq) + Zn(OH)2(s)
- Complete ionic equation: 2Na+(aq) + 2OH-(aq) + Zn2+(aq) + 2NO3-(aq) -> 2Na+(aq) + 2NO3-(aq) + Zn(OH)2(s)
- Net ionic equation: 2OH-(aq) + Zn2+(aq) -> Zn(OH)2(s)

d. Iron (III) nitrate and barium hydroxide:
- Balanced chemical equation: 6Fe(NO3)3(aq) + 8Ba(OH)2(aq) -> Ba3(PO4)2(s) + 18HNO3(aq) + 6Fe(OH)3(s)
- Complete ionic equation: 6Fe3+(aq) + 18NO3-(aq) + 8Ba2+(aq) + 16OH-(aq) -> Ba3(PO4)2(s) + 18NO3-(aq) + 18H+(aq) + 6Fe(OH)3(s)
- Net ionic equation: 6Fe3+(aq) + 16OH-(aq) -> 6Fe(OH)3(s)

As you can see, the molarity of the solutions does not affect the process of writing the net ionic equations. However, it is essential to ensure that the reactions are indeed precipitation reactions, and the solutions are not too dilute for a precipitate to form.