I am currently doing a lab write up and i have a little problem with some ionic equations

the question goes as follows:

You are provided with aqueous solutions of CaCl2, AlCl3, ZnCl2, CuCl2, FeCl2 and FeCl3 in six separate test tubes. To each of the solutions add 2-3 drops of dilute NH3 solution. Record your observations and illustrate changes observed with an ionic equation. Then continue to add NH3 solution until it is in excess.Record any changes observed.

My observations

CaCl2-no ppt
AlCl3-white ppt, insoluble in excess
Zncl2-white ppt, soluble in excess
CuCl2- light blue ppt, soluble in excess
FeCl2-dirty green ppt, insoluble in excess
FeCl3-red brown ppt, insoluble in excess

so the first ionic equation would be
Ca^2+ + 2Cl^- + N^3- +3H^+ yielding the same products right?

but then what will happen to the rest
will the products of the second one for example be AlN or Al(OH)3 ?

Could you please help me with the remaining ionic equations??

thank you
Marysia

By adding the NH3 solution (actually NH4OH), you are creating...

NH4+ and OH-

so you add those two ions to the metal and chloride ions. YOu know the product of ammonium and chloride will be soluble, so any insoluble product will be the metal oxide compound, for instance, aluminum hydroxide.
On your results, calcium hydroxide is not very soluble, so if there were plenty of calcium chloride, you should have gotten a ppt when in excess.

Oh and i have on emore question

for the CuCl2 example i've done some research and what i came up with is
[Cu(NH3)4(H2O)2]^2+, but this product is aqueous and i need to find the temporary ppt that was produced before i added excess NH3 solution

sorry for more complications...

okay thanks bobpursely

that explains a lot
and regarding the CaCl2 and CaOH ppt, i think there was a slight ppt but my teacher let me ignore it so i guess im going to stick to the no ppt result

thanks a lot for helping

love marysia

Yes, you can form that complex, in dilute concentrations. Is Cu(OH)2 the right color?

http://en.wikipedia.org/wiki/Copper_hydroxide

i think i can ignore the complex, because the question is asking for the equation before i add excess and so i only nees to write an equation for the ppt

but i'm just curious could the other metals also form complexes like the Cu does?

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Certainly, Marysia! I can help you with the remaining ionic equations.

Let's start with the first ionic equation you provided: Ca^2+ + 2Cl^- + N^3- + 3H^+ → Ca^2+ + 2Cl^- + N^3- + 3H^+. The products are the same as the reactants because none of them react with each other.

Now, let's move on to the other solutions:

1. AlCl3: You observed a white precipitate that is insoluble in excess ammonia solution. In this case, the precipitate is likely Al(OH)3. The ionic equation would be: Al^3+ + 3Cl^- + 3NH3 + 3H2O → Al(OH)3 + 3Cl^- + 3NH4+.

2. ZnCl2: You observed a white precipitate that is soluble in excess ammonia solution. This suggests the formation of a complex ion. The ionic equation would be: Zn^2+ + 2Cl^- + xNH3 → [Zn(NH3)x]^2+ + 2Cl^-. The value of x depends on the amount of excess ammonia added.

3. CuCl2: You observed a light blue precipitate that is soluble in excess ammonia solution. Again, this indicates the formation of a complex ion. The ionic equation would be: Cu^2+ + 2Cl^- + yNH3 → [Cu(NH3)y]^2+ + 2Cl^-. The value of y depends on the amount of excess ammonia added.

4. FeCl2: You observed a dirty green precipitate that is insoluble in excess ammonia solution. It is likely Fe(OH)2. The ionic equation would be: Fe^2+ + 2Cl^- + 2NH3 + 2H2O → Fe(OH)2 + 2Cl^- + 2NH4+.

5. FeCl3: You observed a red-brown precipitate that is insoluble in excess ammonia solution. This is most likely Fe(OH)3. The ionic equation would be: Fe^3+ + 3Cl^- + 3NH3 + 3H2O → Fe(OH)3 + 3Cl^- + 3NH4+.

Remember, the values of x and y in the complex ion equations depend on the amount of excess ammonia added and may vary.

I hope this helps you with the remaining ionic equations for your lab write-up! If you have any more questions, feel free to ask.