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
see other post.
Sure, Marysia! I'd be happy to help you with the remaining ionic equations.
Let's first discuss the reaction between CaCl2 and NH3. The correct ionic equation for this reaction is:
Ca^2+ + 2NH3 → Ca(NH3)2^2+
So you are right, the products are the same as the reactants.
Now let's move on to the remaining solutions:
1. AlCl3 + NH3:
The ionic equation for this reaction is:
Al^3+ + 3NH3 → Al(NH3)3^3+
The observation you made of a white precipitate that is insoluble in excess indicates the formation of aluminum hydroxide (Al(OH)3).
2. ZnCl2 + NH3:
The ionic equation for this reaction is:
Zn^2+ + 2NH3 → Zn(NH3)2^2+
The observation you made of a white precipitate that is soluble in excess indicates the formation of zinc tetraammine complex ([Zn(NH3)4]^2+).
3. CuCl2 + NH3:
The ionic equation for this reaction is:
Cu^2+ + 4NH3 → [Cu(NH3)4]^2+
The observation of a light blue precipitate that is soluble in excess indicates the formation of a copper(II) tetraammine complex ([Cu(NH3)4]^2+).
4. FeCl2 + NH3:
The ionic equation for this reaction is:
Fe^2+ + 4NH3 → [Fe(NH3)4]^2+
The observation of a dirty green precipitate that is insoluble in excess indicates the formation of iron(II) hydroxide (Fe(OH)2).
5. FeCl3 + NH3:
The ionic equation for this reaction is:
Fe^3+ + 3NH3 → [Fe(NH3)3]^3+
The observation of a red-brown precipitate that is insoluble in excess indicates the formation of iron(III) hydroxide (Fe(OH)3).
It's important to note that the hydroxides formed in reactions 2, 4, and 5 can also exist in amphoteric form, meaning they can react both as bases (forming precipitates) and as acids (dissolving in excess NH3). The specific form that dominates will depend on factors like pH and concentration.
I hope this helps! Let me know if you have any further questions.