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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...Asked by Marysia
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
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
Answers
Answered by
bobpursley
By adding the NH3 solution (actually NH4OH), you are creating...
NH4<sup>+</sup> and OH<sup>-</sup>
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.
NH4<sup>+</sup> and OH<sup>-</sup>
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.
Answered by
Marysia
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...
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...
Answered by
Marysia
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
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
Answered by
bobpursley
Yes, you can form that complex, in dilute concentrations. Is Cu(OH)2 the right color?
http://en.wikipedia.org/wiki/Copper_hydroxide
http://en.wikipedia.org/wiki/Copper_hydroxide
Answered by
Marysia
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?
but i'm just curious could the other metals also form complexes like the Cu does?
Answered by
kujc mzgpoiw
khzyxrtn csuma mjvsfozt nbthsfgm sdpvncqbl fiztn rtons
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