Question

Having issues with this experiment: I am given several solutions, which are mixed with an unknown ionic substance, using the solubility of each of the known substances, I am supposed to determine the identity of the unknown substance.

The aqueous solutions are: the unknown, lead(II) acetate, silver nitrate, sodium hydroxide, copper(II) sulfate, iron(III) nitrate, sodium iodide and sodium phosphate.

It is a virtual lab, so there are videos of each of the substances being mixed-from the video I have observed the following:

Answer 1

silver nitrate+unknown=not soluble
copper(II) sulfate+unknown=not soluble
iron(III) nitrate+unknown=not soluble

sodium hydroxide+unknown=soluble
sodium iodide+unknown=soluble
sodium phosphate+unknown=soluble
lead(II) acetate+unknown=soluble

I am confused, because I thought that nitrate was always soluble?
Also I am given that the unknown is odorless, and has a purple flame when burned

Answer 2

do you know if a precipitation was formed or not?

Answer 3

@drfernandez Are you sure it was silver nitrate & iron nitrate and not silver nitrite & iron nitrate?

Answer 4

because the flame was purple it is definitely potassium but im not sure what the compound would be

Answer 5

yes I just checked, and it was definitely nitrate. It doesn't tell me anything about the precipitates, all I can see in the video is the consistency/color of the liquid after mixing the two solutions. And I know about the potassium, I just don't know how to determine the compound itself. Thank you@

Answer 6

@LeibyStrauss

Answer 7

@drfernandez From the information given, I can't figure out what the unknown compound is.

Answer 8

Thank you for trying @LeibyStrauss

Answer 9

All nitrates are soluble, but not all compounds with silver, or iron (III).

Answer 10

I think it is potassium chloride?

Answer 11

@JoannaBlackwelder

Answer 12

I did this lab in school and it is potassium chloride

Answer 13

If silver nitrate is mixed with sodium iodide a precipitate of silver iodide would form.

Answer 14

Silver nitrate will also produce a white precipitate with potassium chloride.

Answer 15

We can usually find tables with solubility products (kps) and the color of the precipitate that will form. For example silver nitrate in water dissolves into Ag+ and NO3 2- ions, if you add NaCl to the solution, a white precipitate of AgCl will form. But if you add KI, a yellow precipitate of AgI will form.

Answer 16

@drfernandez

In your original post you wrote "The aqueous solutions are: the unknown, lead(II) acetate, silver nitrate, sodium hydroxide, copper(II) sulfate, iron(III) nitrate, sodium iodide and sodium phosphate."

Iron(III) nitrate will not be aqueous, SO3(2-) is insoluble except when combined with group 1A elements and NH4+

If the cation from the nitrate would actually be from group 1A we can figure out what the unknow is.

Answer 17

@mittens23309

These are the precipitates that formed.

silver nitrate+unknown=not soluble
copper(II) sulfate+unknown=not soluble
iron(III) nitrate+unknown=not soluble

If it is potassium chloride KCl why would a precipitate form from copper(II) sulfate (Copper does not form a precipitate with Cl)?

Answer 18

When you mix ionic solutions chances are you will get a precipitate. Whether a precipitate forms or not will depend on the solubility as you said.
But in this case instead of giving you both reactants, you have one reactant and have to guess the other reactant and the probable product.
Let's see if I can do this
1) Each element has an especific color when burned. I'm guessing potassium (K) is the UNKNOWN, because it burned purple.
You didn't write what is the product that you obtained.

2) lead (II) acetate Pb(C2H3O2)2 in water dissociates into both ions Pb2+ and acetate, it is probable that what you obtained is a solution of lead acetate in water.

3) silver nitrate, nitrate is always soluble, however silver as Ag+ tends to precipitate in the presence of chloride, bromide, cianide, cromate, phosphate, end so on.

4) sodium hydroxide, is a strong base that in aqueous solution completely dissociates into Na+ and OH-, if the product is soluble means that the other solution is either HCl, or water, in any case one of the products is water.

5) copper(II) sulfate, is usually soluble, however the Cu2+ ion in large concentrations tends to form copper hydroxide

6) iron(III) nitrate, it is probable that the same happens with iron (III) and forms hydroxide.

In the rest of the cases they are probably soluble because the solvent is water, it's a matter of checking which substance is most likely to precipitate or not.
I'm not sure if this is helping you, could you send us a link to see the videos, maybe.

Answer 19

@LeibyStrauss
look at @tatianagomezb reply, she explains it.

Answer 20

@mittens23309 @tatianagomezb

The original question states:

"The aqueous solutions are: the unknown, lead(II) acetate, silver nitrate, sodium hydroxide, copper(II) sulfate, iron(III) nitrate, sodium iodide and sodium phosphate."

If the copper(II) sulfate initially was in solution, why would it precipitate with Cl-? Cu2+ + Cl- are soluble.

If as proposed the Cu2+ precipitates with OH- why didn't it precipitate in the original solution?

Additionally, if there were OH- ions in the KCl solution why didn'y the potassium precipitate with the OH-?

Answer 21

We can't go on without a table of solubility and Kps.
Each time you're studying a reaction you have to make sure that it is based on accurate information, which is why I looked at my table and found that copper will always precipitate in presence of carbonate, chromate, hydroxide (if pH >7), sulfide and thiocyanate.

Metal ions such as \[Cu ^{+2}\] form a complex with water molecules
\[[Cu \left( H _{2}O \right)_{6}]^{+2}\]
but the water molecules are eventually replaced by charged particules such as \[OH ^{-}\]

So this is what happens
\[[Cu (H _{2} O)_{6}]^{+2} + H _{2}O \rightarrow [Cu(H _{2} O)_{5}(OH)]^{+} + H ^{+}\]
\[[Cu(H _{2} O)_{5}(OH)]^{+} + H _{2} O \rightarrow [Cu(H _{2}O)_{4}(OH)_{2}] _{(s)} + H ^{+}\]

copper doesn't react in the presence of chloride, nor HCl, if you check it it isn't a spontaneus reaction.

If you look at the kps of Cu(OH)2 = \[Kps _{Cu(OH)2} = [Cu ^{+2}] \times [OH ^{-}]\]
\[Kps= 2 \times 10^{-19}\]
such small number suggests that you will see a solid instead of the ions in the solution.

As a chemist I'm guessing, that's what I do, I don't know for sure, until I know, but if you look at my data I'm making a pretty good case.

Answer 22

thank you everyone