Four glass bottles containing clear, colourless solutions are left on a lab bench with the following labels lying nearby in a jumbled pile; Pb 2+(aq), Cl-(aq), Ag+(aq), (SO4)2-(aq). You number...
Four glass bottles containing clear, colourless solutions are left on a lab bench with the following labels lying nearby in a jumbled pile; Pb 2+(aq), Cl-(aq), Ag+(aq), (SO4)2-(aq).
You number the four bottles, 1,2,3,4 and while numbering them, you notice that bottles 1 and 3 have the word "cation" on their lids while bottles 2 and 4 have "anion" on their lids.
You decide to test them by mixing them together in pairs, but you have only three empty beakers.
When the solutions are mixed in pairs.
Three combination give white precipitates;
Bottle 1 + bottle 2. bottle 1 + bottle 4, bottle 2 + bottle 3.
Complete a table like the one that follows to figure out the ions that are in each bottle.
Bottle# _____ Bottle#_____ Total Insoluble:
Anion: _____ Insoluble Insoluble ____________
Anion:______ Insoluble Soluble
Total Insoluble: _______ _________
2 marks for each ion matched to its correct bottle
We know that bottles 1 and 3 are cations. Cations are positively charged ions. Thus, Pb+2 and Ag+ could be bottles 1 and 3.
Likewise, we know that bottle 2 and 4 are anions. Anions are negatively charged ions. Thus, Cl- and SO4-2 could be bottle 2 and 4.
Precipitates are solids that fall out of a liquid solution that is made from two soluble salts. A salt is composed of a cation and anion. Thus, the following salts are possible in this lab with the reagents that are given. A solubility table was used to determine whether or not each possible salt is soluble or insoluble.
- PbCl2 = Soluble
- PbSO4 =Insoluble
- AgCl = Insoluble
- (Ag)2SO4 = Slightly soluble
Therefore, some sort of precipitate will form when lead (II) and sulfate are mixed, silver (I) and chloride are mixed, and when silver (I) and sulfate are mixed.
Thus, the bottles can be identified as follows:
- Bottle 1 must be Ag+ because it is found in two of the insoluble products.
- Bottle 3 must be Pb+2 because it is the only other cation that is left.
- Bottle 2 must be SO4-2 because it is the only anion that is involved in the formation precipitate formation.
- Thus, out of the process of elimination, bottle 4 must be Cl-.
This should be sufficient information to complete your table.