A colourless solution contains one type of ion. When one adds dilute HCl to it, a white precipitate initially forms. However, once more of the HCl solution is added, the precipitate dissolves. Explain why the ion is Pb2+ and not Al3+, Hg22+, Cu2+ nor Ag+.
I know Cu2+ is not an option because it forms a blue solution. Its not Al3+ or Hg22+ because a precipitate wouldn't form in the first place. But out of Ag+ and Pb2+, how do you tell which forms a complex ion and which doesn't?
You are correct about Al+3 and Cu2+.
You need to be aware that Hg2+2 forms Hg2Cl2. In fact, the addition of HCl to form a white ppt is the old Group I of most qualitative analysis schemes because the chlorides of Ag, Hg2^+2, and Pb^+2 are insoluble. To be honest about it, BOTH AgCl and PbCl2 form complex ions. AgCl will form AgCl2^- and AgCl3^= while PbCl2 forms PbCl4^=. The PbCl4^= complex has been fairly well known for some time. The AgCl complexes have, too, but they have not been published that much. I would go with the Pb (since that is the question) and say it forms complex ions and list PbCl4^=.
To determine which ion forms a complex ion, we need to consider the solubility rules and the stability constants of the respective precipitates.
In this case, when dilute HCl is added to the colourless solution, a white precipitate initially forms, indicating the formation of an insoluble chloride salt. However, upon further addition of HCl, the precipitate dissolves.
Pb2+ is the likely ion in this scenario because lead(II) chloride, PbCl2, is relatively insoluble and forms a white precipitate when mixed with dilute HCl. However, when excess HCl is added, a soluble complex ion, [PbCl4]2-, forms. This is due to the formation of strong complexes with chloride ions, which increases the solubility of the lead chloride.
On the other hand, silver chloride, AgCl, also forms a white precipitate when mixed with HCl. However, AgCl remains insoluble even with excess HCl, as it does not form stable complex ions. Therefore, Ag+ can be ruled out in this case.
In summary, based on the solubility behavior and the formation of complex ions, the ion in question is likely Pb2+ and not Al3+, Hg2 2+, Cu2+, or Ag+.
To determine whether Ag+ or Pb2+ forms a complex ion in a given situation, you can consider their respective solubility rules and the formation of complex ions.
Ag+ is known for its tendency to form insoluble compounds with many anions, including chloride (Cl-). When Ag+ ions encounter Cl- ions, they form a white precipitate of silver chloride (AgCl). The precipitate does not dissolve easily.
On the other hand, Pb2+ ions, in the presence of chloride ions (Cl-), form a white precipitate of lead chloride (PbCl2). However, this precipitate is known to be soluble in excess chloride ions.
The key part of this scenario is the addition of dilute HCl. When a small amount of HCl is added to the colourless solution, a white precipitate of PbCl2 is formed. This occurs because the concentration of chloride ions (Cl-) is low and not enough to dissolve the PbCl2. But when more HCl is added, the concentration of chloride ions increases significantly, leading to the formation of the complex ion [PbCl4]2-.
The [PbCl4]2- complex ion is stable and soluble in water, which causes the white precipitate of PbCl2 to dissolve. This is why, in the presence of excess HCl, the precipitate formed initially disappears.
In summary, the reason why the ion is Pb2+ and not Ag+ is due to the formation of a stable and soluble complex ion, [PbCl4]2-, when excess HCl is added. Ag+, however, forms an insoluble precipitate (AgCl) that does not easily dissolve with the addition of more HCl.