Question
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?
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?
Answers
DrBob222
You are correct about Al+3 and Cu2+.
You need to be aware that Hg<sub>2</sub><sup>+2</sup> forms Hg<sub>2</sub>Cl<sub>2</sub>. 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^=.
You need to be aware that Hg<sub>2</sub><sup>+2</sup> forms Hg<sub>2</sub>Cl<sub>2</sub>. 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^=.