Question

how do you know that adding access 6M HCl to a solution of lead (II) ions will not precipitate all the Pb2+ ions from that solution

Answer 1

For you to get a better answer and explanation, I'll put the whole procedure to determinate if there's still a precipitate in the solution:

The Group A cations (Ag+ and Pb2+ ) are the only cations in our qualitative analysis scheme that form insoluble chlorides.

The procedure to follow is:

Measure 10 drops of a solution to be tested for the Group A cations into a test tube. Add 6 M HCl, dropwise, with shaking just until no more precipitate appears to form. Stir thoroughly; then centrifuge. Do not decant yet.

Test for complete precipitation by adding 1 drop of 6 M HCl to the clear supernatant. If cloudiness is observed or a precipitate forms, add another drop, stir and centrifuge. Repeat this process until no new precipitate forms upon addition of the 1 drop of hydrochloric acid.

Wash the precipitate twice by adding 10 drops of cold, deionized water then stir, centrifuge and decant.

The solubility of PbCl2 increases considerable in hot water compared to cold water while the solubility of AgCl remains relatively constant. Hot water can thus be used to separate PbCl2(s) from AgCl(s). The precipitate from step 1 is treated with hot water. After centrifuging and decanting, the resulting supernatant is tested for the presence of Pb2+ by adding KI(aq). The formation a golden yellow solid confirms the presence of Pb2+.

So, if you want to know that the whole Pb ions has not precipitate at all, you should add more HCl to the solution containing that ion, and then follows the procedure above. That's the best way to know it.

Anything else you want more explanation, you may tell me here.