A solution containing a mixture of metal cations was treated as outlined. Dilute HCl was added and a precipitate formed. The precipitate was filtered off.
H2S was bubbled through the acidic solution. No precipitate formed.
The pH was raised to about 9 and H2S was again bubbled through the solution. A precipitate formed and was filtered off.
Finally, sodium carbonate was added to the filtered solution. A precipitate formed and was filtered off.
What can be said about the presence of each of these groups of cations in the original solution?
Options: None of these ions were present
At least one of these ions was present
All of these ions were present
Unknown

The answer is 4 i don't know if its right let me see.

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.