# Question #489d9

Nov 1, 2015

Here's what I got.

#### Explanation:

The idea here is that you need to make sure that each added solution only preicipates one of the cations.

That is, if you add the first solution and more than one cation precipitates, you will loose track of the exact numbers of cations of each type you had in the original solution.

The trick here is to find the solution that only precipitates one cation and the solution that precipitates all three cations and add them first and last, respectively.

Take a look at a table of the solubility rules for ionic compounds

• sodium chloride, $\text{NaCl}$
• sodium sulfate, ${\text{Na"_2"SO}}_{4}$
• sodium sulfide, $\text{Na"_2"S}$

This means that you must look for which pair of cation - anion will produce the desired precipitation sequence.

Notice that chlorides, which are ionic compounds that contain the chloride anion, ${\text{Cl}}^{-}$, are mostly soluble with the exception of those that contain the ${\text{Ag}}^{+}$ and ${\text{Pb}}^{2 +}$ cations.

This means that you cannot add the sodium chloride first, since that would precipitate two cations.

The same is true for sulfates, which are compounds that contain the sulfate anion, ${\text{SO}}_{4}^{2 -}$. Once again, the ${\text{Ag}}^{+}$ and ${\text{Pb}}^{2 +}$ cations would be precipitated if you were to add the sodium sulfate solution first.

On the other hand, the sodium sulfide solution would precipitate all three cations, so you've found the solution to add last.

So the strategy to find the solution that only precipitates one cation is no longer standing.

Now, how would you determine which solutions should come first?

You take a look at the solubilities of silver chloride, $\text{AgCl}$, and of lead(II) chloride, ${\text{PbCl}}_{2}$.

The important thing to notice here is that silver chloride remains insoluble at higher temperatures, but that the solubility of lead(II) chloride actually increases as temperature increases.

This means that if you use the sodium chloride to precipitate both ions at room temperature, then heat the solution, you can dissolve the lead(II) chloride and free up the ${\text{Pb}}^{2 +}$ cations again.

You'd then go on to add the sodium sulfate solution and precipitate the ${\text{Pb}}^{2 +}$ cations, and the sodium sulfide solution to precipitate the ${\text{Ni}}^{2 +}$ cations.

So, the order in which I'd add these solutions is

• sodium chloride, $\text{NaCl}$
• sodium sulfate, ${\text{Na"_2"SO}}_{4}$
• sodium sulfide, $\text{Na"_2"S}$

SIDE NOTE I think that you can use the same technique if you add the sodium sulfate solution first.

Lead(II) sulfate is much less soluble in aqueous solution than silver sulfate as temperature increases, so basically you could

• precipitate both the lead(II) and silver cations by using sodium sulfate
• heat up the solution to dissolve the silver sulfate and free up the silver cations
• add the sodium chloride to precipitate the silver cations

I would still go for the first sequesnce of solutions, though.