Here we're told that a sample of a solution contains 0.405 molar of chromate ion and .628 molar of sulfide ions. These two ions can be precipitated with the use of lead to fluoride. Which ion will precipitate? Precipitate out first and at which concentration? So here the K_sp of lead to chromate is 2.0×10-16, and the K_sp of lead to sulfide is 7.0×10-29.

Well, here we're going to say that this is the smaller K_sp, the lawyer K_sp, the less soluble you will be and therefore the faster you can precipitate. We're going to say lead to sulfide contains the sulfide ions, so we'd say that the sulfide ion will precipitate out first. Now, if we want to think about this visually, we're going to say we have floating around in our solution sulfide ions and chromate ions and I come in and I dissolve or pour in lead to fluoride.

What's going to happen here is that since led to sulfite has a lower cast P, it's going to be the one that forms this solid here first. So we're going to have lead to sulfide precipitating out first. Now how do we determine its concentration? Well, since sulfide precipitates out first, we're going to use its K_sp. So we're going to have PB solid. And with K_sp we talk about how this ionic compound breaks up into its ions to be PB²⁺ aqueous plus S^2- aqueous.

We have initial change equilibrium with a nice chart we look at. We ignore solids and liquids, so the reactant is a solid, so we ignore it. We're going to say here we don't have any initial amount of lead, but we know that initially this is our concentration of sulfide ions. So that's what we're going to plug in, 0.628 molar. They're both products. We're making them so they're both plus X and then bring down everything.

Now remember, when dealing with K_sp, if you have an actual or real number in front of your X variable, you can ignore the X variable. That's because X is going to be so small that it's not going to have a great enough impact on changing this .628 value. Now we're going to say K_sp equals products, so it equals lead to ion times sulfide ion. Here we're going to say that K_sp 4 led to sulfide in 7.0×10-29 lead. Two is equal to X Sulfide is 0.628.

Divide both sides by 0.628. X here is 1.115×10-28 molar. This number here represents the concentration at which our sulfide ion will begin to precipitate. So this will be our final answer.