Can you explain the process for solving this problem (this one is more complicated than the example in our book):
A solution contains both I^- and Cl^-, each at a concentration of 3.0x10^-2M. What concentration of Pb^+2 is required to precipitate the maximum amount of one of these ions without precipitating any of the other? Which ion precipitates first? Ksp for PbI2 is 7.1x10^-9 and that for PbCl2 is 1.6x10^-5.
Thanks
The easiest way is to write the Ksp for the salts.
PbCl2 ==> Pb^+2 + 2Cl^-
and Ksp = (Pb^+)(Cl^-)^2 = 1.6 x 10^-5
PbI2 ==> Pb^+2 + 2I^-
and Ksp = (Pb^+2)(I^-)^2 = 7.1 x 10^-9
Just looking at the Ksp for each salt tells us that PbI2 is much less soluble than PbCl2; therefore, adding Pb^+2 to the solution, say dropwise, means we will exceed Qsp first for PbI2. Calculations follow to prove that.
Calculate (Pb^+) that must be present in the solution to ppt PbCl2. I assume you know how to do that. If not, repost and tell me exactly what you don't understand about it.
Then calculate (Pb^+2) that must be present in the solution to ppt PbI2.
The one that ppts first, of course, will be the one that requires the smaller amount of Pb^+2 since, presumably, we are adding Pb^+2 incrementally. Pb^+2 may continue to be added until the (Pb+2) is 1 ion less than that required to start pptn of the more soluble salt.
Note that the problem did NOT ask for HOW MUCH solution of a certain molarity of lead ion must be added to ppt as much PbI2 as possible before starting to ppt PbCl2. That is a tougher problem. Here the problem is just concerned with concentration and that is easy enough to calculate.
Note, also, that I used Ksp to determine which salt was more soluble. That is possible when the salts have the same formula configuration; i.e., PbCl2, PbI2 have the same type formula. I could not, however, compare PbCl2, PbI2, and PbCO3 that way. BaSO4, BaCrO4, PbCrO4, and BaSO4 may be compared that way but not Ag2CrO4.
To calculate the [Pb^+2] you just divide the Ksp by the concentration of the other element, right? So [Pb][Cl2]=1.6x10^-5 then [Pb]=1.6x10^-5 / 3.0x10^-2 (or would it be 6.0 because it's Cl2 - do you double it since there are 2 chlorines and we're only looking for one?)
No, it isn't doubled. The problem states that THE (Cl^-) = 0.03 so it's 0.03 precisely. The equation, though, says to square it and I don't see that in your post.
For Pb in PbCl2,
(Pb^+2) = Ksp/(Cl^-)^2 = 1.6 x 10^-5/(0.03)^2 = 0.01778. You will need to round to the proper number of significant figures.
Ok, I see what you did - you squared it because of the 2 in the coefficient of the Cl in the balanced equation. Ok, that helps a lot. Thank you!
I found that the concentration to ppt the PbCl2 is 1.8x10^-2 and that to ppt the PbI2 is 7.9x10^-6. Therefore, the one that would ppt first would be the I^- and the concentration of Pb^+2 would be 7.9x10^-6, correct?
yes. Technically, a (Pb^+2)= 7.9 x 10^-6 would make a solution so that
(Pb^+2)(Cl^-)^2 would JUST equal Ksp. One more ion of Pb would cause one molecule of PbCl2 to ppt.
a phatato is chopped
is that a physical of chemical
It's a physical change. You would do better to post your question as a new question by clicking on "post a new question" at the top of the page. Sometimes we are a little slow about checking previous posts, especially when the other person has no further questions. In that case we miss yours but we see it immediately if it is posted as a new question which places it at the top of the heap.
The process for solving the problem is as follows:
1. Write the Ksp expressions for PbCl2 and PbI2:
- PbCl2: Ksp = [Pb^+2][Cl^-]^2 = 1.6x10^-5
- PbI2: Ksp = [Pb^+2][I^-]^2 = 7.1x10^-9
2. Calculate the concentration of Pb^+2 required to precipitate PbCl2:
- Use the Ksp expression for PbCl2 to solve for the concentration of Pb^+2, assuming the concentration of Cl^- is 3.0x10^-2 M.
- (Pb^+2) = Ksp/(Cl^-)^2 = 1.6 x 10^-5/(0.03)^2 = 0.01778 M
3. Calculate the concentration of Pb^+2 required to precipitate PbI2:
- Use the Ksp expression for PbI2 to solve for the concentration of Pb^+2, assuming the concentration of I^- is 3.0x10^-2 M.
- (Pb^+2) = Ksp/(I^-)^2 = 7.1 x 10^-9/(0.03)^2 = 7.9 x 10^-6 M
4. Compare the concentrations calculated in step 3 to determine which ion precipitates first:
- Since the concentration of Pb^+2 required to precipitate PbI2 (7.9 x 10^-6 M) is lower than the concentration required to precipitate PbCl2 (0.01778 M), PbI2 will precipitate first.
Therefore, to precipitate the maximum amount of one ion without precipitating any of the other, a concentration of Pb^+2 at 7.9 x 10^-6 M is required. This would precipitate PbI2 without precipitating any PbCl2.