Chemistry [Saturated solution PbCl2]

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Discussion Overview

The discussion revolves around the precipitation of lead(II) chloride (PbCl2) when an equal volume of lead(II) nitrate (Pb(NO3)2) solution is added to a saturated PbCl2 solution. Participants explore the calculations needed to determine the concentrations of Pb2+ and Cl- ions and whether precipitation occurs based on the solubility product constant (Ks) of PbCl2.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the concentration of Pb2+ and Cl- ions based on the solubility of PbCl2 and the added Pb(NO3)2, suggesting that the total concentration of Pb2+ becomes 0.09 M.
  • Another participant questions the use of the solubility product (Ks) and emphasizes the need to compare calculated values against Ks to determine precipitation.
  • There is a discussion about the correct application of Ksp, with some participants arguing that it is necessary for calculating concentrations in a saturated solution.
  • Confusion arises regarding the mixing of concentrations from different sources and how to account for dilution effects when calculating the equilibrium expression.
  • Participants express differing views on the necessity of using Ksp and the implications of calculated values exceeding Ks.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem. There are multiple competing views on the use of Ksp, the calculation of ion concentrations, and the interpretation of results regarding precipitation.

Contextual Notes

Some participants highlight the importance of considering dilution effects when mixing solutions, which complicates the calculations. There are also references to specific values and methods that are not universally accepted among participants.

Dousin12
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Homework Statement


To a saturated PbCl_2-solution there is a equally big volume of Pb(NO3)_2 added which has the concentration 0.05M. Is there a precipitation of PbCl2 or not? If Ks is 2,54*10^-4 for PbCl2.

The problem here is that I don't know how to find the concentration of Cl2 or Pb. My chem book said that u can solve 11g of PbCl2 per liter. So should I say that

11g/Molar mass ob PbCl2=0.04 moles <=> 0.04 moles Pb2+ <=> 0.08 Moles Cl-
But because there is 0.05M of Pb(NO3)_2 as well so the moles Pb is 0.04+0.05=0.09moles

But if i say:
[CL^-]^2*[Pb^2+]=[0.08]^2*[0.09] I get 5.76*10^-4 which is bigger than ks. But in the book they say:

[CL^-]^2*[Pb2+]=7.16*10^-5

How do they get that value?
 
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Dousin12 said:

Homework Statement


To a saturated PbCl_2-solution there is a equally big volume of Pb(NO3)_2 added which has the concentration 0.05M. Is there a precipitation of PbCl2 or not? If Ks is 2,54*10^-4 for PbCl2.

The problem here is that I don't know how to find the concentration of Cl2 or Pb. My chem book said that u can solve 11g of PbCl2 per liter. So should I say that

11g/Molar mass ob PbCl2=0.04 moles <=> 0.04 moles Pb2+ <=> 0.08 Moles Cl-
But because there is 0.05M of Pb(NO3)_2 as well so the moles Pb is 0.04+0.05=0.09moles

But if i say:
[CL^-]^2*[Pb^2+]=[0.08]^2*[0.09] I get 5.76*10^-4 which is bigger than ks. But in the book they say:

[CL^-]^2*[Pb2+]=7.16*10^-5

How do they get that value?

The problem here more important than getting the answer is "that I don't know how to find the concentration of Cl2 or Pb". Hopefully you have got it from some other information in your book but you ought to be able to get it from the data of the problem, i.e. the solubility product, Ks.

Write the equation for Ks. Then what is the relation between the two variables that are in it? - so that we can reduce it to a single variable.
 
Im not suppose to use Ks I don't think so atleast, Because you want to know how your value is compared torwards Ks!
Because if the value is bigger than ks it will be solid and not liquid!
 
You can't solve this problem without Ksp. However, in the worst case, you can calculate Ksp from the solubility.
 
2,54*10^-4 I wrote in the question!

Is Ks for PbCl2

I wrote if i transform that to gram and such .. i get the values I've gotten. I see no reason why my values are wrong! But they are
 
First of all - do you know how to you use just given Ksp value to calculate concentrations of Pb2+ and Cl- in the saturated solution?
 
Yes I am really sorry, but have u read what i wrote!

11g/Molar mass ob PbCl2=0.04 moles/dm^3 <=> 0.04 moles/dm^3 Pb2+ <=> 0.08 Moles/dm^3 Cl-'

11g is the mass of PbCl2 that is saturated! (per dm^3)
 
Sorry, I misread some of the things you wrote, but you mix everything at the same time and you post things that contradict itself, making your post unnecessarily hard to follow. Why do you use the solubility to calculate concentrations, if you have the Ksp? Why do you state you are not supposed to use Ksp if it is given?

One thing that you definitely did wrong is here:

Dousin12 said:
But because there is 0.05M of Pb(NO3)_2 as well so the moles Pb is 0.04+0.05=0.09moles

Yes, number of moles in the mixture is sum of moles from both sources - PbCl2 and Pb(NO3)2. But it doesn't mean the concentration is the sum of concentrations.
 
Dousin12 said:
But if i say:
[CL^-]^2*[Pb^2+]=[0.08]^2*[0.09] I get 5.76*10^-4 which is bigger than ks. But in the book they say:

[CL^-]^2*[Pb2+]=7.16*10^-5

How do they get that value?

Because they took account of the dilution or mixing factor mentioned by Borek and you didn't. (You have to do it for both ions!)

Apart from errors like that you do seem to know how to calculate an equilibrium constant from a solubility, or more exactly from the saturation concentration. It should not be too difficult then to do the inverse, which I think is what is being asked.

The homework template contains heading:

2 Relevant equations

If you had stated this, which in any case I asked in #2 your discussion would have been easier to follow, and I think you would have found the calculation easier to execute.

From what you quote it sounds like a worked example! :confused:
 

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