How Much Pb(NO3)2 Is Needed to Precipitate PbBr2 from HBr Solution?

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Discussion Overview

The discussion revolves around a homework problem concerning the calculation of the amount of Pb(NO3)2 required to precipitate PbBr2 from a given HBr solution. The focus is on the application of solubility product constant (Ksp) and the relationships between concentrations and moles in a chemical reaction.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates the moles of HBr and uses it to find the concentration of Pb required for precipitation based on the Ksp value.
  • Another participant questions the necessity of converting moles of HBr, suggesting that the molarity of Br- can be directly used in the Ksp expression.
  • A later reply indicates confusion about the conversion process and emphasizes the importance of using the correct volume when calculating moles of Pb.
  • There is a correction regarding the volume used in the calculation of moles of Pb, highlighting that the actual volume of the solution (750 mL) should be used instead of 1 L.

Areas of Agreement / Disagreement

Participants generally agree on the approach to use Ksp for determining the concentration of Pb, but there is disagreement on the necessity of converting moles of HBr and the correct volume to use in calculations. The discussion remains unresolved regarding the best method to calculate the required grams of Pb(NO3)2.

Contextual Notes

Participants express uncertainty about the steps involved in the calculations, particularly regarding the conversion of concentrations and the appropriate volume to use. There are also unresolved questions about the implications of using different volumes in the calculations.

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Homework Statement


How many grams of Pb(NO3)2 must be added to 750.0 mL of 0.10 M HBr in order for PbBr2 to precipitate?

Ksp = 2.1*10^-6

Homework Equations


Ksp = [Pb][Br]^2


The Attempt at a Solution


For PbBr2 to precipitate, [Pb][Br]^2 must be greater than Ksp

0.750 L * 0.10 M HBr = 0.075 moles HBr
0.075 moles HBr/ 1 L = [0.075 HBr] Is this step correct?

Ksp/([HBr]^2) = [Pb] = (2.1*10^-6)/(0.075^2)= 3.73*10^-4 M Pb
so 3.73*10^-4 M Pb * 1 L = 3.73*10^-4 moles Pb

I'm not sure where to go from here, (3.73*10^-4 moles Pb)*(molecular weight of Pb(NO3)2)?


(3.73*10^-4 moles)(331.2098 g/mol) = 0.124 g of Pb(NO3)2.

How did I do?
 
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Seems OK to me. I just don't know what was the conversion for moles of HBr for. You know molarity of the HBr solution, that means you also know molarity of Br- - simply use this value in Ksp.

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Borek said:
Seems OK to me. I just don't know what was the conversion for moles of HBr for. You know molarity of the HBr solution, that means you also know molarity of Br- - simply use this value in Ksp.

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without converting HBr to moles, I get the concentration of Pb.

I need how many grams of Pb(NO3)2 so what Volume would I multiply the concentration of Pb to, so I end up with moles of Pb? Would it be multiplied by 0.750 L the same volume as the HBr?

I need the moles of Pb to multiply with the molecular mass of Pb(NO3)2 to get grams.
 
You have lost me.

Knowing CONCENTRATION of HBr you can calculate CONCENTRATION of Pb2+. Numbe of moles of HBr/Br- is irrelevant. Knowing concentration and volume (which was given in the question - 750 mL) you can calculate number of moles.

I just realized you did it wrong - you correctly calculated concentration of the lead, but you multiplied it by 1L, not by the real volume of the solution.

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methods
 

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