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

[tua28494]

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?

 

[Borek]

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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[tua28494]

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.

 

[Borek]

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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