# Chemistry - PbBr2 Precipitate

## 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

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?

## Answers and Replies

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Borek
Mentor
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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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
Mentor
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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