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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methods

 

[DeeNos]

Your solution is correct. To find the mass of Pb(NO3)2, you multiplied the number of moles of Pb by its molar mass, which is the correct approach. Another way to approach this problem would be to use the molar ratio between Pb(NO3)2 and PbBr2, which is 1:1. This means that for every 1 mole of Pb(NO3)2, 1 mole of PbBr2 will form. Therefore, the number of moles of Pb(NO3)2 needed to form 3.73*10^-4 moles of PbBr2 is also 3.73*10^-4 moles. This can also be converted to grams using the molar mass of Pb(NO3)2. Both approaches will give you the same answer. Overall, you did a great job in solving this problem!