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Chemistry - PbBr2 Precipitate

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    Ksp = [Pb][Br]^2


    3. 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?
     
  2. jcsd
  3. Apr 13, 2010 #2

    Borek

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    Staff: 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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  4. Apr 13, 2010 #3
    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.
     
  5. Apr 13, 2010 #4

    Borek

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    Staff: 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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    methods
     
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