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Waste Water treatment plant problem

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Barium chloride is a toxic compound used in waste water treatment and many manufacturing processes. A 20 mL solution of 0.200 M Na2SO4 is added to a 10.0 mL water sample containing an unknown concentration of BaCl2 that was concentrated by 100 fold after being taken from a waste water treatment plant. After filtration of the solution, 0.7800g of the insoluble BaSO4(s) is obtained. Provide a balanced chemical equation and determine the concentration of the Ba2+ from the sample taken at the waste water treatment plant.

    3. The attempt at a solution

    BaCl2(aq) + Na2SO4 [tex]\rightarrow[/tex] BaSO4(s) + 2NaCl

    Using dilution equation:
    C1V1=C2V2
    C1(0.01)=(0.2)(0.02)
    C1=0.4M = [BaCl2]

    After Barium chloride is concentrated to a 100 fold which means the solution decreased volume. So 0.4M x 100 = 40M

    BUT, this solution isnt re-reacted with sodium sulfate therefore the 100x concentrated part doesn't matter in the problem.

    nBaSO4 = 0.78g / 233.3896 g/mol = 0.00334 mol BaSO4

    [BaSO4] = [Ba2+]

    What I don't understand is why do I need to consider the 100x problem when i am given the mass of the precipitate and therefore just calculate how much barium ions I have ??? The 0.78g is fixed because thats what the experimentalists got. If the concentration of barium chloride is changed to 40M and then used to calculate the amount of Barium sulfate what does 0.78 grams measured from filtration matter then?
     
  2. jcsd
  3. Sep 27, 2009 #2

    Borek

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    Staff: Mentor

    1. Not 40M, recheck your thinking. Which solution has a higher concentration - initial, or concentrated one?

    2. You are asked to calculate concentration in water sample BEFORE it was concentrated, that is, in a sample taken just from the waste water.

    Trick is, in such water concentration of barium was rather low, and determination of low concentrations is often difficult, thus we start concentrating samples, to make determination easier.
     
    Last edited by a moderator: Aug 13, 2013
  4. Sep 27, 2009 #3
    1. I dont understand what you are saying here. Is it Na2SO4 ?

    2. IF thats the case then I dont need to consider the 100x concentrated part. What do I do ?
     
  5. Sep 27, 2009 #4
    [Ba2+] = 0.4M before it was concentrated
     
  6. Sep 27, 2009 #5
    But wait the question asks to determine the concentration of Barium ions which means the only contributer of barium is barium chloride and the only way barium ion can be calculated is from Barium chloride. I know that the two solutions are mixed together to precipitate out Barium sulfate so its a dilution problem? Actually what I mean is 20 mL solution and 10 mL solution are coming together. So the total volume is 30 mL solution containing both compounds. therefore to calculate concentration of barium ions I need volume. I am given the mass and i determined the molar mass I just need volume and I think the only way I can get volume to calculate concentration is from 30 mL because the two solutions are mixed together and filtrated out to get Barium sulfate.

    Does this kinda make sense. I know it sounds a little mixed up but please correct my way of thinking.
     
  7. Sep 27, 2009 #6
    or is it 0.00334 mol BaSO4 x 1 mol BaCl2/1 mol BaSO4 x 1 mol Ba2+ \ 0.01 L = 0.334M Ba2+
     
  8. Sep 27, 2009 #7

    Borek

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    Staff: Mentor

    Seems like you are completely confused. 10 mL sample is the one AFTER concentration. The original volume was 100 times larger.

    You know that 10 mL contained 0.00334 mole Ba2+ - so the same 0.00334 mole of Ba2+ was present in the original volume (100 times 10 mL).
     
    Last edited by a moderator: Aug 13, 2013
  9. Sep 27, 2009 #8
    Is that the answer then ?
     
  10. Sep 27, 2009 #9
    So the concentration of barium ions before the 100x concentration was 3.34 x 10^-6 ????
     
  11. Sep 27, 2009 #10
    IF you say:
    That means 100 x 10 mL = 1000 mL = 1 L

    Before 100x concentration:
    C1 = [Ba2+] = [tex]\frac{\mbox{3.34 x 10^{-3} mol}}{1 L}= 3.34 \mbox{ x } 10^{-3}M[/tex]

    After 100x concentration:
    C2 = [Ba2+] = [tex]\frac{\mbox{3.34 x 10^{-3} mol}}{0.01 L}= 0.334 M[/tex]

    Is this correct ?
     
    Last edited: Sep 27, 2009
  12. Sep 27, 2009 #11

    Borek

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    Staff: Mentor

    Looks OK.
     
    Last edited by a moderator: Aug 13, 2013
  13. Sep 27, 2009 #12
    OK? How would you have done it differently ??
     
  14. Sep 27, 2009 #13
    Is the question asking before 100x or after 100x ?
     
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