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Calculate the molarity of Br2 in the stock solution.

  1. Feb 21, 2005 #1
    The stock solution of bromine is prepared by dissolving 232.5 g of Br2 and 187.5 g of KBr in water and diluting to 750 ml. Calculate the molarity of Br2 in the stock solution.

    how to do this qs?
     
  2. jcsd
  3. Feb 21, 2005 #2

    dextercioby

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    What's the definition of molarity...?Theory must be known b4 attempting to solve exercises...

    Daniel.
     
  4. Feb 22, 2005 #3
    no. of mole / volume?
     
  5. Feb 22, 2005 #4

    dextercioby

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    PER LITER OF SOLLUTION...:wink: Now compute it...

    Daniel.
     
  6. Feb 22, 2005 #5
    From http://www.webelements.com/webelements/elements/text/Br/key.html
    I took the mass of Br. Since you have [itex]\mbox{Br}_{2}[/itex],
    one 1mol of [itex]\mbox{Br}_{2}[/itex] weighs 2* 79.904g

    Divide 232.5 g by (2* 79.904g) , which equals 1.45487.
    So you got 1.45mol of [itex]\mbox{Br}_{2}[/itex].

    Now you need the molarity which is the concentration:

    [itex] c= \frac{1.45mol}{0.75l} = 1.93 \frac{mol}{l} [/itex]

    I hope this is right.

    I don't know what role the KBr plays, I guess that it helps
    dilluting the [itex]Br_{2}[/itex]. Note that I didn't take into account
    the KBr.
     
  7. Feb 22, 2005 #6

    dextercioby

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    Even if this is not posted in the HM section,still solving the problem instead of the OP is not reccomendable.So take this as an advice for future "urges"...:wink:

    Daniel.
     
  8. Feb 22, 2005 #7
    I think there still may be a mistake in my solution.
    The fact that they give the information "187.5 g KBr" makes me a little
    bit doubtful.

    If you have to take the KBr into account, then I hope you know
    how to solve it mousesgr.
     
  9. Feb 22, 2005 #8

    dextercioby

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    That "Br2" (or [itex]Br_{2} [/itex]) is surely misleading.It may still refer only to bromine MOLECULES...But if u read it simply as "bromine" (as in "molarity of bromine"),then,of course,u have to include the bromine from the bromide.

    Danie.
     
  10. Feb 22, 2005 #9

    Gokul43201

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    I've been trying to not answer this because it's not clear what the level of the poster is. At the simplest level, you just do the calculation performed by edgardo.

    At a higher level of sophistication, you need to consider what happens when you dissolve Bromine in water...about the same that happens with chlorine :

    [tex]Br_2 + H_2O \leftrightharpoons HBr + HOBr [/itex] But it's important to note that the products will be present in aqueous solution, in their ionic forms, [itex]H^+,~Br^-,~Br^+,~OH^- [/itex]. The amounts of molecular and ionic bromine can be calculated easily from the equilibrium constant for this reaction. The effect of adding KBr is that it produces [itex]Br^-[/itex] ions in solution, shifting the equilibrium leftwards, resulting in more molecular bromine. Assuming KBr is highly ionic (ie : the dissociation constant is close to 1), one can then calculate [itex][Br_2]_{aq}[/itex].
     
  11. Feb 22, 2005 #10

    Borek

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    [itex]~Br^+[/itex]???? Weak as it is (pKa = 8.7) [itex]~HOBr[/itex] is an acid. Look for [itex]~OBr^-[/itex].


    Chemical calculators for labs and education
    BATE - pH calculations, titration curves, hydrolisis
     
    Last edited: Sep 19, 2011
  12. Feb 22, 2005 #11

    Gokul43201

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    Yes, it is; isn't it (duh) ? :redface: My bad ! :yuck:
     
  13. Feb 22, 2005 #12

    Borek

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    Nobodys perfect... It is www.chembuddy.com, .pl was left from other project :blushing:


    Chemical calculators for labs and education
    BATE - pH calculations, titration curves, hydrolisis
     
    Last edited: Sep 19, 2011
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